Class 11 Medical Chemistry - Classification Of Elements And Periodicity In Properties

A metal 'M' has the electronic configuration 2, 8, 3 and occurs in nature as M $_2$ O $_4$ . It is more reactive than zinc. Write a chemical equation illustrating the use of this metal for joining cracked railway lines.

The reaction of iron (III) oxide (Fe

$_2$ O

$_3$ ) is used to join cracked railway lines. The reaction is highly exothermic and is called thermite reaction. Iron is obtained in the molten state which fills the cracks.

$\underset{Iron (III)\ oxide}{Fe_2 O_3} + \underset{Aluminium}{2 Al} \rightarrow \underset{(molten\ state)}{2 Fe} + \underset{Aluminium\ oxide}{Al_2 O_3} + Heat$

____________ is the electronic configuration of the highest electronegative element.

2,7 is the electronic configuration of the highest electronegative element. Fluorine is the most electronegative element,

Write the full form of IUPAC.

The full form of IUPAC is

$International\ Union\ of\ Pure\ and\ Applied\ Chemistry$ .

Complete the electronic configuration of a chloride ion.
$1s^{2}$ ___________.

Solution:-

Electronic configuration of chlorine atom

$= 1{s}^{2} 2{s}^{2}2{p}^{6} 3{s}^{2} 3{p}^{5}$

Electroni configuration of chloride ion

$\left( {Cl}^{-} \right) = 1{s}^{2} 2{s}^{2}2{p}^{6} 3{s}^{2} 3{p}^{6}$

The elements ( $z = 117$ and $120$ ) have not yet been discovered in which family or group would you place these elements and also give electronic configuration in each case?

Atomic number

$Z=117$

Number of protone

$= 117$

Electronic configuration is

$[Rn]sf^4\, 6d^{10} 7s^2 7p^5$

Belong to group 17 - halogen family

Atomic number

$= z = 120$

Electronic configuration is

$[440]8s^2$

element belong to group 2 - alkali earth family

1116582_1153721_ans_7c3bb947a2384a9f973a72d836d46a05.jpg

Name the type of elements, which have their
(i) outermost shell complete
(ii) outermost shell incomplete
(iii) two outermost shell incomplete
(iv) one electron short of octet
(v) two electrons in the outermost orbit.

1351754_1273053_ans_30e93e17a6164725bf8bdbe6858ecd66.jpg

What is the IUPAC name of an element with atomic number $107$ .

IUPAC
atomic no:

$107$
Un-nil -Heptium

With reference to periodic table, indicate:

An element that is in group IIIA and third period.

AI belong to group IIIA and third period in periodic table.

Explain chemical reactivity in periods?

1376134_1287365_ans_afa5fe540d314f0ab97b897633f0afb2.jpg

Give the electronic configuration of third alkali metal.

$K-1s^2 , 2s^2, 2p^6 ,3s^2,3p^6, 4s^1$

From among the elements, choose the following:
Cl, Br, F, O, Al, C, Li, Cs and Xe

The element whose atom has 6 electrons in the outermost shell.

O is the element whose atom has 6 electrons in the outermost shell.

With reference to periodic table, indicate:

The inert gases placed in the 2nd and 5th period.

Ne and Xe respectively are inert gases placed in the 2nd and 5th period.

Write the names of elements from the following symbols: Zn, Cd, Xe, Br, Ti, Cu, Fe, Si, Ir, Pt.

$\text{Symbol}$	$\text{Name of elements}$
Zn	Zinc
Cd	Cadmium
Xe	Xenon
Br	Bromine
Ti	Titanium
Cu	Copper (Cuprum)
Fe	Iron (Ferrum)
Si	Silicon
Ir	Iridium
Pt	Platinum

Write the electronic configuration of alkali metals.

The general electronic configuration of alkali metals is

$ns ^1$ .

Define the following terms.
Element

Element- It is defined as that pure substance which contains only one type of atoms e.g. hydrogen , chlorine .

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements)
of the modern periodic table with their atomic numbers are given.
3 4 5 6 7 8 9 10
A E G
11 12 13 14 15 16 17 18
B C D F
Which of these is an inert gas ? Name it.

	Atomic No.	Elements	Group
A	3	Lithium	1
B	11	Sodium	1
C	12	Magnesium	2
D	14	Silicon	14
E	8	Oxygen	16
F	17	Chlorine	17
G	10	Neon	18

G is an inert gas (Neon) because its atomic number is 18 and electronic configuration is (2, 8, 8).

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements) of the modern periodic table with their atomic numbers are given.
3 4 5 6 7 8 9 10
A E G
11 12 13 14 15 16 17 18
B C D F
Which of these are metals ? Name them.

	Atomic number	Elements	Group
A	3	Lithium	1
B	11	Sodium	1
C	12	Magnesium	2
D	14	Silicon	14
E	8	Oxygen	16
F	17	Chlorine	17
G	10	Neon	18

$\bullet$ A(Lithium),B (Sodium) and C(Magnesium) are the metals as this belongs to Group 1 and Group 2,

$\text{ With reference to period 3 of the periodic table-State}$
The electronic configuration of the element whose hydroxide is a weak base.

$\bullet$ Hydroxides of the metals are bases.

$\bullet$ In period 3, hydroxide of magnesium

$[Mg(OH)_2]$ forms a weak base.

$\bullet$ The atomic number of Mg is 12 and thus, the electronic configuration of Mg is

$\textbf{2, 8, 2}$ .

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements) of the modern periodic table with their atomic numbers are given.
3 4 5 6 7 8 9 10
A E G
11 12 13 14 15 16 17 18
B C D F
Which of these is a halogen ? Name it.

	Atomic No.	Elements	Group
A	3	Lithium	1
B	11	Sodium	1
C	12	Magnesium	2
D	14	Silicon	14
E	8	Oxygen	`16
F	17	Chlorine	17
G	10	Neon	18

$\bullet$ F is a halogen (chlorine) because its atomic number is 17 and electronic configuration is (2, 8, 7).

State True or False.
The atomic number of Uus is 117.

117 - Ununseptium (Uus)
115 - Ununpentium (Uup)
104 - Rutherfordium, Ununquadrium (Uuq)

Why $BeO$ and $Al_2O_3$ are amphoteric?

Both

$BeO$ and

$Al_2O_3$ are amphoteric due to diagonal relationship.

They can react with both acids and bases.

B and Si resemble each other in their properties. If true enter 1, else enter 0.

Because of the diagonal relationship that exist between Boron and Silicon , both resemble each other in properties.

$BeCl_2$ and $AlCl_3$ are Lewis acids. If true enter 1, else enter 0.

Because of the diagonal relationship that exist between Beryllium and aluminium , both behave as Lewis acid.

"Decreasing lattice energy favours increased solubility while decreasing hydration energy favours decreased solubility."
Answer whether the above statement is true or false.
If true enter 1, else enter 0.

The salt with the smaller cation has the higher lattice energy and high lattice energies tend to decrease the solubility of a salt. However, the solvation energy of the cation is also important in determining solubility and small cations tend to have higher solvation energies. High solvation energies tend to increase the solubility of ionic substances.

But decrease in hydration energy favours decreased solubility. Upon dissolving a salt in water, the outermost ions (those at the edge of the lattice) move away from the lattice and become covered with the neighboring water molecules. If the hydration energy is equal to or greater than the lattice energy, then the salt is water-soluble. In salts for which the hydration energy is higher than the lattice energy, solvation occurs with a release of energy in the form of heat.

As the atomic radius increases, ionisation energy is found to decrease.
If true then enter 1, else enter 0.

As atomic radius increases, ionisation energy decreases because the valence electron is now too far from nucleus and the hold of nucleus is not strong. Hence, it will be easy to remove an electron.

Therefore, the statement is true.

Match the type of elements (in List I) with the corresponding electronic configuration (in List II).

(A) Inert gas elements beongs to zero group. Their valence shell electronic configuration is

$1s^2$ to

$ns^2 np^6$ .
(B) Main group elements belongs to s and p block. Their valence shell electronic configuration is

$ns^1$ to

$ns^2 np^5$ .
(C) Transition elements belongs to d block. Their valence shell electronic configuration is

$(n - 1)d^{1-9} ns^2$ .
(D) Inner transition elements belongs to f block. Their valence shell electronic configuration is

$(n-2)f^{1 - 14} (n - 1)s^2 p^6 d^{10} ns^2$ .

IUPAC representation of the element of atomic number 115 is Uup.
If true enter 1, else enter 0.

IUPAC representation of the element of atomic number 115 is Uup. It is named Ununpentium.

State whether the given statement is true or false:

Maximum ionization energy stands for He.

Maximum ionization energy stands for He due to its lowest atomic size which results in strong electrostatic attraction between nucleus and valence electrons.

Halogen with highest ionization energy is $F$ . State whether the statement is true or false

Halogen with highest ionization energy is

$F$ . The ionization energies (in kJ/mol) of

$F, Cl, Br$ and

$I$ are

$1681, 1255, 1140$ and

$1008$ respectively.

On moving down the group, the ionization energies decrease. Fluorine is smallest among halogens. It has no tendency to lose it's valence electrons. Hence, it has highest ionization energy.

Find the number of element( s) which is/are present in group-I :
Be, Li, Mg, AI, Na, Ca, Cs

The number of elements that are present in group-I is 3.
Li, Na and Cs are present in group IA.
Be, Mg and Ca are present in group IIA
Al is present in group IIIA.

The number of metals from the following :
$As, Fe, Xe, Li, B, Cl, Ba, P, I, Si$ is :

Three elements

$Fe, Li,$ and

$Ba$ are metals.

Identify the elements occupying peaks and bottommost points.

Ionisation potential is the amount of energy required to remove an electron from an isolated atom or molecule. The peak in the graph is occupied by the noble gas; Ar as maximum amount of energy is required to remove an electron from stable fully filled configuration. Bottom most point is occupied by the alkali metal; K as it has only one valence electron which it readily looses to achieve stable noble gas configuration.

The increasing order of reactivity among group 1 element is $Li < Na < Rb < Cs$ , whereas that among group 17 element is $F > CI > Br > I$ . Explain?

Group 1 elements contain one electron in the valence shell. They have strong tendency to lose valence electron. The tendency to loose valence electron depends upon the ionization enthalpy. On moving down the group, the ionization enthalpy decreases. Due to this, the reactivity of the group increases in the same order

$Li < Na < K < Rb < Cs$ .

In the valence shell of group 17 elements, 7 electrons are present. They have strong tendency to accept on more electron. This tendency depends upon the electrode potentials of group 17 elements which decreases from F to I. Their reactivities also decreases in the same order

$F > Cl > Br > I$ .

Arrange the following in order of increasing reducing property.

$NaH$ , $MgH_2$ and $H_2O$

$NaH$ is an ionic hydride and can easily donate its electrons. It is the most reducing.

$\displaystyle MgH_2$ and

$\displaystyle H_2O$ are covalent hydrides.

$\displaystyle H_2O$ has lower reducing power than

$\displaystyle MgH_2$ as it's bond dissociation energy is higher.

Hence, the increasing order of the reducing property is

$\displaystyle H_2O < MgH_2 < NaH$ .

Which important property did Mendeleev use to classify the elements in his Periodic Table, and did he stick to that?

The important property that Mendeleev used for the classification of elements was atomic mass. He arranged all known elements as per atomic masses. Later, he placed the elements with similar properties together instead of in the order of atomic masses.

What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

On moving down the group, in main group elements the ionization energy regularly decreases due to the following factors.
(i) Atomic size: On moving down the group, as additional shells are added, the atomic size increases.
(ii) Shielding effect: As the number of inner electrons increases, the shielding effect on the outermost electrons increases.
(iii) Nuclear charge: On moving from top to down in a group, the nuclear charge increases. The effect of increase in the atomic size and the shielding effect is much more significant than the nuclear charge. The electron becomes less lightly held to the nucleus on moving down the group.

Due to this, the ionization enthalpies gradually decrease on moving down the group.

The first ionization enthalpy values ( in kJ mol $\displaystyle ^{-1}$ ) of group 13 element are :

B Al Ga In Tl
801 577 579 558 589

How would you explain this deviation from the general trend?

On moving down the group, the ionization enthalpy decreases. This is true for B and Al.

The ionization enthalpy of Ga is unexpectedly higher than that of Al. Ga contains 10d electrons in inner shell which are less penetrating. Their shielding is less effective than that of s and p electrons. The outer electron is held fairly strongly by the nucleus. The ionization enthalpy increases slightly.

A similar increase is observed from In to Tl due to presence of 14f electrons in the inner shell of Tl which have poor shielding effect.

The atomic number of elements A, B and C are (Z - 1), Z and (Z + 1) respectively. B is an inert gas (not helium). Which element among A,B and C has the highest ionization energy? Why?

B is an inert gas and thus has a stable fully filled valence configuration.

As ionisation energy is the amount of energy required to remove the valence electron.

Thus, it will be highest for B with atomic number Z.

In the Modern Periodic Table, calcium (atomic number $20$ ) is surrounded by elements with atomic numbers $12, 19, 21$ and $38$ . Which of these have physical and chemical properties resembling calcium?

The electronic configuration of

$Ca(20)$ is

$=2, 8, 8, 2$

Number of valance electrons

$=2$

The elements with atomic number

$12$ and

$38$ both have

$2$ valence electrons in their outer shell but elements with

$19$ has 1 valance electron and

$21$ has

$3$ valance electrons. Therefore elements with atomic number

$12$ and

$38$ have physical and chemical properties resembling calcium.

The electronic configuration of $He$ is $2$ and that of oxygen is ________.

The electronic configuration of

$He$ is K(2) i.e, 2 electrons are present in K shell or 1st shell. So, we will write it as

$2$

The electronic configuration of

$O$ is K(2), L(6) i.e, 2 electrons are present in K shell or 1st shell and 6 electrons are present in L shell or 2nd shell. So we will write it as

$2,6$

True or False
Potassium is the most reactive alkali metals.

Potassium is in the most reactive group of elements, the alkali metals, but it's not the most reactive metal within the group.

The alkali metals, Group 1A, are the most reactive metals because they have one valence or outer electron. They lose this electron very easily, forming ions with a charge of +1. Within the alkali metals family, reactivity increases with increasing atomic number. This makes Francium the most reactive, followed by cesium, rubidium, potassium

Which one has greater value, second ionisation energy or third ionisation energy?

The 3rd ionization energy is greater than 2nd ionization energy. Because, after 2nd ionization, metal M forms M

$^{+2}$ . Protons are more than electrons and nuclear charge attracts the remaining electrons to come closer. To overcome this, some extra energy is to be supplied than 2nd ionization energy.

Which is more metallic: sodium or aluminium?

In the Periodic Table, sodium is placed in the group of alkali metals whereas aluminium is right next to the metalloids, towards the right side of the table. Sodium is more metallic than aluminium because metallic character decreases from left to right in the Periodic Table.

Explain why the elements placed on the extreme left side of the periodic table acts as powerful reducing agents?

The elements on the left side of the periodic table are S-block elements. They have high electropositivity and low ionization energy. Means, they easily lose electrons. They give these electrons to reduce other elements. So they are powerful reducing agents.

Nuclear charge increases in a period as well as group. However, periodic properties such as atomic size and IP show a reverse trend. Why?

Nuclear charge increases but in a period, electrons enter into the same shell but in a group, they enter into different shells. When they are in same shells, protons attract easily because of the uniform distance of all electrons. Thus the ionization enthalpy increases.

But in different shells, shielding effect came into existence which increases the atomic size and decreases IP.

So atomic size and IP show reverse trend along a period and group.

Helium has _____configuration.

Helium, atomic number 2, Thus, He has two electrons in one shell around its nucleus, so its electronic configuration is

$1s^2$ .

An element $Z$ has atomic number $16$ . Answer the following questions on $Z$ :
State the formula between $Z$ and Hydrogen?

The atomic number of

$Z$ is

$16$ . Its electronic configuration is

$2,8,6.$ The valency of

$Z$ is

$2$ whereas the valency of

$H$ is

$1$ . Hence to satisfy the valency of

$Z$ , 2 hydrogen atoms are needed. So,

$H_{2}Z$ .

Without knowing the electronic configurations of the atoms of elements, Mendeleev still could arrange the elements nearly close to the arrangements in the Modern Periodic Table. How can you appreciate this?

Mendeleev arranged the elements on the basis of atomic masses. He arranged the elements with same properties in same group. So, without knowing electronic configuration of atoms of elements, he still could arrange the elements similar to Modern Periodic Table, where periodicity was observed in the properties of the elements he arranged.

Aluminium does not react with water at room temperature but reacts with both dil. $HCl$ and $NaOH$ solutions. Verify these statements experimentally. Write your observations with chemical equations. From these observations, can we conclude that $Al$ is a metalloid?

Aluminium is amphoteric . So it reacts with both acids and bases.

2Al +6HCl (acid)

$\rightarrow$ 2AlCl

$_3$ +3H

$_2$

Al +3NaOH (base)

$\rightarrow$ Al(OH)

$_3$ +3Na

Metals react with only acids. Non-metals with bases.

Metalloids react with both.So Aluminium is a metalloid.

Use the letters only written in the Periodic Table given below to answer the questions that follow :
(i) State the number of valence electrons in atom $J$ .
(ii) Which element shown forms ions with a single negative charge?
(iii) Which metallic element is more reactive than $R$ ?
(iv) Which element has its electrons arranged in four shells?

(i) J belongs to the fifth group hence, it has 5 valence electrons.

(ii) M belongs to group number VII hence, it has 7 valence electrons. To attain stability it needs to gain 1 electron and thereby it forms single negative charged ion.

(iii) T is more reactive than R. The reactivity of an element increases down the group.

(iv) The period number indicates the number of shells present hence, T is the element which is having its electrons arranged in four shells.

Metal $A$ has electronic configuration of $2, 8, 1$ and metal B has electronic configuration of $2, 8, 2$ . Which metal is more reactive and why?

From the configuration we know that both the metal belong to same period but of the adjacent group i.e. A to Group-1 and B to Group-2. Metal

$A$ is more reactive than

$B$ because reactivity of metal goes on decreasing along period from left to right for the metals.

The element eka-silicon in Mendeleev's period table is known as ________ in the modern periodic table.

Eka-silicon in Mendeleev's periodic table is known as germanium.

Identify the following substances which are underlined:
The element which has the highest ionization potential.

The energy which is requires to remove an electron from the atom is called ionization potential. It increases across the period and decreases down the group. Thus, helium has the highest ionization potential.

Arrange the following as per the instructions given in the brackets:
$Na, K, Cl, S, Si$ (increasing order of ionization energy)

Ionization energy increases across the period and decreases down the group.

The order is :

$K<Na<Si<S<Cl$

Complete the following Table:
Element Electronic configuration Bond it forms Valency
Magnesium
Oxygen
Chlorine

Element	Electronic configuration	Bond it forms	Valency
Magnesium	$1s^2 2s^2 2p^6 3s^2$	2	2
Oxygen	$1s^2 2s^2 2p^4$	2	2
Chlorine	$1s^2 2s^2 2p^6 3s^2 3p^5$	1	1

In Period $3$ of the Periodic Table, element B is placed to the left of element A. On the basis of this information, choose the correct word from the brackets of complete the following statement.
(i) The element B would have(lower/higher) metallic character than A.
(ii) The element A would probably have(lesser/higher) electron affinity than B.
(iii) The element A would have(greater/smaller) atomic size than B.

(i) The element B would have higher metallic character than A

(ii) The element A would probably have higher electron affinity than B

(iii) The element A would have smaller atomic size than B.

This is because as we move left to right in any period of periodic table the metallic character increases and electron affinity becomes more and more positive. Also, atomic size decreases along the period as we move left to right.

In the modern periodic table, how does the atomic size of the elements vary along the period and down the group? Explain.

On moving from left to right in a period, the atomic size decreases. More and more electrons are added to same shell. The attraction between nucleus and valence electrons increases with increase in the atomic number.
On moving down the group, from top to bottom, the atomic size increases. This is because with increase in the atomic number, extra shells of electrons are added.

The ionisation potential value of $K$ is less than that of $Na$ . Explain.

The removal of an electron from the valence shell of a neutral gaseous atom requires energy which is termed as the ionization energy or ionization potential.

Due to the greater size of

$K$ as compared to

$Na$ , the electrostatic attraction due to the nucleus is reduced in

$K$ . So the energy required to remove an electron from the outermost shell of

$K$ is less than that of

$Na$ .

How can u predict the group, period and block of an element? Explain with examples.

Prediction of Period

$\longrightarrow$ In the electronic configuration of an atom, the principal quantum number (n) of the valence electrons represents the period of an element.

Prediction of Block

$\longrightarrow$ The orbital out of

$s,p,d$ and

$f$ which receives the last electron represents the block of an element

Prediction of Group

$\longrightarrow$ The group prediction occurs with the help of block as:-

(1) For s block: Group Number = Number of valence

$e^-$

(2) For d block: Group Number = Number of

$e^-$ in

$(n-1)d$ subshell + number of

$e^-$ in

$ns$ subshell

(3) For p block: Group Number = Number of valence

$e^-+10$

(4) For f-block: Group Number =

$3$ (always)

For example for chlorine with

$Z=17$

Electronic configuration =

$1s^22s^22p^63s^23p^5$

Since the last

$e^-$ enters in

$p$ - subshell:

(i)

$\therefore$ Block will be

$'p'$

(ii) For group we have element in

$p$ -block

$\therefore$ Group number =

$10+$ Valence

$e^-$ in

$s$ subshell

$+$ Valence

$e^-$ in

$p$ subshell

Group Number =

$10+2+5=17$

(iii) For period

The quantum number of valence

$e^-=3$

$\therefore$ Period number =

$3$

In the second period, Li to Ne identify the element
(a) Most reactive element
(b) Most non-reactive non-metal
(c) Largest atomic radius
(d) With valency equal to 4

In the second period, there are 8 elements.

a) The most reactive element in period 2 is Fluorine as it has the highest electronegativity.

b) Most non-reactive non-metal in group 2 is neon (Ne). This is because neon is an inert gas and it is stable in its natural monoatomic state.

c) Li has the largest atomic radius in group 2. The atomic radius goes on decreasing from left to right in any period of the periodic table.

What is meant by periodicity in properties of elements with reference to the periodic table?

Repetition of properties after a certain interval is called periodicity of properties. If elements are arranged in increasing order of their atomic number in the periodic table, then elements repeat their properties after a definite interval.

Why do elements in the same group have similar physical and chemical properties ?

Elements in the group have similar chemical properties. This is because their atoms have the same number of electron in the highest occupied energy level.

State the appropriate scientific reasons for the following statements.
The element present in the last shell of an element decides the group in which they are placed.

They are the electrons involved in the chemical bonds with other elements.

Every element in the first coloumn (Group 1) has 1 electron in its outer shell. Every element in the second coloumn (Group 2) has 2 electrons in the outer shell.

Ionisation enthalpies of elements of second period are given below:
Ionisation enthalpy/ $KJ\ mol^{-1} : 520, 899, 801, 1086, 1402, 1314, 1681, 2080$ .

Match the correct enthalpy with the elements and complete the graph given in the figure. Also, write symbols of elements with their atomic number.

Ionisation enthalpy order of second period elements is:

$Li<B<Be<C<O<N<F<Ne$

1381156_1105170_ans_dddbd2c2ceca4c8883f06fbcb7944c14.png

A hydrogen like atom with atomic number Z is in higher excited state of quantum number 'n' . This excited state atom can make a transition to the first excited state by successively emitting two photons of energies 10 eV and 68.2eV respectively. Alternatively the atom from the same the same excited state can make a transition to the 2nd excited state by emitting two photons of energies 4.25eV and 5.95eV respectively. Calculate the value of Z

We have, energy of the electron in the

$n^{th}$ energy state in any atom is given by,

$E=\cfrac {-13.6 z^2}{n^2}eV$

Let the intermediate energy state in the first case be

$x$ . So,

$\cfrac {z^2}{4}-\cfrac {z^2}{n^2}=\cfrac {10.2+17}{13.6}=2 \longrightarrow (1)$

Let the intermediate energy state in the

$2^{nd}$ case be

$y$ . So,

$\cfrac {z^2}{9}-\cfrac {z^2}{n^2}=\cfrac {4.25+5.95}{13.6}=\cfrac {3}{4} \longrightarrow (2)$

$(2)-(1), \cfrac {5}{36}z^2=\cfrac {5}{4}$

$\Rightarrow z^2=9,z=3$

How many electrons,protons and neutrons are present in one molecule of glucose?

A glucose molecule has 96 protons, 84 neutrons and 96 electrons.

A glucose molecule's chemical formula is:

$C_6H_{12}O_6$ .

This is made up of 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms.

For each

$C$ there are - 6 protons, 6 electrons and 6 neutrons

For each

$H$ there is - 1 proton, electron and no neutrons

For each

$O$ there are - 8 protons, 8 electrons and 8 neutrons

So in

$C_6H_{12}O_6$ there are:

$C: 6\times 6=$ 36 protons, 36 electrons and 36 neutrons

$H: 1 \times 12=$ 12 protons, 0 electrons and no neutrons

$O: 8 \times 6=$ 48 protons, 48 electrons and 48 neutrons

An element placed in 2nd Group and 3rd Period of the Periodic Table, burns in presence of oxygen to form a basic oxide.
(a) Identify the element.
(b) Write the electronic configuration.
(c) Write a balanced equation when it burns in the presence of air.
(d) Write a balanced equation when this oxide is dissolved in water.
(e) Draw the electron dot structure for the formation of this oxide.

a) Magnesium (Mg)

b) K,L,M
2,8,2
c)

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$
d)

$MgO(s)+H_2O(I)\rightarrow Mg (OH)_2(aq)$

1646383_1178696_ans_f3c6169341194d5192a40124f498f70a.png

An element $'X'$ has mass number $35$ and number of neutrons $18$ . Write atomic number and electronic configuration of $'X'$ . Also write group number, period number and valency of $'X'$ .

Explanation:

Mass number $={\text{Number of protons}}+{\text{Number of neutrons}}$

An element $X$ has mass number $=35$

The number of neutrons $=18$ .

So, the number of protons $=35-18=17$ .

The atomic number is the number of protons.

So, the atomic number $=17$ .

Electronic configuration is-

$1s^2,2s^2,2p^6,3s^2,3p^5$

The element is Chlorine.

The period number is $3$ and the group number is $17$ .

The valency is $1$ .

The second ionization enthalpy of calcium is more than that of the first and yet calcium form ${ CaCl }_{ 2 }$ not CaCl. Why ?

The higher enthalpy of lattice formation by

$Ca^{2+}$ ions more than compensates the higher ionization enthalpy of calcium.

$H_2O$ is a liquid but $H_2S$ is a gas. Explain why?

In water, Hydrogen bonding is possible resulting in strong inter-nuclear bonding because of high electro-negativity of oxygen while in

$H_{2}S$ , only weak Van Der Waal's or London force of attraction between molecules is possible.

Therefore,

$H_{2}O$ is liquid &

$H_{2}S$ is gas at room temperature.

How does number of valency vary on moving from left to right:
in the third period of a periodic table?

a) On moving from left to right in the second period, the number of valence electrons increases from 1 (Li) to 8 (Ne), whereas the valency increases from 1 (Li) to 4 (C) and then decreases from 4 to 0 (Ne).

b) Similarly, in third period the number of valence electrons increases from 1 (Na) to 8 (Ar) i.e., from left to right, whereas the valency increases from 1 (Na) to 4 (Si) and then decreases from 4 to 0 (Ar).

Define the term ionisation enthalpy.

1337719_1312749_ans_b9aa11564a4849e8875fa28a5c0ed1d3.jpg

Among the element of the second period Li to ne pick out the element:
(i) with the highest first ionisation energy
(ii) with the highest electronegativity

1326093_1312460_ans_aace44f50f674f0a809a009757cad965.jpg

Write a short note on Mendeleev's periodic law.

Mendeleev's periodic law states that " The properties are a periodic function of their atomic mass." This means that the chemical and physical properties of the elements recur periodically when the elements are arranged in the order of their atomic weights. Mendeleev arranged the elements known at that time in order of increasing atomic masses and this arrangement was called the periodic table. Elements with similar properties were grouped into vertical rows called columns. The horizontal rows were known as periods.

Four element $P , Q , R$ and $S$ belong to the third period of the Modern Periodic Table and have respectively $1,3,5$ and $7$ electrons in their outermost shells. Write the electronic configurations of $P, Q$ and $R$ and determine their valences. Write the molecular formula of the compound formed when $P$ and $S$ combine.

$\bullet$ The number of electrons present in the outermost shells of P, Q, R, and S is 1, 3, 5, and 7 respectively.

$\bullet$ Since the elements belong to period 3 thus, they have 3 shells.

$\textbf{Part 1: Electronic configurations of P, Q, R, and their valences}$

$\bullet$ The electronic configuration of P is

$2, 8, 1$ . Since it has 1 electron in its valence shell thus, the valency of P is

$1$ .

$\bullet$ The electronic configuration of Q is

$2, 8, 3$ . Since it has 3 electrons in its valence shell thus, the valency of Q is

$3$ .

$\bullet$ The electronic configuration of R is

$2, 8, 5$ . Since it has 5 electrons in its outermost shell, it has a tendency to accept 3 electrons to complete its octet. Thus, the valency of R is

$3$ .

$\textbf{Part 2: Molecular formula of the compound formed between P and S}$

$\bullet$ The electronic configuration of P is 2, 8, 1 and its valency is 1.

$\bullet$ The electronic configuration of S is 2, 8, 7. It has a tendency to accept an electron to complete its octet. Thus, the valency of S is 1.

$\bullet$ The valencies of P and S are interchanged by using the criss-cross valency method. Therefore, the molecular formula of the compound formed when P and S combine is

$PS$ .

Arrange the following in increasing order of the property indicated:

$I_2,\ F_2,\ Br_2$ and $Cl_2$ (reactivity)

$I_2 < Br_2 < Cl_2 < F_2$

Name the pairs showing a diagonal relationship.

Following pairs show a diagonal relationship.

$Li, Mg;\ Be, Al;\ B , Si$

Arrange the following in increasing order of the property indicated:

Na, Cu and Zn (electropositive character)

$Cu < Zn < Na$

With reference to periodic table, indicate:

Position of the most electropositive element.

IA group and VII th period as on moving down the group electropositive nature increase and on moving right it decrease.

Explain the following:

Why is potassium a strongly electropositive element?

The ionisation potential of potassium is low. Hence, it can lose an electron easily. Therefore, it is strongly electropositive in nature.

The element 119 has not been discovered. What would be IUPAC name and symbol for this element? On the basis of the periodic table, predict the electronic configuration of this element and also the formula of its most stable chloride and oxide.

The IUPAC name and symbol for this element is Ununennium and uue.

The electronic configuration of this element is: 2, 8, 18, 32, 32, 18, 8, 1.

The formula of its most stable chloride and oxid is

$MCl$ and

$M_2O$ respectively.

From among the elements, choose the following:
Cl, Br, F, O, Al, C, Li, Cs and Xe

The element which shows diagonal relationship with Mg.

Li is the element which shows diagonal relationship with Mg.

Give the electronic configuration of ninth element of the first transition series.

Ninth element of the first transition series is Cu.

$Cu-1s^2, 2s^2,2p^6 , 3s^2,3p^6,3d^{10}, 4s^1$

Give the electronic configuration of fifth element of the first transition series.

Fifth element of the first transition series is Mn.

$Mn-1s^2, 2s^2,2p^6, 3s^2,3p^6,3d^5 , 4s^2$

Give the name and atomic number of the inert gas atom in which the total number of d-electrons is equal to the difference in numbers of total p and s-electrons.

Inert gas atom is krypton. Its atomic number is 36. The electronic configuration of Kr-atom is:

$1s^2 , 2s^2 , 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6$

Total s-electrons

$= 8$

Total p-electrons

$= 18$

Difference

$= 18 - 8 =10$

Total d-electrons

$= 10$

The elements Na, Mg, Al, Si, P, S, Cl and Ar are arranged in the periodic table in the increasing order of their atomic numbers.

Which element has the electronic configuration?
$1s^2 , 2s^2 , 2p^6 , 3s^2, 3p^1_x3p^1_y3p^1_z$

P has the electronic configuration

$1s^2 , 2s^2 , 2p^6 , 3s^2, 3p^1_x=3p^1_y=3p^1_z ,$ since it has 5 electrons in the valence shell.

The electronic configurations of some of the elements are given below:
(A) $1s^2 , 2s^2 , 2p^6,3s^2$
(B) $1s^2 , 2s^2 , 2p^6$
(C) $1s^2 , 2s^2 , 2p^3$
(D) $1s^2, 2s^2 ,2p^6,3s^1$
(E) $1s^2 , 2s^2 , 2p^5$

(i) Name the elements.

(ii) Which of these has low ionisation enthalpy?

(iii) Which is a halogen?

(iv) Which is an alkali metal?

(v) Which is an inert gas?

(i) (A) Mg (Z=12) (B) Ne (Z=10) (C) N (Z=7) (D) Na (Z=11) (E) F (Z=9)

(ii) D: Na is alkali metal.

(iii) E: F belongs to the halogen.

(iv) D: Na is alkali metal.

(v) B: Ne an inert gas.

Which elements have the electronic configuration $ns^2\ np^2$ ? Give their names.

$ns^2\ \ np^2$ indicates that elements belong to the p-block element having 2 electrons in the outermost shell.

Hence, elements of

$IVA$ or

$14^{th}$ group have this configuration.

Carbon, Silicon, Germanium, Tin, and Lead.

An element X from group 2 reacts with element Y from group 16 of the periodic table
(a) What is the formula of the compound formed?
(b) What is the nature of the bond in the compound formed?

(a) Since the valency of X and Y are 2 so the formula of the compound by criss cross method is

$XY.$

(b) The bond form by elements X and Y is ionic or electrovalent bond.

A metal X is in the first group of the periodic table. What will be the formula of its oxide?

Valency of group 1 metals is 1 so it will react with oxygen (valency =2 ) to form compound

$X_ 2 O.$

Example:

$Na_2O,\ K_2O, etc$

Name two elements whose properties were predicted on the basis of their positions in Mendeleev's periodic table.

Gallium(eka-Aluminium) and Scandium(eka-Boron) are two elements whose properties were predicted on the basis of their positions in Mendeleev's periodic table.

(a) What is the fundamental difference in the electron configurations between the group 1 and group 2 elements ?
(b) On the basis of electronic configuration how will you identify
(i) Chemically similar electrons ?
(ii) The first elements of a period ?

(a) The fundamental difference between the electron configuration of group 1 and group 2 elements is that group 1 elements have 1 valence electron in their atoms whereas group 2 elements have 2 valence electron in their atoms.

(b) (i) All the chemically similar elements will have same valence electrons.

(ii) The 1st element in a period is determined by the number of valence electrons in its atoms. The 1st element of every period has 1 valence electron.

Atoms of eight elements A, B, C, D, E, F, G and H have the same number of electronic shells but different number of electrons in their outermost shell. It was found that elements A and G combine to form an ionic compound which can also be extracted from sea water. This compound is added in a small amount to almost all vegetable dishes during cooking. Oxides of elements A and B are basic in nature while those of E and F are acidic. The oxide of D is almost neutral. Based on the above information,
Answer the following questions ;
(a) To which group or period of the periodic table do these elements belong?
(b) What would be the nature of compound formed by a combination of elements B and F?
(c) Which two of these elements could definitely be metals?
(d) Which one of the eight elements is most likely to be found in gaseous state at room temperature?
(e) If the number of electrons in the outermost shell of elements C and G be 3 and 7 respectively, write the formula of the compound formed by the combination of C and G.

(a) It belongs to the third period.

Group - 1, 2, 13, 14, 15, 16, 17, 18, respectively.

(b) The nature of the compound formed by a combination of elements B and F is ionic.

(d) H is most likely to be found in the gaseous state at room temperature.

(e) The formula of the compound formed by the combination of C and G is

$CG_3.$

What property do all the elements in the same group of the periodic table as fluorine have in common

The element fluorine is in group 17 of the peiodic table and has a valency of 1 So all the elements in the same group of periodic table as fluorine will have a valency of 1.

Nitrogen (atomic number 7) and phosphorus (atomic number is 15) belong to group 15 of the periodic table . Write the electronic configurations of these two elements. Which of these will be more electronegative why?

Electronic configuration N(2,5) ; P(2,8,5);

Nitrogen will be more electronegative because its atoms has small size due to which the attraction of its nucleus for the incoming electron is more.

How do electronic configurations of elements change in second period of the periodic table with increase in atomic numbers?

Number of shell remain same but number of valence electrons increase as we move from left to right in a period.

Li - 2,1;

Be - 2,2;

B - 2,3;

C - 2,4;

N - 2,5;

O - 2,6;

F - 2,7 and

Ne - 2.8.

How does the electronic configuration of an atom relate to its position in the modern periodic table?

The modern periodic table contains groups and periods. So, the electronic configuration of an atom relates to its position in the modern periodic table as:

1) The atoms with identical outer shell electronic cell electronic configuration are in the same groups.

2) The atoms with the same number of shells are in the same period.

This question refers to the elements of the periodic table with atomic numbers $3$ to $18$ .
What is the electronic configuration of an element with atomic number $10$ ?

1693291_1773973_ans_5a772d4fdd1b4a6d87b72253c3471466.jpg

Study the periodic table and answer the following questions:
(i) Na has physical and chemical properties similar to which element(s) and why ?
(ii) Write the electronic configuration of N and P. Which one of these will be more electronegative and why ?

(i) Na has similar properties as of the same group members as K , Rb , Cs . They contain 1 valence in the outermost shell so all of the, are reactive.
(ii) Nitrogen :- 2,5
Phosphorus :- 2,8,5
Electronegativity is measure of how strongly atoms attract bonding electrons to themselves.
Nitrogen is more electronegative because in nitrogen outermost shell is nearer to nucleus and nucleus will attract electrons easily.

Show the formation of $Na_2O$ by the transfer of electrons.

Sodium loses one electron to have a completely filled valence shell , where as oxygen gains two electrons to complete the valence shell.

$Na \rightarrow Na^+ + e^-$

$O + 2e^- \rightarrow O^{2-}$

1622104_1772211_ans_1e06859170b54b2c9362bb37de5e87da.png

An element has electronic configuration $2,8,7$ .
What is atomic number of this element?

1691205_1774054_ans_050dd8d4b8fc4c2191b93b3f62f6f847.jpg

Name any two elements of group one and write their electronic configurations.What similarity do you observe in their electronic configurations? Write the formula of oxide of any of the aforesaid element.

Two elements of group I are Sodium (Na) and potassium (k).

Their electronic configurations: Na=2,8,1; K=2,8,8,1

Similarly: Both have one valence electron.

Formula of their oxides:

$Na_2O;K_2O$

In nature, metal A is found in a free state state while metal B is found in the from of its compounds. Which of these two will be nearer to the top of the activity series of metals?

Metal B will be nearer to the top of the activity series of metals as it is so reactive that it is found in combined state

An element has electronic configuration $2,8,7$ .
Identify the element.

1698535_1774062_ans_de8b2336af6c4f2b83f7d6fedef64995.jpg

In the following table are given eight elements, $A, B, C, D, E, F, G$ and $H$ (here letters are not the usual symbols of the elements) of the Modern Periodic Table with the atomic numbers of the elements in parenthesis.
Period Group $1$ Group $2$
$2$ $A(3)$ $E(4)$
$3$ $B(11)$ $F(12)$
$4$ $C(19)$ $G(20)$
$5$ $D(37)$ $H(38)$
What is the electronic configuration of $F$ ?

1705137_1774153_ans_940441d0250f462ca24137171a7660f7.jpg

An element has electronic configuration $2,8,7$ .
Name an element chemically similar to this element

1698536_1774064_ans_6765a2347f7b45f8953cf46acbf8347d.jpg

Atoms of seven elements $A, B, C, D, E, F$ and $G$ have a different number of electronic shells but have the same number of electron in their outermost shell. The elements $A$ and $C$ combine with chlorine to form an acid and common salt respectively. The oxide of element $A$ is liquid at room temperature and is a neutral substance, while the oxide of the remaining six elements are basic in nature.
Based on the above information, answer the following question:
Show the formation of the compound by a combination of element $C$ with chlorine with the help of electronic structure.

1710153_1774115_ans_68b86c76e34f41e7971c15c380952b07.jpg

Two elements $X$ and $Y$ have atomic number $12$ and $16$ respectively. To which period of the modern periodic table do these two elements belong? What type of bond will be formed between them and why? Also give the chemical formula of the compound formed.

1705276_1774174_ans_4fd22fe134d343c5a6f42f6d71fe0b35.jpg

The atomic number of Lithium is $3$ . On the basis of this information answer the question that follows:
Write the electronic configuration of $Li$

1691217_1774176_ans_f9f2246278134f4aa843b2379ba0546c.jpg

In the following table, sic element $A, B, C, D, E$ and $F$ (here letter are not the usual symbols of the elements) of the modern Periodic Table with atomic numbers $3$ to $18$ are given:
$3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$
$A$ $E$
$G$
$11$ $12$ $13$ $14$ $15$ $16$ $17$ $18$
$B$ $C$ $D$ $F$
Write the electronic configuration of $C$ and $E$ .

1705136_1774142_ans_afa2e143b28542608d3518eee6b1b5ba.jpg

Who classified the elements on the basis of fundamental properties of elements like atomic mass?

Mendeleev classified the elements on the basis of fundamental properties of elements like atomic mass.

The position of eight elements in the Modern Periodic Table is given below where atomic numbers of elements are given in the parenthesis.
Period No
2 Li(3) Be(4)
3. Na(11) Mg(12)
4. k(19) Ca (20)
5. Rb(37) Sr(38)
i) Write the electronic configuration of Ca
ii) Predict the number of valence electrons in Rb
iii) What is the number of shells in Sr?
iv) Predict whether k is a metal or a non-metal.
v) Which one of these elements has the largest atom in size?
vi) Arrange Be,Ca,Mg and Rb in the increasing order of the size of their respective atoms.

Electronic configuration of Ca-2,8,8,2

i) Valence electrons in Rb is 1.

ii) Five

iv) Metal

v) Rb is biggest in size.

vi) Be < Mg < Ca < Rb

The atomic number of an element 'X' is 19
a) Write its electronic configuration.
b) To which period of the Modern Periodic Table does it belong and what is its valency?
c) If 'X' burns in oxygen to form its oxide,what will be its nature-acidic,basic or neutral?
d) Write balanced chemical equation for the reaction when this oxide is dissolved in water.

a) Configuration of X(19)=2,8,8,1.

b) It belongs to fourth period and its valency is 1.

c) Basic oxide

$(X_2O)$

$X_2O+H_2O\rightarrow 2XOH$

Identify and name the metals out of the following elements whose electronic configurations are given below.
(a) 2, 8, 2
(b) 2, 8, 1
(c) 2, 8,7
(d) 2, 1

(a) Metal; Magnesium
(b)Metal; Sodium
(c) Non metal : chlorine
(d) Metal; Lithium

In Mendeleev's Periodic Table the elements were arranged in the increasing order of their atomic masses. However, cobalt with atomic mass of 58.93 amu was placed before nickel having an atomic mass of 58.71 amu. Give reason for the same.

There are instances in Mendeleev's Periodic Table when an element with higher atomic mass is placed before and element with lower atomic mass. This was done to ensure that elements with similar chemical properties were put in the same group. Due to this, cobalt was placed before nickel in spite of the fact that cobalt has higher atomic mass than nickel.

'M' is an element which is out of Cu, Fe, Al, Na. It shows the following properties:
(i) One of its ore is rich in $M_{2}O_{3}$ .
(ii) $M_{2}O_{3}$ is not affected by water.
(iii) It corrodes easily.
(iv) It forms two chlorides $MCl_{2}$ and $MCl_{3}$ .
Identify 'M'.

The metal 'M' forms oxide

$M_{2}O_{3}$ it is trivalent. Out of the metals listed, only Fe and Al are trivalent.

$M_{2}O_{3}$ is not affected by water, so 'M' can be out of Fe or Al. Fe and Al both corrode easily.

An element X which is a yellow solid at room temperature shows catenation and allotropy.X forms two oxides which are also formed during the thermal decomposition of ferrous sulphate crystals and are the major air pollutants.
a) Identify the element X
b) Write the electroic configuration of X.
c) Write the balanced chemical equation for the thermal decomposition of ferrous sulphate crystals.
d) What would be the nature (acidic/basic) of oxides formed?
e) Locate the position of the element in the Modern Periodic Table.

a) Element X is sulphur (atomic no 16)

b) K,L,M

2,8,6

$2FeSO_4(s)\overset{Heat}{\rightarrow}Fe_2O_3(s)+SO_2(g)+SO_3(g)$

d) Acidic

e) 3rd period,group 16

How do you classify elements into metals and non-metals on the basis of their electronic configuration? Choose metal and non-metal out of the following:

For obtaing metals that are in the middle of reactivity series, oxides of such metals can be reduced with coke 9carbon) which acts as a reducing agent.

Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the elements predicted by Mendeleev.

Chloride of Eka-silicon:

$ECl_4$
Chloride of Eka-aluminium:

$ECl_3$

An element X (atomic number 17) reacts with an element 1' (atomic number 20) to form a divalent halide.
(a) Where in the periodic table are elements X and Y placed?
(b) Classify X and Y as metal (s), non-metal (s) or metalloid (s)
(c) What will be the nature of oxide of element Y? Identity the nature of bonding in the compound formed
(d) Draw the electron dot structure of the divalent halide

(a) X is in Group 17, period 3 and Y is in group 2. period 4
(b) X is a non-metal, While Y is a metal.
(c) Oxide of element Y would be basic. This compound is formed by ionic bond.
(d)
1651515_1791654_ans_a47955b4ac904dfa9644c30787881e52.png

1651515_1791654_ans_a47955b4ac904dfa9644c30787881e52.png

(a) In this crossword puzzle (Figure), the names of $11$ elements are hidden. Symbols of these elements are given below. Complete the puzzle.
(1.) Cl (2.) H (3.) Ar (4.) O (5.) Xe (6.) N (7.) He (8.) F (9.) Kr (10.) Rn (11.) Ne
(b) Identify the total number of inert gases, their names, and symbols from this crossword puzzle.

(b) There are 5 inert gases in the crossword puzzle. Their names and symbols are Argon(Ar), Xenon(Xe), Helium(He), Krypton(Kr) and Radon(Rn).
1652811_1793299_ans_ae0032de7fb54c3fb411246467ecbc37.png

1652811_1793299_ans_ae0032de7fb54c3fb411246467ecbc37.png

Is the following symbol of element incorrect? Give the correct symbol.
Sodium So

Incorrect, the correct symbol of sodium is Na. (The symbol of sodium has been derived
from the Latin word - Natrium.)

Is the following symbol of element incorrect? Give the correct symbol.
Aluminum AL

Incorrect, the correct symbol of aluminium is Al and not AL.

Is the following symbol of element incorrect? Give the correct symbol.
Helium He

Correct, He is the correct symbol for Helium. (The first alphabet in the symbol He is a
capital letter and the second alphabet is a small letter.)

Mendeleev predicted the existence of certain elements not known at that time and named two of them as Eka-silicon and Eka-aluminium.
(a) Name the elements which have taken the place of these elements
(b) Mention the group and the period of these elements in the Modern Periodic Table.
(c) Classify these elements as metals, non-metals or metalloids
(d) How many valence electrons are present in each one of them?

(a) Eka-silicon was replaced by Germanium and Eka-aluminium was replaced by Gallium.
(b) Gallium: Group 13 Period 5
Germanium: Group 14 period S
(c) Gallium is metal, while Germanium is metalloid
(d) Gallium has 3 valence electrons, while Germanium has 4.

Is the following symbol of element incorrect? Give the correct symbol.
Cobalt CO

Incorrect, the correct symbol of cobalt is Co and not CO.(By convention, the second
alphabet in a symbol is a small letter.)

Is the following symbol of element incorrect? Give the correct symbol.
Carbon c

Incorrect, the correct symbol of carbon is C (By convention, the first alphabet in a symbol is always a capital letter.)

Give an account of the process adopted by Mendeleev for the classification of elements. How did he arrive at "Periodic Law"?

During Mendeleev's time, only 63 elements were known. Mendeleev examined the correlation between chemical properties and Atomic masses of elements. For this, he concentrated on compounds which were formed by the elements with hydrogen and oxygen. He selected hydrogen and oxygen, because these elements make compound with most of the other elements. Mendeleev made 63 cards and wrote the name of an element on each card; along with the chemical properties of a particular element. Then. he pinned those cards on a wall. He could observe that most of the elements could be arranged in periodic table in increasing order of their atomic masses. That is how Mendeleev arrived at "Periodic law". Mendeleev's Periodic Law states that properties of elements are periodic function of their atomic masses.

Fill in the blanks
_________ refers to the number of atoms in the molecule of an element.

$\underline{\textbf{Atomicity}}$ refers to the number of atoms in the molecule of an element.
For example, Water

$H_2O$ contains two atoms of hydrogen and one atom of oxygen and hence its atomicity is

$2+1= 3$ i.e triatomic.

Name the scientists who discovered the following:
Elements

Antoiene Lavoisier was a French nobleman.He revolutionized chemistry.Lavoiser named the elements carbon,hydrogen and oxygen and discovered the role of oxygen in combustion and respiration for which he is most noted.He established that water is a compound and helped to continue the transformation of chemistry from a quantitative science to a quantitative one.

State the contributions of :
Dimitri Mendeleev

He formulated the -Periodic Table of elements.
1) He systematically arranged the dozens of known elements by atomic weights and could even predict the properties of the still unknown elements.
2) He devised the Periodic Table. He was best known for his discovery of the Periodic Law.

Fill in the blanks
The most abundant element in the earth's crust is __________.

The most abundant element in the earths crust is

$\underline{\textbf{oxygen(nearly 46.1%)}}$ .

Explain the term 'combining capacity' of an element i.e. valency.

The combining power of an element especially as measured by the number of hydrogen atoms can combine or displace with the element.

State the trends in chemical reactivity : down the group

Down the group , chemical reactivity increases as the tendency to lose electrons increases down the group.
The farther to the left and down the periodic chart you go, the easier it is for electrons to be given or taken away, resulting in higher reactivity.

The element X has the electronic configuration 2, 8, 18, 8,Without identifying x,
write if X be an oxidizing agent or reducing agent and why.

The oxidizing agent, because it can donate electrons. as only one electron is present in the outermost shell.

How does the electronic configuration in atoms change in a group top to bottom ?

While going from top to bottom in a group, the number of shells increases successively i.e., one by one but the number of valence electrons remains the same.

How does the electronic configuration in atoms change in a period from left to right ?

Though the number of shells remain the same , number of valence electrons increases by one , as we move across any given period from left to right.

The element X has the electronic configuration 2, 8, 18, 8,Without identifying x,
predict the sigh and change on a simple ion of X.

$X^{+}$

as element X has only one electron in the outermost shell so by donating it acquire stable configuration.

How is sodium chloride different from its constituent elements.

The properties of sodium chloride are completely different from those of sodium and chlorine. Sodium is a soft, highly reactive metal. Chlorine is a poisonous non-metallic gas while sodium chloride is a very useful non poisonous compound which Is added to our food to get minerals and also to add taste to it.

Define:
(a) Elements
(b) Compounds

(a) Elements: Element is a substance which cannot be broken further into simpler substances and has a definite set of properties. Elements are made up of only one kind of atoms.

(b) Compounds: Compounds are pure substances composed of two or more elements in definite proportion by mass and has properties. entirely different from those of its constituents elements. Compound. are made up of different types of atoms combined chemically.

An element barium has atomic number $56$ . Look up its position in the periodic table and answer the following questions :
Is it more or less reactive than calcium ?

$Barium$ is placed below

$Cesium$ in the group. Since , it has more number of shells ; it is easier for it to lose its valence electrons to complete its octet configuration . Hence it is

$\textbf {more reactive}$ than calcium.

State the trends in chemical reactivity : across the periods left to right

Across a period, the chemical reactivity of elements first decreases and then increases.

What is the common feature of the electronic configuration of the elements at the end of period $2$ and period $3$ ?

The element at the end of period

$2$ and Period

$3$ have

$8$ electrons in its outermost shell.

A group of elements in the periodic table are given below ( boron is the first member of the group and thallium is the last).
Boron , Aluminium , Gallium , Indium , Thallium.
Answer the following question in relation to the above group of elements :
If the electronic configuration of aluminium is $2 , 8 , 3$ how many electrons are there in the outers shell of thallium.

$3$ . Since all the elements in a group have same number of valence electrons.

The electronic configuration of nitrogen is $(2, 5)$ . How many electrons in the outer shell of a nitrogen atom are not involved in the formation of a nitrogen molecule?

Two electron

Elements X, Y and Z have atomic numbers 6, 9 and 12 respectively. Which one:
Stat type of bond between Y and Z and give its molecular formula.

Molecular formula is X

How do the following change on moving from left to right in a period of the periodic table ? Give examples in support of your answer.
Chemical reactivity of elements .

On moving from left to right in a period of the periodic table chemical reactivity of elements first decreases and then increases.

Example : In the 3rd period of elements, Sodium is a very reactive element, magnesium is less reactive whereas aluminium is still less reactive in 3rd period. Now the phosphorus is quite reactive, sulphur is still more reactive whereas chlorine is very reactive.

Arrange the following metals in the decreasing order of reactivity.
Copper, iron, magnesium, sodium and zinc.

Arrangement of metal in decreasing order of reactivity are:

$Sodium > Magnesium > Zinc > Iron > Copper$

How does the chemical reactivity of alkali metals vary ?

Since , the distance of the valence electrons from the nucleus keeps on increasing down the group , therefore the ionization energy keeps on decreasing . Hence the reactivity of alkali metals increases from lithium to francium.

Which element has :
The electronic configuration $2 , 8 , 3$ ?

Aluminum

Which element from the following has the highest ionization energy ?
Explain your choice.
Ne , He , Ar

$He$

Ionization energy decreases with an increase in the atomic size , i.e., it decreases as one moves down a group. Ne , He and Ar are inert gases . He - Period

$1$ , Ne - Period

$2$ and Ar - Period

$3$ .

Explain why the following is not correct :
Reactivity increases with atomic number in a group as well as in a period.

On moving from left to right in a period , the reactivity first decreases and then increases since the tendency to lose electrons first decreases on going from left to right and then from P to Cl tendency to gain electrons increases , so reactivity increases then. In case of a group, reactivity increases on going down since the tendency to lose electrons increases but for nonmetals , reactivity decreases on going down the group as the tendency to gain electrons decreases down the group.

Write down the electronic configuration of the : $^{35}_{17}Y$

Electronic configuration

$2,8,7$
Number of electrons

$=17$
Number of neutrons

$=35-17=18$

Write down the electronic configuration of the following:
$^{27}_{13}X$

Electronic configuration

$2,8,3$
Number of electrons

$=13$
Number of neutrons

$=27-13=14$

In which part of a group would you separately expect the elements to have the greatest metallic character?

Greatest metallic characters are shown in the elements at the bottom of the group. Because of their large size, they can easily lose electrons.

From the standpoint of atomic structure, what determines which element will be the first and which the last in a period of the periodic table?

From the standpoint of atomic structure, atomic number determines which element will be the first and which the last in a period of the periodic table.

State the merits of Mendeleev's classification of elements.

Merits of Mendeleev’s classification of elements are as follows:

1. It generalised the study of the elements.

2. Mendeleev left some gaps in his periodic table for subsequent inclusion of elements not known at that time.

3. He predicted the properties of then-unknown elements on the basis of the property of elements lying adjacent to the vacant slots.

From the symbol $\dfrac {24}{12}Mg$ state the mass number, and atomic configuration of magnesium.

Mass number

$=24$
Atomic number

$=12$
No. of electrons

$=24-12=12$
Electrons configuration

$=2,8,2$

This question refers to elements of the periodic table with atomic numbers from $3$ to $18$ . In the table below, some elements are shown by letters, even though the letters are not the usual symbols of the elements.What is the electronic configuration of $G$ ?

Atomic number of

$G$ is

$9.$ Hence electronic configuration is

$2,7.$

Name two elements whose atomic weights were corrected on the basis of their positions in Mendeleev's periodic table.

Gold and Platinum are two elements whose atomic weights were corrected on the basis of their positions in Mendeleev's periodic table.

What are the following groups known as?
Group $1$

(i) Group

$1-$ are knowns alkali metals

(ii) Lithium and sodium two elements of group-1.

This question refers to elements of the periodic table with atomic numbers from $3$ to $18$ . In the table below, some elements are shown by letters, even though the letters are not the usual symbols of the elements.
$3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$
$A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$
$11$ $12$ $13$ $14$ $15$ $16$
$17$
$18$
$I$ $J$ $K$ $L$ $M$ $N$ $O$ $P$

Which of these is an alkali metal?

$A$ and

$I$ are alkali metals as atomic number 3 and 11 are Lithium and Sodium respectively.

Write electronic configuration of element $\ ^{35}_{17}T$
State number of protons and neutrons in $T$ .

Electronic configuration of

$T=2,8,7$

Number of protons

$=17$ , Number of neutrons

$=18$

Name the first three alkaline earth metals.
Write their reactions with dil. hydrochloride acid.

Beryllium, Magnesium and Calcium are the first three alkaline earth metals.

Their reactions with dil. hydrochloride acid are-

$Be+2HCl\to BeCl_2+H_2$

$Mg+2HCl \to MgCl_2+H_2$

$Ca+2HCl \to CaCl_2+H_2$

An element $A$ has $2$ electrons in its, fourth shell. State: is it a metal or a non-metal.

Element

$A$ is an alkaline earth metal.

An element $A$ has $2$ electrons in its fourth shell. State: its electronic configuration.

Electronic configuration of A is

$1s^2 2s^2 2p^63s^2 3p^6 4s^2.$

Name the alkali metals. How many electrons do they have in their outermost orbit?

Lithium, sodium, potassium, rubidium, caesium and francium are alkali metals.

They have

$1$ electron in their outermost orbit.

An element $A$ has $2$ electrons in its fourth shell. State: is it an oxidizing or reducing agent?

Element

$A$ is a reducing agent since it can easily lose two electrons.

Write the electronic configuration of the first two alkaline earth metals.

The electronic configuration of the first two alkaline earth metals are -

$_4Be\to 1s^2 2s^2$

$_{12}Mg\to 1s^2 2s^2 2p^6 3s^2$

An element $A$ has $2$ electrons in its fourth shell. State: its atomic number.

An element

$A$ has

$2$ electrons in its fourth shell and its atomic number is

$20.$

Write electronic configuration of element $\ ^{35}_{17}T.$
Is it a metal or a non-metal?

Electronic configuration of

$T=2,8,7.$

Element

$T$ is a non-metal.

Arrange the following:
Elements of group $1$ , in increasing order of reactivity.

$Li < Na < K < Rb < Cs:$ Reactivity increase down the group as it can easily lose the electron since electrons are loosely bonded to the nucleus.

Write the name from the description.
The period with electrons in the shells $K$ , $L$ and $M$

The period with electrons in the shells

$K$ ,

$L$ and

$M = 3$ period.

An element' X has its electronic configuration of 'K shell is $(n - 5)s^2$ and it has the total number of electrons in its outermost, penultimate and antepenultimate shell are $2, 8$ and $25$ respectively, then find out the total number of unpaired electrons in element 'X in their ground state.

1751648_1818620_ans_1108a6da064e4031b0a191bf220d9c1c.jpg

Write down the electronic configuration of the following elements from the given atomic numbers. Answer the following question with an explanation.
$_3Li$ , $_{14}Si$ , $_{11}Na$ , $_{15}P$
Which of these elements belong to be period $3$ ?

$_3Li$ ,

$_{14}Si$ ,

$_{11}Na$ ,

$_{15}P$

Electronic configuration of the following elements is:

$_3Li\ =\ 2,1$

$_{14}Si\ =\ 2,8,4$

$_{11}Na\ =\ 2,8,1$ ,

$_{15}P\ =\ 2,8,5$

$_{14}Si$ ,

$_{15}P$ belong to the third period because according to their electronic configuration , each element contains three shell i.e.

$K$ ,

$L$ ,

$M$ .

Write down the electronic configuration of the following elements from the given atomic numbers. Answer the following question with explanation.
$_{1}H$ , $_7N$ , $_{29}Ca$ , $_{16}S$ , $_4Be$ , $_{18}Ar$ .
Which of these elements belong to the second group?

$_{1}H$ ,

$_7N$ ,

$_{29}Ca$ ,

$_{16}S$ ,

$_4Be$ ,

$_{18}Ar$ .

Electronic configuration of the following elements is:

$_{1}H = 1$

$_7N = 2,5$

$_{29}Ca = 2,8,8,2$

$_{16}S = 2,8,6$

$_4Be = 2,2$

$_{18}Ar = 2,8,8$ .

$_{29}Ca$ , and

$_4Be$ belong to the second group because these elements have

$2$ electrons in its outermost shell.

Comment on the reactivity of Halogens.

Halogens are the most reactive non-metals and their reactivity decreases down the group. Fluorine is the most reactive halogen and iodine is the least reactive halogens.

Answer the following on the basis of the modern periodic table.

(I) Group no. of the elements with the valence shell ground-state electron configuration $ns^2np^5$
(II) Group no. of the elements with the valence shell ground-state electron configuration $ns^2np^3$
(Ill) Group no. of the elements that have only three unpaired p electron in the ground state
(IV) Alkaline earth metals
(V) Group 3A elements

(I) Group no. of the elements with the valence shell ground-state electron configuration

$ns^2np^5$ is

$17$

(II) Group no. of the elements with the valence shell ground-state electron configuration

$ns^2np^3$ is

$15$

(Ill) Group no. of the elements that have only three unpaired p electron in the ground state is

$15$

(IV) Alkaline earth metals are

$2$

(V) Group 3A elements in group

$3$

Entries of Column-I are to be matched with entries of Column-Il. Each entry of Column-I may have the matching with one or more than one entries of Column-lI.

1769048_1818726_ans_d00195bb623d424facd237ea955c0731.jpg

How do you think the tendency to lose electrons will change in a group ?

Down the group, the effective nuclear charge experienced by valence electron decreases, hence they can easily lose electrons.

Consider the following elements A, B, C and D and their outer electronic configurations are $ns^2np, ns^2np^3, ns^2np^4$ and $ns^2np^5$ respectively. Element E also has same outer electronic configuration like D but shows only single oxidation state (-1). If element A, B, C and D belong to same period as that of sodium. Consider the following compounds.
(i) $CE_4$
(ii) $BD_2E_3$
(iii) $DE_3$
(iv) $CE_2$
(v) $BD_3E_2$
(vi) $C_2E_2$
(vii) $DE$
(viii) $A_2D_6$
Then calculate the value of " $x \div y$ ", (where x and y are total number of polar and non-polar compounds.)

$\frac{6}{2} = 3$ , Elements A = Al, B = P , C= S, D = Cl, E = F

Compounds Polarity

(i)

$CE_4(SF_4)$ P

(ii)

$BD_2E_3(PCl_2F_3)$ P

(iii)

$DE_3(ClF_3)$ P

(iv)

$CE_2(SF_2)$ P

(v)

$BD_3E_2(PCl_3F_2)$ NP

(vi)

$C_2E_2(S_2F_2)$ P

(vii)

$DE(ClF)$ P

(viii)

$A_2D_6(Al_2Cl_6)$ NP

The following elements() belong to same period arrange them in order of atomic radius $\rightarrow 231 , 262, 242$ .

The atomic size decreases in a period.
1724186_1822709_ans_66e0fd123b07494b91e9b34755d43b2d.png

The electronic configuration of an atom is 2, 8,Give its atomic number and the nature of oxide.

Electronic configuration

$= 2, 8, 7$

$\therefore$ Atomic number

$= 17$
Nature of oxide = Acidic oxide

Which element has the electronic configuration $2, 8, 2 ?$

Magnesium has the electronic configuration

$2, 8, 2$ .

How does electronic configuration of atoms change in a period with increase in atomic number ?

In a period, the inner core shell remains same and the valence electrons in valence shell goes on increasing from left to right. The number of shells is same.

Why do we classify elements?

As more and more elements were discovered through the course of time, there was a need to categorise them in some way so that we can organise their information. Scientists have always made attempts to classify elements according to their properties. We classify elements to make it easier to study them and their properties.

Name two elements you would expect to show chemical reactions similar to magnesium.

As elements of a group show a similar reaction.

So, Calcium and strontium are the two elements we would expect to show chemical reactions similar to magnesium.

Name three elements with filled outermost shells.

Elements of group 18 have electrons filled completely in their outermost shells. So, Helium, Neon, and Argon are the 3 elements that have electrons completely filled in their outermost orbit.

An atom has electronic configuration 2, 8,What is the atomic number of this element?

An atom has electronic configuration 2, 8, 7 implies that there are 17 electrons in its ground state which is equal to the no. of protons. So, the Atomic number of the given element is

$17.$

How many shells are present in all the elements that belong to period 3?

All elements present in 3rd period contain 3 shells

$(K, L, M)$ in which the electrons are distributed.

Write the general outer electronic configurations of the following elements,

a) alkali metals
b) alkaline earth metals
c) halogens
d) noble gases

(a) alkali metals -

$ns^{-1}$

(b) alkaline earth metals -

$ns^{2}$

$ns^{2}np^{5}$

(d) noble gases -

$ns^{2}np^{6}$

Name three elements that have two electrons in their outermost shells.

Elements of group II have two electrons in their outermost shells. So, Magnesium, Calcium, and Strontium are the 3 elements that have two electrons in their outermost orbit.

Name three elements that have a single electron in their outermost shells.

Elements of group I have one electron in their outermost shells. So, Lithium, Sodium, and Potassium are the 3 elements that have a single electron in their outermost orbit.

Amongst the elements $B, Al, C\, and\,Si$ : (a) Which has the highest first ionization enthalpy? (b) Which has most negative electron gain enthalpy? (c) Which has the largest atomic radius? (d) Which has the most metallic character?

a) Carbon, (b) Carbon, (c) Al. (d) Al.

Explain the features that influence/affect the ionization energy.

Effective Nuclear charge:- If an effective nuclear charge increases Ionization energy decreases.
Atomic Size:- If atomic size increases Ionization energy increases.
Half or completely filled orbitals:- From Half or completely filled orbitals, starting high IE...
Orbital of same energy: If orbitals having the same principle quantum number n,
Shielding (Screening) Effect:- Inner electrons repel the outer valence electrons. They reduce the nuclear force.

(a) Account for the following: (i) $Mg^{+2}$ ion is smaller than $O^{-2}$ ion although both have same electronic structure, (ii) Ionisation enthalpy of nitrogen is more than that of oxygen,
(b) Write the IUPAC name and the symbol for the element with at.no.118.

(a) (i) It is due to greater nuclear charge and greater effective nuclear charge in

$Mg^{2+}$ and

$O^{2-}$

(ii) It is due to stability of electronic configuration of nitrogen.

(b) Uuo (Ununoctium).

Write the electronic configuration of $Cl^{-}$ ion.

Electronic configuration of

$Cl^{-} \ is \ 1s^2, 2s^2, 2p^6, 3s^2, 3p^6$

Write the electronic Configuration of cation $Fe^{+}$ .

Electronic configuration of Iron Fe $(Z = 26)$	$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^6$
Electronic configuration of $Fe^{+}$ (by removing two electrons)	$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1, 3d^6$

What do you mean by electronic configuration ?

Electronic Configuration is the distribution of electrons in the various available orbitals of an atom of an element.

The sequenced in which the electron occupy the various orbitals is as follows:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p...

Write the electron configuration of $Na^{+}$ ion.

Electronic configuration of

$Na^{+} \ is \ 1s^2, 2s^2, 2p^6.$

On the basis of their electronic configurations, explain why alkali metals are highly reactive.

Alkali metals have general electronic configuration ns1. They can lose 1 electron to acquire stable electronic configuration. They have large atomic size, therefore, they can lose electron easily and they are most reactive.

Among the elements $Li$ , $K$ , $Ca$ , $S$ and $Kr$ , which one is expected to have the lowest first ionization enthalpy and which one has the highest first ionization enthalpy?

$K$ has lowest first ionization enthalpy whereas

$Kr$ has highest first ionization enthalpy.

Explain electronic configuration of monoanion of N.

As in formation of anions, electrons are added to the element. So, if a element contains p electrons in its elemental state then no. of electrons in monoanion of that element will be p+1.

Electronic configuration of Nitrogen $(Z = 7)$	$1s^2, 2s^2, 2p^3$
Electronic configuration of $N^{-}$ (by adding electron)	$1s^2, 2s^2, 2p^4$

What is the valency of alkaline earth metal ?

Alkaline earth metals have a valency of

$2$ . Alkaline earth metals are placed in group 2 of the periodic table ,which implies that they each have 2 electrons in their outermost shells .

Elements X is placed at first position in 15 Group . Write outer most electronic configuration of this element .

Nitrogen is placed at first position in 15 group . Atomic number is N = 7

Electronic configuration is

$1s^{2}, 2s ^{2}, 2p^{3}$ so outer most electronic configuration is [

$2s ^{2}, 2p^{3}$ ]

In a periodic table , the penultimate shell of which of the elements is incomplete ?

Transition elements because last two shells of these elements namely outermost and penultimate shells are incomplete .

Why second ionization energy of alkali metals are higher than first ionization energy?

The second ionization energy of alkali metals are higher than first ionization energy is due to following reason.

1.The electrons configuration of

$M^+ ions$ is

$ns^2 np^6$ i.e., stable noble gas configuration.

2.The size of

$M^+ ions$ is smaller due to increase in nuclear size.

In what way the electronic configuration of hydrogen is similar to that of the electronic configuration of alkali metals?

Hydrogen and alkali metals both have one electron in the outermost orbital.

Write any two factors which affect ionization.

$\bullet$ Concentration : In a solution, on increasing the concentration of electrolyte, ionization decreases. The extent of ionization of an electrolyte is inversely proportional to the concentration of its solution.

$\bullet$ Temperature : The degree of ionization increases with increase in temperature.

Why alkali metals do not from di-positive ions ?

Alkali metals have only one electron in their outermost energy levels , so (they can loose one electron to) they form a mono positive ion only . That's why alkali metal do not from dispositive ions .

What is the IUPAC name and symbol of elements with atomic number 120 ?

The IUPAC name for atomic number 120 is Unbinilium and symbol is Ubn .

What is lonization enthalpy of an element ? Explain the factors affecting it and it's periodicity?

lonisation enthalpy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so as the to convert it into a gaseous cation . It is respectively as IE . The process can be represented as :

$M\left ( g \right ) \rightarrow M^{+}\left ( g \right ) + e^{-}$

Where M (g) and M are gaseous atom resultant gaseous cation .Unit of ionisation enthalpy is kJ .

The value of ionisation enthalpy depends upon the

$\textbf {size of atom }$ , because with increase in size , the distance between nucleus and valence electrons increase and hence force of attraction , between nucleus and valence electron decrease . As a result of which the valence electrons are loosely held and smaller energy is required to remove an electron . Thus lonization enthalpy decreases with increase in size and increase with decrease in size .

$\textbf {Magnitude of nuclear charge}$ : As nuclear charge increases , the force of attraction between nucleus and valency electronic increases and hence it is difficult to remove an electron from the valence shell . Thus with increase in nuclear charge , the lonization enthalpy increases .

Pick out the wrong ones fro, the subshell electronic configuration given below.
$1s^2 2s^2 2p^5 3s^2$

$1s^2 2s^2 2p^5 3s^1$

$\text{incomplete p orbital}$ $\because$ electrons are filled in $3s$ only after filling $6$ electrons in $2p$ )

Correct electronic configuration is $1s^2 2s^2 2p^6 3s^0$

Pick out the wrong ones from, the subshell electronic configuration given below.
$1s^2 2s^2 2p^7$

Wrong electronic configuration

$1s^2 2s^2 2p^7$

as maximum

$6$ electron can accomodate in

$p$ orbital.

Write electronic configuration of sodium and complete the table

1755975_1833509_ans_22559c020c3d4b9d89e90cbadf83ddd8.png

Pick out the wrong ones from the subshell electronic configuration given below.
$1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2$

$Error: 4s\text{ has lower energy than }3d.$
Correct electronic configuration is $1s^22s^22p^63s^23p^64s^23d^2$

1908370_1833519_ans_bd56e5209ff34a7a9e3851d805856c66.png

Two compound of iron are given below.
$FeSO_{4}Fe_{2}(SO_{4})_{3}$
(The oxidation state of sulfate radical is $-2$ )
Write the subshell electronic configuration of $Fe^{3+}$ ion.

the subshell electronic configuration of

$Fe^{3+}$ ion. is

$\rightarrow1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}$

Pick out the wrong ones from the subshell electronic configuration given below.
$1s^2 2s^2 2p^2$

$\text{No error}$
Given electronic configuration is correct since it is in accordance with the rules for filling up of various orbitals.

Pick out the wrong ones fro, the subshell electronic configuration given below.
$1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2$

$1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^1$

(electrons are filled in $3d$ only after filling $2$ electrons in $4s$ )

correct electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2$

The electronic configuration of the elements $A, B, C, D$ are given below.
$A-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{4}$
$B-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}$
$C-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{5}$
$D-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{-1}$
Which metal belongs to $17^{th}$ group ?

As element

$C$ has 7 electrons in the outermost shell, thus it belongs to

$17^{th}$ group.

Complete the table of details about shells and subshells.
Shell $K_{1}$ $L_{2}$ $M_{3}$ $N_{4}$
Subshell
No. of Electrons

1756016_1833514_ans_72cba27399a645d7a6a9bd2133e1acdd.png

What is the maximum number of electrons that can be accommodated in the $M$ shell?

The maximum number of electrons that can be accommodated in the

$M$ shell is

$18$

How many electrons are present ?

A hydrogen atom is made of one proton and

$\text{one electron.}$ Because hydrogen has one proton, it has an atomic number of one.

Identify the incorrect subshell electronic configuration.
$-1s^{2}$
$-1s^{2}2p^{6}$
$-1s^{2}2s^{2}2p^{6}$
$-1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}$

incorrect electronic configuration is

$1s^{2}2p^{6}$

$\because$ 2p orbital is filled before 2s as 2p have higher energy than 2s

correct electronic configuration is

$1s^{2}2s^2p^{4}$

1757088_1833556_ans_42258fd48e1a47c08bba1f6a45548969.png

The element $Cu$ with atomic number $29$ undergoes chemical reaction to formation with oxidation number $+2$
Write down the subshell electronic configuration of this ion.

$_{29}Cu -1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^1$

$Cu\rightarrow Cu^{2+}+ 2e^-$

thus, subshell electronic configuration is

$Cu^{2+}=1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^9$

Identify the incorrect electronic configurations and correct them.
i) $1s^{2}\ 2s^{2}\ 2p^{3}$
ii) $1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{1}$
iii) $1s^{2}\ 2s^{2}\ 2p^{6}\ 2d^{7}$
iv) $1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 3d^{4}$

incorrect electronic configurations

iii)

$1s^{2}\ 2s^{2}\ 2p^{6}\ 2d^{7}$

iv)

$1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 3d^{4}$

correct electronic configurations

iii)

$1s^{2}\ 2s^{2}\ 2p^{6}3s^23p^5$

iv)

$1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 4s^2\ 3d^2$

1757992_1833531_ans_08d391454b9142068ec76bac9a34360b.png

Sub-shell electronic configuration of $X$ is given below. $1s^2,2s^2,2p^5$
Write the sub-shell electronic configuration of the element next to $X$ in the same period.

Sub-shell electronic configuration of the element next to X in the same period is

$1s^2,2s^2,2p^6$

Prepare the comperhensive table which indicates the name, symbol electron configuration, subshell configuration of elements having atomic number $1$ to $36$ ?

1758019_1833551_ans_908e083738994eb8a386f568a76ce0f6.png

Sub-shell electronic configuration of $X$ is given below. $1s^2,2s^2,2p^5$
The element $Y$ is coming just below the element in the same group.
Then write the sub-shell electronic configuration of $Y$ .

$X- 1s^2,2s^2,2p^5$

X is fluorine thus, Y must be

$Cl$

Hence, the Sub-shell electronic configuration of

$Cl$ is

$1s^2,2s^2,2p^6,3s^2,3p^5$

Find the wrong electronic configurations from the following. What is wrong in these?
a. $1s^2,2s^2,2p^6,3s^2,3p^6,3d^9,4s^2$
b. $1s^2$
c. $1s^2,2s^1,2p^6$
d. $1s^2,2s^2,2p^6,3s^2,3p^6$

(a) and (c) are wrong electronic configurations.

In (a) one electron from

$4s$ is to be trans-ferred to

$3d$ completely filled configu-ration are more stable. So the correct electronic configuration is

$1s^2,2s^2,2p^6,3s^2,3p^6,3d^{10},4s^1$

in (c) electrons are filled in

$2p$ only after filling electrons in

$2s$ .

Find the wrong electronic configurations from the following. What is wrong in these?
$1s^2$

$\text{No error}$
Given electronic configuration is correct since it is in accordance with the rules for filling up of various orbitals.

The atomic number of four elements are given below. (The symbols ore not real)
$A-8$
$B-10$
$C-12$
$D-18$
Write the sub-shell electronic configuration of the elements.

$A-1s^{2}2s^{2}2p^{4}$

$B-1s^{2}2s^{2}2p^{6}$

$C-1s^{2}2s^{2}2p^{6}3s^{2}$

$D-1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$

1757131_1833564_ans_9bf96422abfe4dd0ba0e1b6f8aada7c2.png

The atomic number of two elements are given below.
$Si-14\ Ni-28$
Write the subshell electronic configuration of these elements.

the subshell electronic configuration

$Si-14\rightarrow1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}$
$Ni-28 \rightarrow 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}4s^{2}$

1757366_1833600_ans_bf7e4278caa94cb5803abb94b8bbc484.png

How many electrons are present in helium $(_2He)$ ?

there are 2 electrons in a helium atom. Looking at the picture, you can see there are two electrons in shell one. That means the first shell is full.
1757269_1833584_ans_205c14aef6894603808e20caa8d33ab0.png

1757269_1833584_ans_205c14aef6894603808e20caa8d33ab0.png

The subshell electronic configuration of two elements as follows. (Symbols are not real) $P-3s^{2}\ Q-3p^{4}$
Write the complete subshell electronic configuration.

$concept:$

$\text{period number}\Rightarrow \text{no. of shells}$

the complete subshell electronic configuration.

$P-1s^{2}2s^{2}2p^{6}3s^{2}$
$Q-1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$

1757230_1833577_ans_657e561a17c04b64ae65e9c0ec2c54c7.png

Match the following
Block Outer most electronic configuration Characteristics
$s$ $3p^{5}$ Most of the compound are coloured.
$p$ $3s^{4}4s^{2}$ Includes Lanthanoids ( $6^{th}$ period)
$d$ $4f^{1}5d^{1}6s^{2}$ Highest atomic radius in the respective period.
$f$ $3s^{1}$ Highelectro negativity

Block	Outer most electronic configuration	Characteristics
$s$	$3s^{1}$	Highest atomic radius in the respective period.
$p$	$3p^{5}$	High electronegativity
$d$	$3d^{4}4s^{2}$	Most of the compounds are coloured.
$f$	$4f^{1}5d^{1}6s^{2}$	Includes Lanthanoids ( $6^{th}$ period)

In which subshell?

Hydrogen has 1 electron in the first shell and filled in

$s$ sub shell.

Complete the electronic configuration of beryllium ?

the electronic configuration of beryllium is

$Be[Z=4] - 1s^2\ 2s^2$

The elements $'X'$ has $4$ shells and its $3d$ subshell has $6$ electrons. (Symbol is not real)
Write the complete electronic configuration of the element.

$concept:$

$\text{group number} \Rightarrow \text{no. of valence electron}$

$\text{period number}\Rightarrow \text{no. of shells}$

the subshell electronic configuration is

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^{2}$
1757419_1833605_ans_cd36df88c3d8486a9d163d2b3d170c02.png

In which shell is the electron filled?

Hydrogen has 1 electron in the first shell and filled in K shell.

Atomic number of iron is $26.$ It exhibits $Fe^{2+},Fe^{3+}$ oxidation state. Write the subshell electronic configuration.
Subshell electronic configuration.
$Fe$
$Fe^{2+}$
$Fe^{3+}$

	Subshell electronic configuration.
$Fe$ $Fe^{2+}$ $Fe^{3+}$	$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^{2}$ $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}$ $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}$

Which element has the highest ionization energy?

$Fluorine(F)$

Fluorine has the highest ionization energy among halogen family as it is at a lower energy level and does not receive the same intensity of shielding as Iodine. Thus Fluorine has a stronger nuclear force on the surrounding electrons and requires more energy to remove electrons.

Which element is having last electron in $3p$ ?

the answer is

$C, D$ ;

$(B)-1s^2,2s^2,2p^3$

$(C)-1s^2,2s^2,2p^6,3s^2,3p^2$

$(D)-1s^2,2s^2,2p^6,3s^2,3p^6$

$(A)-1s^2,2s^2,2p^6,3s^2,3p^6,4s^2$

$(E)- [Kr]5s^1,4d^5$

1758065_1833625_ans_03ab31ade5104a7bbb9e6549dcc80ec2.png

The outermost electronic configuration of the elements $A$ is $2s^{2}\ 2p^{2}$ . (Symbol is not real)
Write the complete electronic configuration of the element just below $A$ in the periodic table.

$concept:$

$\text{group number} \Rightarrow \text{no. of valence electron}$

$\text{period number}\Rightarrow \text{no. of shells}$

the subshell electronic configuration is

$1s^22s^22p^2$ which is

$carbon$ and belongs to

$14$ group. element just below carbon is

$silicon$ and its electronic configuration is

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}$

Write the subshell wise electronic configu-ration of potassium .

1758352_1833642_ans_c8a28a13633c4ef9962ed2c092dc8d6a.png

The figure of an incomplete periodic table is given below.
Which of these elements have $3$ electrons in their outermost $p$ subshell?

$E$ element have

$3$ electrons in its outermost

$p-$ subshell

$E:$ Atomic number

$=7$

$\text{Electronic Configuration}:\ 1s^2,2s^2,2p^3$

Complete the table
Element No. of electrons Subshells electronic configuration
$_7N$ $7$ $1s^2\ 2s^2\ 2p^3$
$_9F$ $9$ $1s^{....}2s^{.....}2p^{....}$
$_{11}Na$
- $1s^{....}2s^{.....}2p^{....}\ 3s^{.....}$
$_{13}Al$
- $1s^{....}2s^{.....}2p^{....} 3s^{....} 3p^{.....}$
$_{17}Cl$
-
$_{18}Ar$ -

Element	No. of electrons	Subshells electronic configuration
$_7N$	$7$	$1s^2\ 2s^2\ 2sp^3$
$_9F$	$9$	$1s^2 2s^2 2p^5$
$_{11}Na$	$11$	$1s^2 2s^2 2p^6\ 3s^1$
$_{13}Al$	$13$	$1s^2 2s^2 2p^6 3s^23p^1$
$_{17}Cl$	$17$	$1s^2 2s^2 2p^6 3s^23p^5$
$_{18}Ar$	$18$	$1s^2 2s^2 2p^6 3s^23p^6$

Examine the given electronic configurations.
$A-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 3d^{10}\ 4s^{2}$
$B-1s^{2}\ 2s^{6}\ 3s^{1}$
$C-1s^{2}\ 2s^{2}\ 2p^{1}\ 3s^{2}\ 3p^{6}$
$D-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}$
$E-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 4s^{2}$
Which element has highest metallic character?

$A_{30}-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 3d^{10}\ 4s^{2}$ (group-12,period-4)

$B_{10}-1s^{2}\ 2s^{2}\ 2p^{6}$ (group-18,period-2)

$C_{18}-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}$ (group-18,period-3)

$D_{12}-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}$ (group-2,period-3)

$E_{20}-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 4s^{2}$ (group-2,period-4)

Metallic character increases as we move down the group because the atomic size increases which leads to easy loss of electrons. On the other hand, it decreases across a period as we move from left to right.

thus, from the above elements

$E$ has the highest metallic character.

1759097_1833647_ans_76c9b4ec1f314a66a0ca35944eb63fe5.png

Which element ends with electronic configuration $4d^5,5s^1$ ?

the answer is

$E$ ;

$(B)-1s^2,2s^2,2p^3$

$(C)-1s^2,2s^2,2p^6,3s^2,3p^2$

$(D)-1s^2,2s^2,2p^6,3s^2,3p^6$

$(A)-1s^2,2s^2,2p^6,3s^2,3p^6,4s^2$

$(E)- [Kr]5s^1,4d^5$

The electronic configuration of scandium $(21Sc)$ is

1758378_1833645_ans_a4ffd4f1d6e44cfba64769a0a033996c.png

How was the shell wise electronic configuration of potassium written ?

1757656_1833620_ans_9474bcc65d404d269673be31192405d8.png

Examine the given electronic configurations.
$A-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 3d^{10}\ 4s^{2}$
$B-1s^{2}\ 2s^{6}\ 3s^{1}$
$C-1s^{2}\ 2s^{2}\ 2p^{1}\ 3s^{2}\ 3p^{6}$
$D-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}$
$E-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 4s^{2}$
Which elements belong to the same group?

$A-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 3d^{10}\ 4s^{2}(period-4)$

$B-1s^{2}\ 2s^{6}\ 3s^{1}(period-3)$

$C-1s^{2}\ 2s^{2}\ 2p^{1}\ 3s^{2}\ 3p^{6}(period-3)$

$D-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}(period-3)$

$E-1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 4s^{2}(period-4)$

$A,E$ in period

$4$ and

$B,C,D$ in period

$3$

$\because \text{period number }\Rightarrow \text{highest shell number in electronic configuration}$

Write its subshell electronic configuration of neon .

subshell electronic configuration of neon is

$_{10}Ne - 1s^2\ 2s^2\ 2p^6$
1758410_1833651_ans_22237743dacd43a09f2fae4053e5a821.png

How can the subshell electronic configuration be written in a short form ?

the subshell electronic configuration be written in a short form as

$[Ne]3s^2$

$(\because [Ne]_{10}-1s^2 ,2s^2, 2p^6 )$

Complete the Table
Elements Subshell electronic configuration
$_{21}Sc$ $[Ar]\ 3d^1\ 4s^2$
$_{20}Ca$ ......
$_{12}Mg$ ......
$_{27}Co$ ......
$_{30}Zn$ .......

Elements	Subshell electronic configuration
$_{21}Sc$	$[Ar]\ 3d^1\ 4s^2$
$_{20}Ca$	$[Ar] \ 4s^2$
$_{12}Mg$	$[Ne]\ 3s^2$
$_{27}Co$	$[Ar]\ 3d^7\ 4s^2$
$_{30}Zn$	$[Ar]\ 3d^{10}\ 4s^2$

Write the subshell electronic configuration of $_{24}Cr$

The atomic number of chromium is 24. Its electronic configuration in ground state is

$_{24}Cr - 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^5\ 4s^1$

Using the symbol of neon, write the subshell electronic configuration of sodium ?

the subshell electronic configuration of sodium is

$[Ne]\ 3s^1$

identify the correct electronic configuration of $29Cu$ from those given below :
$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^9\ 4s^2$
$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^{10}\ 4s^1$

$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^9\ 4s^2$ -incorrect
$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^{10}\ 4s^1$ -correct

This electronic configuration is exceptional because electrons entered in

$3-d$ orbitals without filling the

$4s$ orbitals complete.

Reason for the Exceptions ⇒ It is said that d orbitals can be stable if it is half-filled or full filled.

Thus, In the given case, electrons reach near the half-filled and full filled thus due to the attraction one electron from the

$s-$ orbitals came in

$d-$ orbitals.

Subshell electronic configuration of sodium ?

$_{11}Na - 1s^2\ 2s^2\ 2p^6\ 3s^1$
1758415_1833653_ans_0bfe7bb4384347f8805f4e0b42474b8a.png

Write the electronic configu ration of $22Ti,\ 23V$ , the two elements after $Sc$

$_{22}Ti - 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^2\ 4s^2$

$_{23}V-1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^3\ 4s^2$

Electronic configuration of same elements are given. Write answers to the following questions.
i. $[Ne]3s^2$
ii. $[Ar]3d^2,4s^2$
iii. $[Xe]6s^2$
iv. $[Ne]3s^2,3p^2$
Which is the non-metal?

$[Ne]3s^2\rightarrow \ _{12}Mg$

ii.

$[Ar]3d^2,4s^2 \rightarrow \ _{22}Ti$

iii.

$[Xe]6s^2\rightarrow\ _{56}Ba$

iv.

$[Ne]3s^2,3p^2\rightarrow\ _{14}Si$

among all

$[Ne]3s^2,3p^2\rightarrow\ _{14}Si$ is non-metal.

Write down subshell electronic configuration of $Cu^{1+}$ and $Cu^{2+}$

1760028_1833677_ans_9bbff18a9a9d4d9f96ee4d94d99ecec9.png

Write the subshell electronic configuration of the following elements and find the blocks to which they belong
$_4Be$

$_4Be$ : Atomic number $=4$
$E.C :$ $1s^2\ 2s^2$
$block:\ -s\ block$

Write the subshell electronic configuration of the following elements and find the blocks to which they belong
$26Fe$ :

$26Fe$ : Atomic number $=26$

$E.C:1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^6\ 4s^2$

$block:-d\ block$

Write the subshell electronic configuration of the following elements and find the blocks to which they belong
$18Ar$ :

$18Ar$ : Atomic number $=18$

$E.C:1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6$

$block:-p\ block$

How many electrons are donated by the first group elements in a chemical reaction?

$one$ electron is donated by the first group elements in a chemical reaction.

$X\rightarrow X^++e^-$

In which subshell did the filing of the last electron takes place in p block elements?

The filing of the last electron takes place in p block elements in

$p$ subshell

Which element shows metallic character ?

$X$ shows the metallic character as it belongs to s block.

Complete subshell electronic configuration

$:1s^2,2s^2,2p^6,3s^2$

Group number

$:2$

Block

$:s-$ block

$\ _{24}Cr-[Ar] 3d^{5}4s^{1}$
$\ _{29}Cu-[Ar] 3d^{10}4s^{1}$
Why chromium and copper exhibits such electronic configuration ?

$\text{Electron orbitals are most stable when they are fully filled or half-filled.}$ In the case of Chromium, and Copper after the

$4s^2 ,3d^4\ and\ 4s^2,3d^9$ respectively configuration is attained, an electron from the

$4s$ orbital jumps to

$3d$ subshell because

$3d^5 \ and \ 3d^{10}$ is a much more stable configuration than

$3d^4\ and \ 3d^9$ . That's why chromium and copper exhibits such electronic configuration

Element $'X'$ is having atomic number $28,$ it gives two electrons to element $'Y'.$
Write down the electronic configuration of $'X'$ and its ion

$X_{28}-1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}4s^{2}$

$X\rightarrow X^{2+}+2e^-$

$X^{2+}-1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}$

Write down the subshell electronic configuration of $Fe_{21}$ .

1760503_1834359_ans_0f4e8b4c2c7a4732b4764b02099d5b3c.png

How many electrons are donated by the second group element in a chemical reaction?

$Two$ electrons are donated by the second group elements in a chemical reaction in order to acquire noble gas configuration.

$X\rightarrow X^{2+}+2e^-$

Which is the element ?

the element is

$H$

$H:$ Atomic number $=10$
$E.C:\ 1s^2,2s^2,2p^6$

Write electronic configuration of group I elements of the periodic table.

Element	Atomic Number	Electronic configuration	Group	Period
$H$	$1$	$1$	$1$	$1$
$Li$	$3$	$2, 1$	$1$	$2$
$Na$	$11$	$2, 8, 1$	$1$	$3$
$K$	$19$	$2, 8, 8, 1$	$1$	$4$
$Rb$	$37$	$2, 8, 18,8, 1$	$1$	$5$
$Cs$	$55$	$2, 8, 18, 18, 8, 1$	$1$	$6$
$Fr$	$87$	$2, 8, 18, 32, 18, 8, 1$	$1$	$7$

What are the factors affecting the ionization energy?

Factors affecting the ionization energy are:

$\text{Size of the positive nuclear charge}$

As the nuclear charge increases, its attraction for the outermost electron increases, and more energy is required to remove an electron. This means that the ionization energy increases.

$\text{Size of an atom }$ (the distance of the outermost electron from the nucleus)

As atomic size increases, the attraction of the positive nucleus for the negative electron decreases, and less energy is required to remove an electron. This means that the ionization energy decreases.

$\text{Screening (shielding) effect of inner shell electrons}$

The outermost electron is screened (shielded) from the attraction of the nucleus by the repelling effect of the inner electrons.

As shielding increases, the attraction of the positive nucleus for the negative electron decreases, and less energy is required to remove an electron. This means that the ionization energy decreases.

List the elements in group $18$

Now try to write their electronic configuration.

$_{2}He - 2$

$_{10}Ne - 2, 8$

$_{18}Ar - 2, 8, 8$

$_{36}Kr - 2, 8, 18, 8$

$_{54}Xe - 2, 8, 18, 18, 8$

$_{86}Rn - 2, 8, 18, 32, 18, 8$

Write down the complete subshell electronic configuration ?

the element is

$H$

$H:$ Atomic number $=10$
$E.C:\ 1s^2,2s^2,2p^6$

List the elements in group $18$

How many electrons are there in the outermost shell of each element ?

18 Group elements are called as noble gas elements.

Every Noble Gas has completed its octet, which means it has 8 electrons in its valence shell. (outermost shell)

Write the electronic configuration of elements with atomic number $1 - 10$

Element	Atomic number	Electronic configuration
Hydrogen	$1$	1
Helium	$2$	2
Lithium	$3$	$2,1$
Beryllium	$4$	$2, 2$
Boron	$5$	$2, 3$
Carbon	$6$	$2, 4$
Nitrogen	$7$	$2, 5$
Oxygen	$8$	$2, 6$
Fluorine	$9$	$2, 7$
Neon	$10$	$2, 8$

The elements of these groups are called representative elements.

Write any two characteristics of this element ?

the element is

$H$

$H:$ Atomic number

$=10$

$E.C:\ 1s^2,2s^2,2p^6$

$\bullet$ Noble element / gases.

$\bullet$ The outermost shell is completely filled

Which is highly reactive nonmetal ?

$G$ is highly reactive nonmetal as the reactivity of non-metal increases from left to right along the period.

Why inert gases do not possess reactivity?

The

$18^{th}$ group of the periodic table contains the inert gas elements. The elements in this group show a stable outer shell configuration. As a result of this closed shell configuration, inert gases do not possess any chemical reactivity.

Find how ionization energy changes with increase in nuclear charge ?

On moving from left to right in a period, as nuclear charge increases, the size of the atom decrease hence ionization energy increase.

What is doublet closed configuration? Which atom occupies this structure.

Doublet closed configuration is a state of stable configuration with only two electrons in the valence shell of an atom. Doublet closed configuration is shown by helium gas.

The ionisation enthalpy decreases as we go from top to bottom in the group, while it increases as we move from left to right in the period?

The minimum amount of energy required to remove an electron from the outermost orbit of a gaseous atom in its ground state is called ionisation energy.

Ionisation energy of elements depends on various factors like nuclear charge, shielding effect, stability of electronic configuration and atomic radius.

Ionisation energy decreases as we go down a group. As we go down a group, the atomic radii of the elements increase, the number of electrons shells also increases, thus, the nuclear attraction over a valence electron decreases and ionisation energy also decreases.

Ionisation energy generally increases across a period from left to right. As we move across a period from left to right, the atomic radii decreases, and the nuclear attraction over a valence electron increases. Thus, ionisation energy increases.

How many electrons are possessed by outermost orbits in the elements of $17^{th}$ group?

There are

$7$ electrons that are possessed by outermost orbits in the elements of

$17^{th}$ group.

In which element except the inert gas elements have $7$ electrons in the outermost orbit?

The elements of the

$17^{th}$ group have

$7$ electrons in the outermost orbit.

What would IUPAC names and symbols for elements with atomic numbers $122, 127 , 35 , 149$ and $150$ ?

The roots for

$2 , 7, 5 , 9$ and

$0$ are bi , sept , pent , enn and nil respectively. Therefore their names and symbols are:

Z Name Symbol

$122$ Unbibium Ubb

$127$ Unbiseptium Ubs

$135$ Untripentium Utp

$149$ Unquadennium Uqe

$150$ Unpetnilium Upn

Name the elements in the periodic table which has the highest and lowest first ionization enthalpies.

Highest : Helium and Lowest : Francium

Which of the elements $Na , Mg , Si$ and $P$ would have the greatest difference between the first and the second ionization enthalpies. Briefly explain your answer.

Among

$Na , Mg , Si$ and

$P$ ,

$Na$ is an alkali metal. It has only one electron in the valence shell, therefore , its

$IE_{1}$ is very low. However , after removal of one electron, it acquires neon gas configuration, i.e.,

$Na^{+} (1s^2 2s^2 2p^6)$ .

Therefore , its

$IE_{2}$ is expected to be very high. Consequently, the difference in first and second ionization enthalpies would be greatest is case of Na.

However, it mat be noted here that in case of

$Mg , Si$ and

$P$ , although their

$IE_{1}$ will be much higher than that of

$Na$ but their

$IE_{2}$ will be lower than that of

$Na$ . As a result , the difference in their respective

$IE_{1}$ and

$IE_{2}$ would be much lower than that of

$Na$ .

What would be the atomic number of the next (i) alkali metal (ii) alkaline earth metal (iii) halogen and (iv) inert gas, if discovered in future?

(i) The next alkali metal if discovered will have to be placed in the eighth period and hence its outer electronic configuration will be

$8s^{1}$ . Therefore its atomic number will be

$(118 + 1) = 119$ .

(ii) Similarly, the next alkaline earth metal, if discovered, will have

$8s^2$ an its outer configuration and hence its atomic number will be

$(118 + 2) = 120$ .

(iii) The next halogen, if discovered , will have

$7s^2 7p^5$ as it outer electronic configuration. Since the filling of

$7p$ -orbitals will begin after

$6d$ -orbitals and up to

$6d$ orbitals, in accordance with Aufbau principle, we can accommodate only

$112$ elements, therefore the atomic number of the next page halogen, if discovered , will be

$(112 + 5) = 117$ .

(iv) In a similar way we can easily explain that the inert gas, if discovered will have

$7s^2 7p^6$ as its outer electronic configuration and its atomic number will be

$(112 + 6) = 118$ .

The values for three of the quantum numbers for the last electron in A and B are listed below. In what families of the periodic table are these elements present ?
$l$ $m$ $s$
A $1$ $+1$ $+1/2$
B $3$ $-2$ $-1/2$

2038487_1868354_ans_85ea3a1260cd4c35b9e7cb22ea4e9e00.jpg

Why ionization enthalpy of nitrogen is greater than that of oxygen ?

Nitrogen has exactly half-filled p-orbitals.

Explain why ionization decreases down a group of the periodic table ?

The decrease in ionization enthalpies down any group is because of the other following factors:

Increase in size of atom as an increase in the number of the main energy shells $(n)$ in moving from one element to the other.
There is also an increase in the magnitude of the screening effect due to the gradual increase in the number of inner electrons.

The position of certain elements in the modern Periodic Table are shown below.

$\downarrow$ Period /Group $\rightarrow$ 1 2 3 to 12 13 14 15 16 17 18
1 G H
2 A I B C
3 D E F

Using the above table answer the following questions giving reasons in each case:
Which element will form only covalent compounds?

"E" element will form only covalent compounds.
"E" element has 4 electrons in the valence shell which need 4 electrons to complete its octet.

$\therefore$ It can't gain or loss 4 electrons.

$\therefore$ It completes its octet by forming covalent bond. (sharing electrons)

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements) of the modern periodic table with their atomic numbers are given.
3 4 5 6 7 8 9 10
A E G
11 12 13 14 15 16 17 18
B C D F
If B combines with F, what would be the formula of the compound formed? if B & F are replaced by their respective metal what will be the formula of the compound formed.

	Atomic number	Elements	Group
A	3	Lithium	1
B	11	Sodium	1
C	12	Magnesium	2
D	14	Silicon	14
E	8	Oxygen	16
F	17	Chlorine	17
G	10	Neon	18

$\bullet$ Electronic configuration of B is (2, 8, 1)

$\bullet$ Electronic configuration of F is (2, 8, 7)

$\bullet$ Therefore, the valency of B and F is 1.$

$\bullet$ So, the formula of the compound formed will be

$BF$ .

$\bullet$ If B and F replace by thei respective metal, then the formula of the compound formed will be

$NaCl$ .

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements) of the modern periodic table with their atomic numbers are given.
3 4 5 6 7 8 9 10
A E G
11 12 13 14 15 16 17 18
B C D F
Write the electronic configuration of C and E and type of bond formed by their combination.

	Atomic number	Elements	Group
A	3	Lithium	1
B	11	Sodium	1
C	12	Magnesium	2
D	14	Silicon	14
E	8	Oxygen	16
F	17	Chlorine	17
G	10	Neon	18

$\bullet$ Electronic configuration of C is (2. 8. 2).

$\bullet$ Electronic configuration of E is (2. 6).

$\bullet$ Type of bond formed by their combination is ionic bond because the compound formed by the transfer of electrons.

Alkaline earth metals always from divalent cations even though their second ionization enthalpies are almost double than their first ionization enthalpies . Explain.

The higher enthalpy of lattice formation by

$M^{2+}$ ions more than compensates the higher second ionization enthalpies of metals.

Explain the following:
Ionization enthalpy of Mg is more than that of Na and Al.

First we shall write the electronic configuration of given atom:

$Mg_{12}=1s^2,\ 2s^2,\ 2p^6,\ 3s^2$

$Na_{11}=1s^2,\ 2s^2,\ 2p^6,\ 3s^1$

$Al_{13}=1s^2,\ 2s^2,\ 2p^6,\ 3s^2,\ 3p^1$

The valence electron of aluminum is in

$3p$ subshell which is farther than

$3s$ electron from the nucleus. Therefore, it is easy to remove this electron. Due to this, the first ionization energy of

$Al$ is less than

$Mg$ . In other words, the first ionization energy of

$Mg$ is greater than that of both

$Na$ and

$Al$ taken separately. Moreover,

$Mg$ has a full-filled

$s$ subshell so a large amount of energy will be required to remove an electron from

$Mg$ .

The electrons in the atoms of four elements A, B, C and D are distributed in three shells having 1, 3, 5 and 7 electrons respectively in their outermost shells.
(i) Write the group numbers in which these elements are placed in the Modern
Periodic Table.
(ii) Write the electronic configuration of the atoms of B and D and the molecular formula of the compound formed when B and D combine.

$\textbf{(i)}$

Elements	Electrons present in their outermost shell or in the 3rd shell	Group number
A	1	1st group
B	3	13th group
C	5	15th group
D	7	17th group

$\textbf{(ii)}$

$\textbf{Part 1: Determine the electronic configuration of B and D:}$

$\textbf{Electronic configuration of B:}$

K shell or I shell = 2 electrons

L shell or II shell = 8 electrons

M shell or III shell = 3 electrons

So, its electronic configuration of B is 2, 8, 3.

Valency of B = 3

$\textbf{Electronic configuration of D:}$

K shell or I shell = 2 electrons

L shell or II shell = 8 electrons

M shell or III shell = 7 electrons

So, its electronic configuration of B is 2, 8, 7.

Valency of D = 1

$\textbf{Part 2: Determine the molecular formula of the compound formed when B and D combine.}$

$Formula: BD_3$

Write the formulae and names of the oxides of second-period elements ( $Li$ to $N$ ). Identify the oxides as acidic, basic or amphoteric.

$Li_2O$ basic

$BeO$ amphoteric

$B_2O_3$ acidic

$CO_2$ acidic

$N_2O_5$ acidic

Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increases in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

The option (d) represents incorrect statement. It states that
"Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value".
The correct statement is "Removal of electron from orbitals bearing lower n value is difficult than from orbital having higher n value".
The attraction between the nucleus and the electron present in orbital having lower n value is higher than the attraction between the nucleus and the electron present in orbital having higher n value.

Match the atomic number (in List I) with its IUPAC nomenclature (in List II).

(A) Unnilpentium

$(Z=105)$ with symbol

$Unp$ has official IUPAC name Dubnium

$(Db)$ .
(B) Unnilseptium

$(Z=107)$ with symbol

$Uns$ has official IUPAC name Bohrium

$(Bh)$
(C) Unnilennium

$(Z=109)$ with symbol has

$Une$ official IUPAC name Meitnerium

$(Mt)$
(D) Ununnilium

$(Z=110)$ with symbol Uun has official IUPAC name Darmstadtium

$(Ds)$
(E) Ununhexium

$(Z=116)$ has symbol

$Unh$ .

Considering the elements F, Cl, O and N the correct order of their chemical reactivity in terms of oxidizing property is:
(a) F > Cl > O > N (b) F > O > Cl > N
(c) Cl > F > O > N (d) O > F > N > Cl

In a period, from left to right, with an increase in the atomic number, the oxidizing character increases due to the presence of vacant d orbitals in their valence shells.
Hence, the decreasing order of oxidizing character is F > O > N.
On moving down the group, the oxidizing character decreases. hence, F > Cl.
But O has higher oxidizing character than Cl thus, O > Cl.
Thus, the correct order of chemical reactivity of F, Cl, O, and N in terms of their oxidizing property is F > O > Cl > N.

What is ionization potential? Explain the factors which affect ionization potential.

Ionization potential:- the amount of energy required to remove a

$e^-$ from the outermost shell of the neutral atom.

Factors affecting:-

(i) Atomic radii:-

as atomic radii increase

$\longrightarrow$ the outermost

$e^-$ is loosely held by the nucleus and hence ionization potential decreases and vice versa.

(ii) Effective nuclear charge:-

as effective nuclear charge increases the

$e^-s$ are held more tightly with the nucleus and hence ionization potential increases.

(iii) Shielding effect:-

with an increase in shielding effect

$\longrightarrow$ ionization potential decreases. Due to this shielding,

$e^-s$ are loosely held and hence low ionization potential.

(iv) Penetrating effect:-

Penetration effect is the ability of an orbital to attract a

$e^-$ . So as penetrating effect increases ionization potential increases.

(v) Stability of half-filled and full-filled orbitals

ionization potential has maximum value due to the extra stability of the half-filled and full filled electron configuration.

Arrange the following in the given order.

Increasing first ionization energy: $Mg,\ Al,\ Si, \ Na.$

The increasing order of first ionization energy is

$Na < Al < Mg < Si$

From left to right, the ionization energy increases. But in magnesium, stable

$3s$ configuration is present. Therefore, its ionization energy is greater than that of aluminium.

$Mg : 1s^2 2s^2 2p^6 3s^2$

Among the elements B, Al, C, and Si,
(i) Which element has the highest first ionization enthalpy?
(ii) Which element has the most metallic character? Justify your answer in each case.

1979884_1725177_ans_02267076f9764bf1b3118296a55d893d.jpg

Write down the outermost electronic configuration of alkali metals. How will you justify their placement in group 1 of the periodic table?

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1986847_1725306_ans_9f73733a066147599fb89f5dc9faba91.jpg

Arrange the elements N, P, O and S in the order of :
(i) Increasing first ionization enthalpy.
(ii) Increasing non-metallic character.

1986845_1725217_ans_1d7e022bb0e34686bb89501d54fa56ad.jpg

Define ionization enthalpy. Discuss the factors affecting ionization enthalpy of the elements and its trends in the periodic table.

Ionization Enthalpy:- ionization enthalpy of an element can be defined as amount of energy required to remove an e from an isolated gaseous atom in its gaseous state.

factors affecting ionization enthalpy are:-

(i) Size of atom :- The larger the atomic size, smaller is the value of ionization enthalpy. In large atom,

the outer

$e^{-}$ are for away

from the nucleus and thus force of attraction with which they are attracted by nucleus is less and hence can be easily removed.

Size of atom

$\alpha \dfrac{1}{\text { Atomic size. }}$

(ii)Screening Effect :-

Screening effect

$\alpha \dfrac{1}{\text { Ionization enthalpy. }}$

Screening effect reduces the force of attraction towards nucleus and hence the outer

$e^{-}$ can be easily removed.

(iii) Nuclear charge:-

Ionization enthalpy

$\alpha$ nuclear charge

As nuclear charge increases among atoms having same number of energy shells, the Ionization enthalpy increases because force of attraction towards nucleus increases.

(iv) Shape of orbital:-

Shape of ionization enthalpy for orbitals :-

$s>p>d>f$

Match the column :

1553392_1725786_ans_ef46e7bd2043427db3f03caa62b2d3f0.jpeg

Match the following:

1716368_1725773_ans_3a29b3adc1954aa8afe9d9ac2524e613.jpg

Using the part of the periodic Table given below,answer the questions that flows :
i) Na has physical and chemical properties similar to which element (s) and why?
Groups $\rightarrow$
Periods $\downarrow$ 1 2 13 14 15 16 17 18
1 H He
2 Li Be B C N O F Ne
3 Na Mg Al Si P S Cl Ar
4 K Ca
ii) Write the electronic configure of N and P, which one of these will be more electronegative and why?
iii) State a chemical property common to fluorine and chlorine.

i) Lithium and potassium,due to same number of valence electrons.

ii)

$N\rightarrow 2,5$

$P\rightarrow 2,8,5$

N is more electroegative element as electronegative decreases on moving down the group.

iii) Both fluorine and chlorine form their hydracids on reacting with hydorgen.

$H_2+F_2\rightarrow 3HF$

$H_2+Cl_2\rightarrow 2HCl$

Answer the following:
Why caesium is used in photoelectric cell while lithium cannot be?

Caesium has low ionisation enthalpy in comparison to lithium

Caesium can lose electrons very easily.

From the data given, calculate the energy deficit in the formation of $BeCl_2$ (it is a stable molecule).
$I.E._1$ for $Be = 899\ kJ/mol$
$I.E._2$ for $Be = 1757\ kJ/mol$
$I.A.$ for $Cl = -348\ kJ/mol$

$Be \longrightarrow Be^+$ energy required

$= 899$ kJ/mol

$Be^+ \longrightarrow Be^{2+}$ energy required

$= 1757$ kJ/mol

___________________________________

$Be \longrightarrow Be^{2+}$ energy required

$= 2656$ kJ/mol

$Cl \longrightarrow Cl^-$ energy released

$= -348$ kJ/mol

For

$2Cl^-$ , energy released

$= -696$ kJ

Therefore, the energy deficit for the process

$Be + 2Cl \longrightarrow BeCl_2$ is

$2656 - 696 = 1960 \ kJ/mol$

What happens when?

Aluminium is added to copper sulphate solution.

1621509_1728049_ans_91a4441cf2db4936b3968ebdc86ef975.jpg

You are given Avogadro's no. of $' X '$ atoms . In half of the atoms of X transfer one electron to the other half of $' X '$ atoms , $409 \,kJ$ must be added . If these $X^-$ ions are subsequently converted to $X^+$ , an additional $733 \,kJ$ must be added . Calculate IP and EA of $X$ in $eV$ .

Use ( $1 \,eV = 1.602\times10^{-19} \,J$ and $N = 6.023\times10^{23}$ ).

Assume that ,

$X \rightarrow X^+ + e ;$

$\Delta H = IP_1 = a\,eV$

$X + e \rightarrow X^- ;$

$\Delta H = -EA_1 = -b\,eV$

$N/2$ atoms of

$X$ lose electrons which are taken up by remaining

$N/2$ of

$X$ to give

$X^-$ , then

$a \times \dfrac{N}{2} - b \times \dfrac{N}{2} = \dfrac{409 \times 10^3}{1.602 \times 10^{-19}}\,eV$

$a - b = \dfrac{409\times 10^3 \times 2}{1.602\times 10^{-19} \times 6.023 \times 10^{23}}$

$\therefore$

$a - b = 8.477$

Now ,

$N/2$ of

$X^-$ lose two electrons to give

$X^+$

$X^- \rightarrow X + e ;$

$\Delta H = +EA_1 = +b$

$X \rightarrow X^+ + e ;$

$\Delta H = + IP_1 = +a$

$\displaystyle \therefore$

$a \times \dfrac{N}{2} + b\times\dfrac{N}{2} = \dfrac{733 \times 10^3}{1.602 \times 10^{-19}} \,eV$

or ,

$a + b = \dfrac{733 \times 10^3 \times 2}{1.602 \times 10^{-19} \times 6.023 \times 10^{23}}$

$a + b = 15.194$

$\displaystyle \therefore$

$a = 11.835 \,eV$

and

$b = 3.358 \,eV$

A mixture contains atoms of fluorine and chlorine . The removal of an electron from each atom of sample absorbs $284 \,kJ$ while the addition of an electron to each atom of mixture releases $68.8\,kJ$ .Determine the percentage composition of mixture . Given $IE_1$ for $F$ and $Cl$ are $27.91\times 10^{-22}$ and $20.77\times 10 ^{-22} \,kJ/atom$ respectively and $EA_1$ for $F$ and $Cl^-$ are $-5.53\times10^{-22}$ and $-5.78\times10^{-22} \,kJ/atom$ respectively .

Let the mixture contains

$a, b$ atoms of

$F$ and

$Cl$ respectively .

Thus , total energy absorbed is:

$284 = a \times 27.91 \times 10^{-22} + b \times 20.77\times 10^{-22}$ ......(1)

Also total energy released is :

$-68.8 = a \times (-5.53 \times 10^{-22}) + b\times (-5.78 \times 10^{-22})$

$68.8 = 5.53 \times 10^{-22} \times a + 5.78 \times 10^{-22} \times b$ ..... (2)

By Eqs. (1) and (2) ,

$a = 4.57 \times 10^{22}$

$b = 7.53 \times 10^{22}$

% of

$F = \left [\dfrac{4.57 \times10^{22}}{4.57 \times10^{22} + 7.53 \times 10^{22}} \right ] \times 100 = 37.76$ %

% of

$Cl = 62.24$ %

$Na_{2}O$ is basic wheres $Cl_{2}O_{7}$ is acidic . Explain with reason.

Sodium oxide

$\left ( Na_{2}O \right )$ is a strongly basic oxide. It reacts with water to from

$NaOH$ which is a strong base .

$Na_{2}O + H_{2}O \rightarrow 2 NaOH$

$Cl_{2}O_{7}$ reacts with water to from

$HClO_{4}$ which is a strong acid . Therefore

$Cl_{2}O_{7}$ is acidic .

The ionisation enthalpy of noble gases is very high . Explain

The ionisation enthalpy of noble gases is very high as noble gases posses complete valence shell configurations . Hence energy required to remove an electron from valence shell of noble is the highest in a period .

Thin strips of magnesium, copper and iron are taken.Write down what happens when these metals are treated as follows:
Arrange these metals in decreasing order of reactivity

$Mg > Fe > Cu$

Match the Column $A$ with Column $B$

Column $A$	Column $B$
1. Elements short by $1$ electrons in octet	Halogen
2. Highly reactive metals	Alkali metals
3. Non-reactive element	Noble gases
4. Elements of group $3$ to $12$	Transition elements
5. Radioactive elements	Actinides
6. Elements with $2$ electrons in the outermost orbit.	Alkali earth metals

Which of the two elements given below would be chemically more reactive, 'X' of atomic number 18 or element 'Z' of atomic number 16 and why?

Explanation:

An element with a complete octet is less reactive.

The element with the atomic number $18$ has the electronic configuration $[Ne]3{s^2}3{p^6}$ , so there are $8$ electrons in the valence shell. Hence, it has a complete octet configuration.

The element with the atomic number $16$ has the electronic configuration $[Ne]3{s^2}3{p^4}$ , so there are $6$ electrons in the valence shell. Hence, it has an incomplete octet configuration.

Final Answer: So the element with the atomic number $16$ will be more reactive.

What element has the atom whose K, L, and M shells and $4s$ subshell are filled completely and $4p$ subshell is half-filled ?

1879968_1855536_ans_ba3ebf0931b44a1bb2fea8f6eef5b5b6.jpeg

Period	Group $1$	Group $2$
$2$	$A(3)$	$E(4)$
$3$	$B(11)$	$F(12)$
$4$	$C(19)$	$G(20)$
$5$	$D(37)$	$H(38)$

$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$A$					$E$		$G$
$11$	$12$	$13$	$14$	$15$	$16$	$17$	$18$
$B$	$C$		$D$			$F$

Period No
2	Li(3)	Be(4)
3.	Na(11)	Mg(12)
4.	k(19)	Ca (20)
5.	Rb(37)	Sr(38)

$\downarrow$ Period /Group $\rightarrow$	1	2	3 to 12	13	14	15	16	17	18
1	G								H
2	A			I			B		C
3		D			E				F

Groups $\rightarrow$ Periods $\downarrow$	1	2	13	14	15	16	17	18
1	H							He
2	Li	Be	B	C	N	O	F	Ne
3	Na	Mg	Al	Si	P	S	Cl	Ar
4	K	Ca

Classification Of Elements And Periodicity In Properties - Class 11 Medical Chemistry - Extra Questions

A metal 'M' has the electronic configuration 2, 8, 3 and occurs in nature as M2_2O4_4. It is more reactive than zinc. Write a chemical equation illustrating the use of this metal for joining cracked railway lines.

____________ is the electronic configuration of the highest electronegative element.

Write the full form of IUPAC.

Complete the electronic configuration of a chloride ion.1s21s^{2} ___________.

The elements (z=117z = 117 and 120120) have not yet been discovered in which family or group would you place these elements and also give electronic configuration in each case?

Name the type of elements, which have their (i) outermost shell complete(ii) outermost shell incomplete(iii) two outermost shell incomplete(iv) one electron short of octet(v) two electrons in the outermost orbit.

What is the IUPAC name of an element with atomic number 107107.

With reference to periodic table, indicate:An element that is in group IIIA and third period.

Explain chemical reactivity in periods?

Give the electronic configuration of third alkali metal.

From among the elements, choose the following:Cl, Br, F, O, Al, C, Li, Cs and XeThe element whose atom has 6 electrons in the outermost shell.

With reference to periodic table, indicate:The inert gases placed in the 2nd and 5th period.

Write the names of elements from the following symbols: Zn, Cd, Xe, Br, Ti, Cu, Fe, Si, Ir, Pt.

Write the electronic configuration of alkali metals.

Define the following terms.Element

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements)of the modern periodic table with their atomic numbers are given. 345678910AEG1112131415161718BCDFWhich of these is an inert gas ? Name it.

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements) of the modern periodic table with their atomic numbers are given.345678910AEG1112131415161718BCDFWhich of these are metals ? Name them.

With reference to period 3 of the periodic table-State\text{ With reference to period 3 of the periodic table-State}The electronic configuration of the element whose hydroxide is a weak base.

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements) of the modern periodic table with their atomic numbers are given.345678910AEG1112131415161718BCDFWhich of these is a halogen ? Name it.

State True or False.The atomic number of Uus is 117.

Why BeOBeO and Al2O3Al_2O_3 are amphoteric?

B and Si resemble each other in their properties. If true enter 1, else enter 0.

BeCl2BeCl_2 and AlCl3AlCl_3 are Lewis acids. If true enter 1, else enter 0.

"Decreasing lattice energy favours increased solubility while decreasing hydration energy favours decreased solubility."Answer whether the above statement is true or false.If true enter 1, else enter 0.

As the atomic radius increases, ionisation energy is found to decrease.If true then enter 1, else enter 0.

Match the type of elements (in List I) with the corresponding electronic configuration (in List II).

IUPAC representation of the element of atomic number 115 is Uup.If true enter 1, else enter 0.

State whether the given statement is true or false:Maximum ionization energy stands for He.

Halogen with highest ionization energy is FF. State whether the statement is true or false

Find the number of element( s) which is/are present in group-I :Be, Li, Mg, AI, Na, Ca, Cs

The number of metals from the following : As,Fe,Xe,Li,B,Cl,Ba,P,I,SiAs, Fe, Xe, Li, B, Cl, Ba, P, I, Si is :

Identify the elements occupying peaks and bottommost points.

The increasing order of reactivity among group 1 element is Li<Na<Rb<CsLi < Na < Rb < Cs, whereas that among group 17 element is F>CI>Br>IF > CI > Br > I. Explain?

Arrange the following in order of increasing reducing property.NaHNaH, MgH2MgH_2 and H2OH_2O

Which important property did Mendeleev use to classify the elements in his Periodic Table, and did he stick to that?

What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

The first ionization enthalpy values ( in kJ mol−1 \displaystyle ^{-1} ) of group 13 element are :B Al Ga In Tl801 577 579 558 589How would you explain this deviation from the general trend?

The atomic number of elements A, B and C are (Z - 1), Z and (Z + 1) respectively. B is an inert gas (not helium). Which element among A,B and C has the highest ionization energy? Why?

In the Modern Periodic Table, calcium (atomic number 2020) is surrounded by elements with atomic numbers 12,19,2112, 19, 21 and 3838. Which of these have physical and chemical properties resembling calcium?

The electronic configuration of HeHe is 22 and that of oxygen is ________.

True or FalsePotassium is the most reactive alkali metals.

Which one has greater value, second ionisation energy or third ionisation energy?

Which is more metallic: sodium or aluminium?

Explain why the elements placed on the extreme left side of the periodic table acts as powerful reducing agents?

Nuclear charge increases in a period as well as group. However, periodic properties such as atomic size and IP show a reverse trend. Why?

Helium has _____configuration.

An element ZZ has atomic number 1616. Answer the following questions on ZZ:State the formula between ZZ and Hydrogen?

Without knowing the electronic configurations of the atoms of elements, Mendeleev still could arrange the elements nearly close to the arrangements in the Modern Periodic Table. How can you appreciate this?

Aluminium does not react with water at room temperature but reacts with both dil. HClHCl and NaOHNaOH solutions. Verify these statements experimentally. Write your observations with chemical equations. From these observations, can we conclude that AlAl is a metalloid?

Metal AA has electronic configuration of 2,8,12, 8, 1 and metal B has electronic configuration of 2,8,22, 8, 2. Which metal is more reactive and why?

The element eka-silicon in Mendeleev's period table is known as ________ in the modern periodic table.

Identify the following substances which are underlined:The element which has the highest ionization potential.

Arrange the following as per the instructions given in the brackets: Na,K,Cl,S,SiNa, K, Cl, S, Si (increasing order of ionization energy)

Complete the following Table:ElementElectronic configurationBond it formsValencyMagnesiumOxygenChlorine

In the modern periodic table, how does the atomic size of the elements vary along the period and down the group? Explain.

The ionisation potential value of KK is less than that of NaNa. Explain.

How can u predict the group, period and block of an element? Explain with examples.

In the second period, Li to Ne identify the element(a) Most reactive element(b) Most non-reactive non-metal(c) Largest atomic radius(d) With valency equal to 4

What is meant by periodicity in properties of elements with reference to the periodic table?

Why do elements in the same group have similar physical and chemical properties ?

State the appropriate scientific reasons for the following statements.The element present in the last shell of an element decides the group in which they are placed.

How many electrons,protons and neutrons are present in one molecule of glucose?

An element ′X′'X' has mass number 3535 and number of neutrons 1818. Write atomic number and electronic configuration of ′X′'X'. Also write group number, period number and valency of ′X′'X'.

The second ionization enthalpy of calcium is more than that of the first and yet calcium form CaCl2{ CaCl }_{ 2 } not CaCl. Why ?

H2OH_2O is a liquid but H2SH_2S is a gas. Explain why?

How does number of valency vary on moving from left to right: in the third period of a periodic table?

Define the term ionisation enthalpy.

Among the element of the second period Li to ne pick out the element:(i) with the highest first ionisation energy(ii) with the highest electronegativity

Write a short note on Mendeleev's periodic law.

Arrange the following in increasing order of the property indicated: I2, F2, Br2 I_2,\ F_2,\ Br_2 and Cl2Cl_2 (reactivity)

Name the pairs showing a diagonal relationship.

Arrange the following in increasing order of the property indicated: Na, Cu and Zn (electropositive character)

With reference to periodic table, indicate:Position of the most electropositive element.

Explain the following: Why is potassium a strongly electropositive element?

The element 119 has not been discovered. What would be IUPAC name and symbol for this element? On the basis of the periodic table, predict the electronic configuration of this element and also the formula of its most stable chloride and oxide.

From among the elements, choose the following:Cl, Br, F, O, Al, C, Li, Cs and Xe The element which shows diagonal relationship with Mg.

Give the electronic configuration of ninth element of the first transition series.

Give the electronic configuration of fifth element of the first transition series.

Give the name and atomic number of the inert gas atom in which the total number of d-electrons is equal to the difference in numbers of total p and s-electrons.

A metal 'M' has the electronic configuration 2, 8, 3 and occurs in nature as M $_2$ O $_4$ . It is more reactive than zinc. Write a chemical equation illustrating the use of this metal for joining cracked railway lines.

Complete the electronic configuration of a chloride ion.
$1s^{2}$ ___________.

The elements ( $z = 117$ and $120$ ) have not yet been discovered in which family or group would you place these elements and also give electronic configuration in each case?

Name the type of elements, which have their
(i) outermost shell complete
(ii) outermost shell incomplete
(iii) two outermost shell incomplete
(iv) one electron short of octet
(v) two electrons in the outermost orbit.

What is the IUPAC name of an element with atomic number $107$ .

With reference to periodic table, indicate:

An element that is in group IIIA and third period.

From among the elements, choose the following:
Cl, Br, F, O, Al, C, Li, Cs and Xe

The element whose atom has 6 electrons in the outermost shell.

With reference to periodic table, indicate:

The inert gases placed in the 2nd and 5th period.

Define the following terms.
Element

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements)
of the modern periodic table with their atomic numbers are given.
3 4 5 6 7 8 9 10
A E G
11 12 13 14 15 16 17 18
B C D F
Which of these is an inert gas ? Name it.

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements) of the modern periodic table with their atomic numbers are given.
3 4 5 6 7 8 9 10
A E G
11 12 13 14 15 16 17 18
B C D F
Which of these are metals ? Name them.

$\text{ With reference to period 3 of the periodic table-State}$
The electronic configuration of the element whose hydroxide is a weak base.

In the following table, seven elements A, B, C, D, E, F and G (here letters are not the usual symbols of the elements) of the modern periodic table with their atomic numbers are given.
3 4 5 6 7 8 9 10
A E G
11 12 13 14 15 16 17 18
B C D F
Which of these is a halogen ? Name it.

State True or False.
The atomic number of Uus is 117.

Why $BeO$ and $Al_2O_3$ are amphoteric?

$BeCl_2$ and $AlCl_3$ are Lewis acids. If true enter 1, else enter 0.

"Decreasing lattice energy favours increased solubility while decreasing hydration energy favours decreased solubility."
Answer whether the above statement is true or false.
If true enter 1, else enter 0.

As the atomic radius increases, ionisation energy is found to decrease.
If true then enter 1, else enter 0.

IUPAC representation of the element of atomic number 115 is Uup.
If true enter 1, else enter 0.

State whether the given statement is true or false:

Maximum ionization energy stands for He.

Halogen with highest ionization energy is $F$ . State whether the statement is true or false

Find the number of element( s) which is/are present in group-I :
Be, Li, Mg, AI, Na, Ca, Cs

The number of metals from the following :
$As, Fe, Xe, Li, B, Cl, Ba, P, I, Si$ is :

The increasing order of reactivity among group 1 element is $Li < Na < Rb < Cs$ , whereas that among group 17 element is $F > CI > Br > I$ . Explain?

Arrange the following in order of increasing reducing property.

$NaH$ , $MgH_2$ and $H_2O$

The first ionization enthalpy values ( in kJ mol $\displaystyle ^{-1}$ ) of group 13 element are :

B Al Ga In Tl
801 577 579 558 589

How would you explain this deviation from the general trend?

In the Modern Periodic Table, calcium (atomic number $20$ ) is surrounded by elements with atomic numbers $12, 19, 21$ and $38$ . Which of these have physical and chemical properties resembling calcium?

The electronic configuration of $He$ is $2$ and that of oxygen is ________.

True or False
Potassium is the most reactive alkali metals.

An element $Z$ has atomic number $16$ . Answer the following questions on $Z$ :
State the formula between $Z$ and Hydrogen?

Aluminium does not react with water at room temperature but reacts with both dil. $HCl$ and $NaOH$ solutions. Verify these statements experimentally. Write your observations with chemical equations. From these observations, can we conclude that $Al$ is a metalloid?

Metal $A$ has electronic configuration of $2, 8, 1$ and metal B has electronic configuration of $2, 8, 2$ . Which metal is more reactive and why?

Identify the following substances which are underlined:
The element which has the highest ionization potential.

Arrange the following as per the instructions given in the brackets:
$Na, K, Cl, S, Si$ (increasing order of ionization energy)

Complete the following Table:
Element Electronic configuration Bond it forms Valency
Magnesium
Oxygen
Chlorine

The ionisation potential value of $K$ is less than that of $Na$ . Explain.

In the second period, Li to Ne identify the element
(a) Most reactive element
(b) Most non-reactive non-metal
(c) Largest atomic radius
(d) With valency equal to 4

State the appropriate scientific reasons for the following statements.
The element present in the last shell of an element decides the group in which they are placed.

An element $'X'$ has mass number $35$ and number of neutrons $18$ . Write atomic number and electronic configuration of $'X'$ . Also write group number, period number and valency of $'X'$ .

The second ionization enthalpy of calcium is more than that of the first and yet calcium form ${ CaCl }_{ 2 }$ not CaCl. Why ?

$H_2O$ is a liquid but $H_2S$ is a gas. Explain why?

How does number of valency vary on moving from left to right:
in the third period of a periodic table?

Among the element of the second period Li to ne pick out the element:
(i) with the highest first ionisation energy
(ii) with the highest electronegativity

Arrange the following in increasing order of the property indicated:

$I_2,\ F_2,\ Br_2$ and $Cl_2$ (reactivity)

Arrange the following in increasing order of the property indicated:

Na, Cu and Zn (electropositive character)

With reference to periodic table, indicate:

Position of the most electropositive element.

Explain the following:

Why is potassium a strongly electropositive element?

From among the elements, choose the following:
Cl, Br, F, O, Al, C, Li, Cs and Xe

The element which shows diagonal relationship with Mg.

The elements Na, Mg, Al, Si, P, S, Cl and Ar are arranged in the periodic table in the increasing order of their atomic numbers.

Which element has the electronic configuration?
$1s^2 , 2s^2 , 2p^6 , 3s^2, 3p^1_x3p^1_y3p^1_z$

Which elements have the electronic configuration $ns^2\ np^2$ ? Give their names.

An element X from group 2 reacts with element Y from group 16 of the periodic table
(a) What is the formula of the compound formed?
(b) What is the nature of the bond in the compound formed?

(a) What is the fundamental difference in the electron configurations between the group 1 and group 2 elements ?
(b) On the basis of electronic configuration how will you identify
(i) Chemically similar electrons ?
(ii) The first elements of a period ?

This question refers to the elements of the periodic table with atomic numbers $3$ to $18$ .
What is the electronic configuration of an element with atomic number $10$ ?

Study the periodic table and answer the following questions:
(i) Na has physical and chemical properties similar to which element(s) and why ?
(ii) Write the electronic configuration of N and P. Which one of these will be more electronegative and why ?

Show the formation of $Na_2O$ by the transfer of electrons.

An element has electronic configuration $2,8,7$ .
What is atomic number of this element?

An element has electronic configuration $2,8,7$ .
Identify the element.

An element has electronic configuration $2,8,7$ .
Name an element chemically similar to this element

Two elements $X$ and $Y$ have atomic number $12$ and $16$ respectively. To which period of the modern periodic table do these two elements belong? What type of bond will be formed between them and why? Also give the chemical formula of the compound formed.

The atomic number of Lithium is $3$ . On the basis of this information answer the question that follows:
Write the electronic configuration of $Li$

Identify and name the metals out of the following elements whose electronic configurations are given below.
(a) 2, 8, 2
(b) 2, 8, 1
(c) 2, 8,7
(d) 2, 1

'M' is an element which is out of Cu, Fe, Al, Na. It shows the following properties:
(i) One of its ore is rich in $M_{2}O_{3}$ .
(ii) $M_{2}O_{3}$ is not affected by water.
(iii) It corrodes easily.
(iv) It forms two chlorides $MCl_{2}$ and $MCl_{3}$ .
Identify 'M'.

Is the following symbol of element incorrect? Give the correct symbol.
Sodium So

Is the following symbol of element incorrect? Give the correct symbol.
Aluminum AL

Is the following symbol of element incorrect? Give the correct symbol.
Helium He

Is the following symbol of element incorrect? Give the correct symbol.
Cobalt CO

Is the following symbol of element incorrect? Give the correct symbol.
Carbon c

Fill in the blanks
_________ refers to the number of atoms in the molecule of an element.

Name the scientists who discovered the following:
Elements

State the contributions of :
Dimitri Mendeleev

Fill in the blanks
The most abundant element in the earth's crust is __________.

The element X has the electronic configuration 2, 8, 18, 8,Without identifying x,
write if X be an oxidizing agent or reducing agent and why.

The element X has the electronic configuration 2, 8, 18, 8,Without identifying x,
predict the sigh and change on a simple ion of X.

Define:
(a) Elements
(b) Compounds

An element barium has atomic number $56$ . Look up its position in the periodic table and answer the following questions :
Is it more or less reactive than calcium ?

What is the common feature of the electronic configuration of the elements at the end of period $2$ and period $3$ ?

The electronic configuration of nitrogen is $(2, 5)$ . How many electrons in the outer shell of a nitrogen atom are not involved in the formation of a nitrogen molecule?

Elements X, Y and Z have atomic numbers 6, 9 and 12 respectively. Which one:
Stat type of bond between Y and Z and give its molecular formula.

How do the following change on moving from left to right in a period of the periodic table ? Give examples in support of your answer.
Chemical reactivity of elements .

Arrange the following metals in the decreasing order of reactivity.
Copper, iron, magnesium, sodium and zinc.