MCQ Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 8

Given the PES from the previous question, explain why the ionization energy for the valence electrons of

$Ne$ is higher than that of

$Ar$ .

0%
$Ne$ has more protons than $Ar$ which requires more energy to remove the valence electron.
0%
$Ne$ has more nuclear shielding than $Ar$ which requires less energy to remove the valence electron.
0%
$Ar$ has more protons than $Ne$ which requires more energy to remove the valence electrons.
0%
$Ar$ has more nuclear shielding than $Ne$ which requires less energy to remove the valence electron.

Which of these metals would be most reactive?

0%
$K$
0%
$Ca$
0%
$Ti$
0%
$Mn$
0%
$Cu$

Identify the correct ground state electron configuration for

$Cr$ .

0%
$[Ar] 3s^23d^4$
0%
$[Ar]3s^2 3d^5$
0%
$[Ar] 4s^2 3d^5$
0%
$[Ar] 4s^2 3d^4$
0%
$[Ar] 4s^1 3d^5$

Explanation

The expected ground state electron configuration for

$Cr (Z=24)$ is

$[Ar] 4s^2 3d^4$ . But the observed ground state electron configuration is

$[Ar] 4s^1 3d^5$ . This is due to extra stability of half filled 3d and 4s subshells.

An ion is an atom with an electric charge due to a loss or gain of an electron.
Which of the following is the electron configuration of a ground state of the most common alkali metal ion?

0%
$1s^{2}2s^{2}$
0%
$1s^{2}2s^{1}$
0%
$1s^{2}2s^{2}2p^{6}$
0%
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{1}$

Explanation

Alkali metals are the group one

$s$ block elements in which last electrons enter in nS energy level.

The general electronic configuration of

$s-block$ elements is

$ns^1$ where

$n= 2$ to

$7.$

Which element has the lowest first ionization energy?

0%
$As$
0%
$P$
0%
$N$
0%
$Bi$
0%
$Sb$

Explanation

All the given options of elements belong to the same group

${ 5 }^{ th }$ . The value of ionization energy (IE) decreases down the group due to the increasing size as the valence electrons are more loosely bound. Thus lowest IE in this case is of

$Bi$ as it is bottom most element.

The electron configuration for

$Cu$ is

$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1} 3d^{10}$ .
Based on the electron configuration, what can be said about the relative ionization energies of the electrons in

$n = 3$ ?

0%
The ionization energies gradually decrease from s to p to d subshells.
0%
The ionization energies gradually increase from s to p to d subshells.
0%
The ionization energies are relatively the same from s to p to d subshells.
0%
The ionization energies are unpredictable at this energy level.

Explanation

Thus, ionization energy increases from left to right on the periodic table. Another factor that affects ionization energy is electron shielding. Electron shielding describes the ability of an atoms's inner electrons to shield its positively-charged nucleus from its valence electrons.

Thus the ionization energy decreases from

$s$ to

$p$ to

$d$ sub shells.

Which of the following statement best describes the first ionization energies of sodium compare of potassium?

0%
The first ionization energy for sodium is greater than the first ionization energy for potassium due to a decrease in the effective nuclear charge.
0%
The first ionization energy for sodium is less than the first ionization energy for potassium due to a increase in the effective nuclear charge.
0%
The first ionization energy for sodium is greater than the first ionization energy for potassium due to a decrease in nuclear shielding.
0%
The first ionization energy for sodium is less than the first ionization energy for potassium due to a increase in nuclear shielding.

A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion

$(Y^{–})$ will be :

0%
275.1 pm
0%
322.5 pm
0%
241.5 pm
0%
165.7 pm

Explanation

Given, radius of the cation is 100 pm .

As we know :

$\frac{r^+}{r^-}=0.414$ for NaCl

Substituting value of

$r^+$ in above equation , we get

$r^-=\frac{100}{0.414}$

$\therefore\ r^-=241.5\ pm$

Hence, option C is correct .

Which element of the following has the smallest first ionization energy?

0%
$Li$
0%
$Mg$
0%
$K$
0%
$Be$

Explanation

The ionization energy decreases down the group and increases along the period towards right. Thus the comparisons are

$Li>K,Li<Be,Be>Mg,Mg>K$ .

Thus the least value is of

$K$ . Hence answer is option C.

The first ionisation potentials (eV) of Be and B respectively are

0%
8.29 eV, 9.32 eV
0%
9.32 eV, 9.32 eV
0%
8.29 eV, 8.29 eV
0%
9.32 eV, 8.29 eV

Explanation

electronic configuration for B and Be as follows:

Be -

$1s^{2} 2s^{2}$

B -

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

Beryllium has more first ionisation potential because electron is being removed from full filled 2s orbital.

What is the symbol of atomic number 117 ?

0%
Uus
0%
Uns
0%
Uno
0%
Uut

The first ionization for phosphorus is

$1060 kJ/mol$ , and that for sulfur is

$1005 kJ/mol$ .
Which is the best explanation for this difference in first ionization energy?

0%
Sulfur has a larger nuclear charge than phosphorus.
0%
Sulfur has a larger number of inner electrons that create a greater shielding effect.
0%
Electron repulsion in the $3p^{4}$ in sulfur makes the electron easier to remove than the $3p^{3}$ electron in phosphorus.
0%
A sulfur atom is larger than a phosphorus atom.

Explanation

Phosphorous has half filled

$3p$ subshell and hence has extra paramagnetic stability, however Sulphur has

$3{ p }^{ 4 }$ configuration hence one electron pair is present which results in repulsion, so removal of electron becomes easier. Thus correct option is C.

Consider the table below, showing several properties of four representative elements, potassium, calcium, sulfur, and chlorine, all from period

$4$ of the periodic table. Notice that the ionization energies of the nonmetals and the electron affinities of the metals have not bee included.

Atom	Ionization Energy $\left (\dfrac {kJ}{mol}\right )$	Electron Affinity $\left (\dfrac {kJ}{mol}\right )$	Atomic Radius (pm)	Most Stable Ion	Ionic Radius (pm)
$K$	$419$	N/A	$227$	$K^{+}$	$133$
$Ca$	$(1) 589$ $(2) 1145$	N/A	$197$	$Ca^{2+}$	$99$
$S$	N/A	$(1) - 201$ $(2) + 532$	$104$	$S^{2-}$	$184$
$Cl$	N/A	$-348$	$99$	$Cl^{-}$	$181$

Note:

$1 pm = 10^{-10}cm$
Which of the four compounds listed would have the least exothermic lattice energy?

0%
$KCl$
0%
$K_{2}S$
0%
$MgCl_{2}$
0%
$MgS$

Elements in the same column/group of the periodic table have similar chemical properties because their atoms have the same_________.

0%
number of protons.
0%
number electrons.
0%
number of valence electrons.
0%
atomic mass.

Explanation

The elements of same group have similar chemical properties because they have same number of valence electrons. For eg: Li has

$1$ valence electron and so does

$NA,K$ and

$Cs$ . Hence they have similar properties.

Which term describes the tendency of elements to participate in a chemical change process?

0%
Ionic radius
0%
Reactivity
0%
Ionization enthalpy
0%
Atomic radius

In chemistry lab class, your teacher asks you to find magnesium oxide,

$MgO$ , to use in the experiment. When you go to the stockroom, you realized there is no magnesium oxide available.
Which of the following would be the most suitable replacement?

0%
Sodium oxide, $Na_{2}O$
0%
Calcium oxide, $CaO$
0%
Carbon dioxide, $CO_{2}$
0%
Silicon dioxide, $SiO_{2}$

Explanation

$Mg$ is a s block metal. It is placed above to

$Ca$ in the periodic table. So it has similar properties as of

$Mg$ . Also from the given option,

$CaO$ is the only s-block metals oxide. So it most closely resembles properties of

$MgO$ . Hence most suitable replacement of

$MgO$ would be Calcium Oxide

$CaO$ (option B)

The first ionization energy for

$Be$ is

$899 kJ/mol$ while B has a first ionization energy of

$800 kJ/mol$ . This data does not fit the usual trend of ionization energies.
Select the BEST statement to justify the data.

0%
The valence electrons in $Be$ are removed from a full $2s$ subshell with less shielding compared to a not full $2p$ subshell in $B$ .
0%
Be has valence electrons in the $2s$ subshell which is much closer to the nucleus and less difficult to remove than the $2p$ valence electrons in $B$ .
0%
Moving left to right on the periodic table, there are more protons in the nucleus attracting the electrons thus making it more difficult to remove electrons.
0%
$B$ has more electrons than $Be$ , increasing the electron-electron repulsions and lowering the ionization energy for $B$ .

Explanation

The valence electron of

$Be$ is in

$2s$ subshell which faces less shielding compared to valence electron of

$B$ in

$2p$ subshell. Thus IE of

$B$ is less than that of

$Be$ . Thus the solution for the answer is option A.

What is the atomic number of the element Ununnilium?

0%
$199$
0%
$191$
0%
$119$
0%
$110$

The correct order in which the first ionisation potential increases is:

0%
$Na,K,Be$
0%
$K,Na,Be$
0%
$K,Be,Na$
0%
$Be,Na,K$

Explanation

Ionisation energy decreases down the group with increase in size. These elements belong to same group (1st). The correct order of first ionization energy should be: $K < Na$ .

Ionisation energy increases across a period from left to right. Be lies on the right of Lithium. So Be has the highest ionisation enthalpy as the number of shell is less than Na and K. So the correct order is : $K<Na<Be$

What is the IUPAC symbol of Rutherfordium?

0%
Rt
0%
Ru
0%
Rf
0%
Rd

What would be the IUPAC name of element with a atomic number 120?

0%
Unbinilium
0%
Unnilhexium
0%
Unnilseptium
0%
Ununnilium

What is the IUPAC symbol for Mendelevium ?

0%
Md
0%
Mn
0%
Me
0%
Mv

For the properties mentioned, the correct trend for the different species is :

0%
inert pair effect - $Al > Ga > In$
0%
first ionization enthalpy - $B > Al > Tl$
0%
strength as Lewis acid - $B{ Cl }_{ 3 } > Al{ Cl }_{ 3 } > Ga{ Cl }_{ 3 }$
0%
oxidising property - ${ Al }^{ +3 } > { In }^{ +3 } > { Tl }^{ +3 }$

Explanation

On moving down in group tendency to accept lone pair decreases. Therefore, lewis acid strength decreases

$BCl_3>AlCl_3>GaCl_3$

An element with atomic number

$17$ is placed in the group

$17$ of the long from periodic table. Element with atomic number

$9$ is placed above and with atomic number

$35$ is placed below it. Element with atomic number

$16$ is placed left and with atomic number

$18$ is placed right to it. Which of the following statements are correct?

a) Valency of the element with atomic number

$18$ is zero.

b) Elements with same valency will have atomic number

$16$ ,

$17$ and

$18$ .

c) Valency of element with atomic number

$9$ ,

$17$ and

$35$ is one.

d) Element with atomic number

$17$ is more electronegative than element with atomic number

$16$ and

$35$ .

0%
(a), (b) and (c)
0%
(a), (c) and (d)
0%
(b), (c) and (d)
0%
(a), (b) and (d)

Explanation

The element with the atomic number 18 is argon. It is a noble gas and belongs to zero group or 18th group. It has its octet complete and so the valency is zero.
Elements with atomic umber 9, 17 and 35 are F, Cl and Br. They are all halogens and belongs to group 17. All halogens have valency 1.
As the atomic size increases, electronegativity decreases. In a period electronegativity increases and in a group it decreases. Thus 17 is more electronegative than 16 and 35.
Hence, the correct option is $B$ .

Which of the following element will show almost same reactivity like

$Ca$ ?

0%
$Cr$
0%
$Mn$
0%
$Sc$
0%
$Na$

Explanation

A diagonal relationship is said to exist between certain pairs of diagonally adjacent elements in the second and third periods of the periodic table. Diagonal relationships occur because of the directions in the trends of various properties as you move across or down the periodic table. Many of the chemical properties of an element are related to the size of the atom.

$Ca$ and $Na$ show diagonal relationship hence, similar properties. Hence, they show almost same reactivity.

The atomic number of magnesium isThen its electron distribution is __________.

0%
2,2,8
0%
2,8,2
0%
8,2,2
0%
None of the above

What is the atomic number of the element with symbol Uus?

0%
117
0%
116
0%
115
0%
114

Explanation

Ununseptium have atomic number 117 having symbol

$Uus$ .

U stands for 1,

u stands for 1,

s stands for 7.

Hence, the correct option is

$A$ .

What is the electronic configuration of neutral

$Mg$ and

$Cl$ ?

0%
$Mg ; [Ne]3s^2$ ; $Cl : [Ne]3s^2 3p^5$
0%
$Mg ; [Ne]3s^1$ ; $Cl : [Ne]3s^2 3p^5$
0%
$Mg ; [Ne]3s^2$ ; $Cl : [Ne]3s^2 3p^4$
0%
$Mg ; [Ne]3s^2$ ; $Cl : [Ne]3s^2 3p^6$

Explanation

The electron configuration is the distribution of electrons of an atom or molecule (or other physical structure) in atomic or molecular orbitals.

For example:

$[Ne]=1{ s }^{ 2 }2{ s }^{ 3 }2{ p }^{ 6 }$

Therefore, the electronic configuration of

$Mg$ =

$[Ne]3{ s }^{ 2 }$

The electronic configuration of

$Cl$ =

$[Ne]3{ s }^{ 2 }3{ p }^{ 5 }$

Arrange the elements

$Se, Cl$ and

$S$ in the increasing order of ionisation energy.

0%
$Se > S > Cl$
0%
$Se < S < Cl$
0%
$Se < S > Cl$
0%
None of the above

Explanation

Ionisation energy is the energy required to remove one electron from a neutral atom of an element.

$S$ and

$Se$ belong to VIA group elements. Down the group ionisation energy decreases so

$S$ is having more IE value than

$Se.$

$S$ and

$Cl$ are the elements of the same period the IE value increases from left to right. So,

$Cl$ is having more IE value than

$S$ .

So the correct order of IP is

$Se<S<Cl$

Hence option

$B$ is correct.

Which of the following is most polarised?

0%
$Kr$
0%
$Ar$
0%
$He$
0%
$Xe$

Explanation

$Xe$ is highly polarised because of its large size and the ionisation potential of xenon is quite close to the ionisation of oxygen.

The spin only magnetic moment of

$[ZCl_{4}]^{2-}$ is

$3.87\ BM$ where

$Z$ is:

0%
$Mn$
0%
$Ni$
0%
$Co$
0%
$Cu$

Explanation

$\mu=3.87=\sqrt{n(n+2)}\quad \Rightarrow n=3$

$[ZCl_{4}]^{2-}$ ; oxidation state of

$Z$ is

$+2$

No. of unpaired electrons is

$3$ .

Hence,

$Z^{2+} = d^{7}$

So, element is

$Co$ .

What is the name of the element with atomic number

$105$ ?

0%
Dubnium
0%
Holmium
0%
Kurchatovium
0%
Nobelium

Explanation

The element with atomic number

$105$ is Dubnium (

$Db$ ). Its IUPAC name is Un-nil pentium.

Amongst the followings, select the element having highest ionization enthalpy.

0%
Sodium
0%
Potassium
0%
Beryllium
0%
Magnesium

Which of the following configurations has the highest ionization energy?

0%
$1s^22s^22p^1$
0%
$1s^22s^22p^3$
0%
$1s^22s^22p^2$
0%
$1s^22s^22p^4$

Explanation

The electron will move from a lower stationary state to a higher stationary state when the required amount of energy is absorbed by the electron or energy is emitted when the electron moves from a higher stationary state to lower stationary state. The energy change does not continuously take place.

Hence, the correct option is

$\text{B}$

The correct order of

$2$ nd ionisation energy for the elements carbon, nitrogen, oxygen and fluroine is:

0%
$F > O > N > C$
0%
$C > N > O > F$
0%
$O> F > N > C$
0%
$O > N > F > C$

Explanation

The second ionisation energy means the energy required to remove the electron from the corresponding monovalent cation of the respective atom. For

$IE_2$ , electronic configuration of

$C^+, N^+, O^+$ and

$F^+$ -ions which are as follows.

$C^+=2s^2, 2p1$

$N^+=2s^2, 2p^2$

$O^+=2s^2, 2p^3$

$F^+=2s^2, 2p^4$ .

$\because$

$O^+$ is half-filed configuration, obviously it is most stable and greater ionisation energy than

$F^+$ as well as

$N^+$ .
The ionisation energy of an element depends upon its size as. Hence, correct order will be,

$O > F > N > C.$

The correct order in which the first ionisation potential increases is :

0%
$Na, K, Be$
0%
$K, Na, Be$
0%
$K, Be, Na$
0%
$Be, Na, K$

Explanation

The atomic radius of elements increases across a period and decrease down the group as the nuclear charge increases across the period and decrease down the group respectively.

Exception: But elements with half-filled and completely filled valence orbitals are more stable hence they have more ionization energy.

So, Beryllium has completeley filled

$s$ orbital (

$1s^22s^2$ ).

Hence it has higher ionization potential than

$Na, K$ .

So, the order is

$Be>Na>K$ .

Hence the correct option is D.

The decreasing order

$IE_2$ of K, Cr and Ba is:

0%
$K > Ca > Ba$
0%
$Ca > Ba > K$
0%
$Ba > K > Ca$
0%
$K > Ba > Ca$

Explanation

The decreasing order IE2 of K, Cr and Ba is:

$\displaystyle K > Ca > Ba$

K (alkali metal) loses one electron to attain stable electronic configuration of noble gas Ar. The removal of second electron will break this stable electronic configuration and will require highest energy.

Ca and Ba (alkaline earth metals) loses 2 electrons to attain stable electronic configuration of noble gas. Hence, they have lower IE2 compared to that of K.

Ca has higher IE2 than Ba because on moving down the group of alkaline earth metals, the ionization energy decreases.

Electronic configuration of

$Gd(64)$ is written as:

0%
$[Xe]4f^{7} 5d^{1} 6s^{2}$
0%
$[Xe]4f^{8} 6s^{2}$
0%
$[Xe]4f^{9} 6s^{1}$
0%
$[Xe]4f^{10}$

Explanation

$Gd. (64) = [Xe]4f^{7} 5d^{1}$
All electrons are unpaired, hence greater stability.

The decreasing order of the ionisation potential in the following elements is :

0%
$Ne > Cl > P > S > Al > Mg$
0%
$Ne > Cl > P > S > Mg > Al$
0%
$Ne > Cl > S > P > Mg > Al$
0%
$Ne > Cl > S > P > Al > Mg$

Explanation

Along the period from left to right ionisation potential increases.

1st ionisation potential of group IIA elements is always higher than the 1st ionisation potential of group IIIA elements.

1st ionisation potential of group VA elements is always higher than the 1st ionisation potential of group VIA elements.

So, the sequence is :

$Ne > Cl > P > S > Mg > Al$

Considering the element

$F, Cl, O$ and

$N,$ the correct order in their chemical reactivity In terms of oxidising property is :

0%
$F > Cl > O > N$
0%
$F > O > Cl > N$
0%
$Cl > F > O > N$
0%
$O > F > N > Cl$

Explanation

In a group oxidising power decreases from top to bottom. as the size increases but when we move from left to right in a period, it increases because site decreases. Therefore among

$F, Cl, O$ and

$N,$ the oxidising power decreases in the order.

$F > O > CI > N$
Note oxygen is more electronegative than chlorine,
Hence,

$O$ is stronger oxidising agent than

$Cl.$

Which of the following represents correct acidity order

$Li_2O, BeO$ and

$B_2O_3$ ?

0%
$Li_2O$ < $BeO$ < $B_2O_3$
0%
$B_2O_3 < BeO < Li_2O$
0%
$BeO < Li_2O < B_2O_3$
0%
$BeO < B_2O_3 < Li_2O$

Explanation

The chemical nature of oxides of elements changes across the period as
Basic oxide

$\rightarrow$ Amphoteric oxide

$\rightarrow$ Acidic oxide Thus, the correct increasing order of acidity is

$Li_2O < Be0 < B_2O_3$

Which of the following electronic configuration is not possible?

0%
$2p^{6}$
0%
$3s^{1}$
0%
$2p^{5}$
0%
$3f^{12}$

Explanation

Correct Option: D

Explanation

Electrons are filled in orbitals in accordance with three rules:
1. Hund’s rule: It states that each orbital of a subshell should be singly occupied before pairing starts.
2. Pauli’s exclusion principle: It states that no two electrons in an atom can have all four quantum numbers the same.
3. Aufbau’s principle: According to this, electrons are filled in orbitals in order of their increasing energy, i.e., orbitals with lower energy must be filled before those having higher energy.
Number of maximum electrons in each subshell:
1. $s= 2$
2. $p= 6$
3. $d= 10$
4. $f = 14$
The shell number from which each sub-shell begins
1. $s \rightarrow n = 1$
2. $p \rightarrow n = 2$
3. $d \rightarrow n = 3$
4. $f \rightarrow n = 4$
Therefore, $3f$ is not possible as $f-subshell$ begins from $n=4$ . Only $4f$ and further subshells exist.

Hence, the electronic configuration which is not possible is $3f^{12}$

Which one of K, I, Cl and Li Will display the highest first ionisation energy ?

0%
K
0%
I
0%
Cl
0%
Li

Which of the following processes involves absorption of energy?

0%
Cl (g) + e $^{-}$ $\rightarrow$ Cl $^{-}$ (g)
0%
O $^{-}$ + $e^{-}$ $\rightarrow$ O $^{2-}$ (g)
0%
O (g) + e $^{-}$ $\rightarrow$ O $^{-}$ (g)
0%
S (g) + e $^{-}$ $\rightarrow$ S $^{-}$ (g)

Explanation

$O^-$ to $O^{2-}$ because of interelectronic repulsion, it will require energy to add an electron to its outermost shell.
Hence option B is correct answer.

Which of the following species has the highest electron affinity?

0%
${F}$
0%
$O$
0%
${O}^{-}$
0%
${Na}^{+}$

Explanation

Energy is released when an electron is added to the atom. Hence 1st electron affinity

$\left( E{ A }_{ 1 } \right)$ is always taken as negative. The addition of second electron to the negativity charged ions is opposed by Coulombic repulsion, hence energy has to be supplied for the addition of second electron. Hence

$\left( E{ A }_{ 2 } \right)$ is taken as positive.

Addition of an electron to

${F}^{-}$ and

${O}^{-}$ ions will involve repulsions, hence their

$\left( E{ A }_{ 2 } \right)$ will be positive.
Due to stable fully filled configuration of

${Na}^{+}$

$\left( 1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 } \right)$ it has lower value of electron affinity than

$O$ . Hence

$O$ has highest electron affinity.

Which electronic configuration of an element has abnormally high difference between second and third ionisation energy?

0%
$1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 6 },3{ s }^{ 1 }$
0%
$1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 6 },3{ s }^{ 2 },3{ p }^{ 1 }$
0%
$1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 6 }$
0%
$1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 6 },3{ s }^{ 2 }$

Explanation

(a)

$1 s^2 \,2s^2 \,2p^6 \,3s^1 \xrightarrow{+E_1} 1s^2 \,2s^2 \,2p^6$

$\xrightarrow{+1E_2} 1s^2 \,2s^2 \,2p^5 \xrightarrow{+1E_3}1s^2 \,2s^2 \,2p^4$

(b)

$1 s^2 \,2s^2 \,2p^6 \,3s^2 \,3p^1 \xrightarrow{+E_1} 1s^2 \,2s^2 \,2p^6 \,3s^2$

$\xrightarrow{+1E_2} 1s^2 \,2s^2 \,2p^6 \,3s^1 \xrightarrow{+1E_3}1s^2 \,2s^2 \,2p^6$

(c)

$1 s^2 \,2s^2 \,2p^6 \xrightarrow{+E_1} 1s^2 \,2s^2 \,2p^5 \xrightarrow{+1E_2} 1s^2 \,2s^2 \,2p^4$

$\xrightarrow{1E_3} 1s^2 \,2s^2 \,2p^3$

(d)

$1 s^2 \,2s^2 \,2p^6 \,3s^2 \xrightarrow{+E_1} 1s^2 \,2s^2 \,2p^6 \,3s^1$

$\xrightarrow{+1E_2} 1s^2 \,2s^2 \,2p^6 \xrightarrow{+1E_3} 1s^2 \,2s^2 \,2p^5$

In option D, after removal of second valence electron from 3s orbital, the ion formed achieves noble gas configuration. Therefore, to remove the third electron from 2p orbital, a lot of energy is required. Thus, there is an abnormally high difference between second and third ionization enthalpies.

The chemical activity of an atom is dependent on the number of __________.

0%
protons
0%
neutrons
0%
orbits
0%
valence electrons

Which of the following species have the same number of electrons in its outermost as well as penultimate shell?

0%
$Cl^-$
0%
$O^{2-}$
0%
$Na^+$
0%
$Mg^{2+}$

Explanation

$Cl^-=2, 8, 8$

$O^{2-}=2, 8$

$Na^+=2, 8$

$Mg^{2+}=2, 8$

$Cl^-$ contains same number of electrons in outermost and penultimate shells.

Few general names are given along with their valence shell configurations. Mark the incorrect name.

0%
$ns^{2} np^{6}$ - Noble gases
0%
$ns^{2} np^{5}$ - Halogens
0%
$ns^{1}$ - Alkali metals
0%
$ns^{2} np^{2}$ - Chalcogens

Explanation

Chalcogens are group 16 elements with valence shell electronic configuration

$ns^2np^4$ . They belong to oxygen family. Group 14 element (Carbon family) have valence shell electronic configuration

$ns^2np^2$ .

One mole of magnesium in the vapour state absorbed 1200 kJ of energy. If the first and second ionization potentials of magnesium are 750 and 1450 kJ/mole respectively, the final composition of the mixture is:

0%
$69%{ Mg }^{ + }$ , $31%{ Mg }^{ 2+ }$
0%
$59%{ Mg }^{ + }$ , $41%{ Mg }^{ 2+ }$
0%
$49%{ Mg }^{ + }$ , $51%{ Mg }^{2 + }$
0%
$29%{ Mg }^{ + }$ , $71%{ Mg }^{ 2+ }$

Explanation

Let mass of

$Mg^{+}$ ion=

$x$

$g$

$Mg^{2+}$ ion=

$y$

$g$

and total

$=1$

$x+y=1$

Molar mass of

$Mg=24.g/mol$

No. of moles of

$Mg^+=\cfrac {x}{24} g/mol$

No. of mole of

$Mg^{2+}=\cfrac {y}{24}g/mol$

Energy absorbed to convert to

$Mg^+=\cfrac {x}{24}\times 750$

Energy absorbed to convert to

$Mg^{2+}=\cfrac {y}{24}\times (1450+750)$

Total energy absorbed=

$\cfrac {x}{24}\times 740+\cfrac {y}{24}\times 2200=50$

$24\times 50=740x+2200y$

$=740x+2200-2200x$

$24\times 50=2200-1460x$

$1200-2200=-1460x$

$x=\cfrac {-1000}{1460}$

$=0.6849$

$y=1-0.6849=0.3151$

Mass % of

$Mg^{2+}$ ion=

$\times0.3151\times 100$

$=31.51$

Mass% of

$Mg^+$ ion=

$0.6849\times 100$

$=68.49$

Mass of

$Mg^+$ ion=

$68.49g$

Mass of

$Mg^{2+}$ ion=

$31.51g$

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Number	Root
0	nil
1	un
2	bi
3	tri
4	quad
5	pent
6	hex
7	sept
8	oct
9	enn

Number	Root
0	nil
1	un
2	bi
3	tri
4	quad
5	pent
6	hex
7	sept
8	oct
9	enn

Number	Root
0	nil
1	un
2	bi
3	tri
4	quad
5	pent
6	hex
7	sept
8	oct
9	enn

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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