MCQ Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 5

State the electronic configuration for Boron [p=5, n=6].

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2, 3
0%
5, 10
0%
5, 12
0%
5, 11

Explanation

As we know, atomic number is the total no of proton present in nucleus of an atom and atomic mass number is the total no of proton and neutron.
So boron have atomic number 5 and mass number

5 + 6 = 11

$5+6=11$ . Therefore, K shell will accommodate 2 electrons and L shell will accommodate remaining electrons i.e 3. The electronic configuration is 2, 3.

Name the following with reference to the elements of the modern periodic table.

The alkali metal in period 2

0%
H
0%
Li
0%
Rb
0%
Cs

Which element is considered radioactive among the options given below?

0%
Sodium
0%
Gold
0%
Thorium
0%
Neon

Explanation

an element subject to spontaneous degeneration of its nucleus accompanied by the emission of alpha particles, beta particles, or gamma rays. All elements with atomic numbers greater than

83

$83$ are radioactive. Naturally occurring radioactive elements include radium, thorium, and uranium. Several radioactive elements not found in nature have been produced by the bombardment of stable elements with subatomic particles in a cyclotron.

What is the electronic configuration of nitrogen [Atomic number=7, Atomic mass=14]?

0%
$2, 3$
0%
$2, 5$
0%
$2, 2$
0%
$7$

Refer to changes in properties of elements on moving left to right across a period of the periodic table. For each property, choose the letter corresponding to the correct answer from A, B, C and D

The ionization potential _______ .

0%
goes up and down
0%
decreases
0%
increases
0%
remains the same

Electronic configuration of Mg is:

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$1s^2 2s^1p^5 3s^23d^2$
0%
$1s^2 2s^2p^6 3s^2$
0%
$1s^2 2s^2p^7 3s^1$
0%
$1s^2 2s^2p^6 3s^2p^2$

Explanation

Electronic configuration of Mg is

[N e] 3 s^{2}

$[Ne]3s^2$

Ne is

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

$1s^22s^2p^6$ .

What is the electronic configuration of silicon [Atomic no.=14, Mass number=28]?

0%
$2, 8, 1$
0%
$2, 8, 2$
0%
$2, 8, 3$
0%
$2, 8, 4$

What is the electronic configuration of oxygen [Atomic no.=8, Atomic mass=16]?

0%
$2, 3$
0%
$2, 4$
0%
$2, 5$
0%
$2, 6$

Ionisation potential of elements on moving down a group:

0%
increases
0%
decreases
0%
constant
0%
none of these

The electronic configuration,

1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}, 3 d^{9}

$1s^2, 2s^22p^6, 3s^23p^6, 3d^9$ represents a

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Metal atoms
0%
non metals atom
0%
non-metallic anionic
0%
metallic cation

What is the electronic configuration of carbon [Atomic mass=12, Atomic number=6]?

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$2, 3$
0%
$2, 5$
0%
$2, 6$
0%
$2, 4$

Which is the metallic element?

0%
$Y$
0%
$X$
0%
Both $A$ and $B$
0%
None of the above

Explanation

Y : 2, 8, 2

$Y: \ \ 2, 8, 2$ is a metallic element as it has a tendency to loose an electron and get stable octet configuration.

Atom with ____atomic radii and ___ionisation potential tends to gain electrons

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large , high
0%
large , low
0%
small , low
0%
small , high

Compare the properties of a typical metal and a non-metal on the basis of the following.

Electronic configuration :

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Metal have 1, 2, 3, 4 valence electrons while non metals have 5, 6, 7, 8 valence electrons.
0%
Metal have 1, 2 valence electrons while non metals have 4, 5, 6, 7, 8 valence electrons.
0%
Metal have 1, 2, 3 valence electrons while non metals have 6, 7, 8 valence electrons.
0%
Metal have 1, 2, 3 valence electrons while non metals have 4, 5, 6, 7 valence electrons.

Explanation

$\begin{array}{l}\text { Metals have } 1,2,3 \text { valence electrons because } \text { they can Loose either of these electrons to } \\\text { gain octet configuration. } \text { Wheras Non metals can gain remaining electrons } \\\text { to get octet configuration. }\end{array}$

Hence, option D is correct.

Compare the properties of metals and non-metals with respect to electronic configuration.

0%
Metals have 1,2 valence electrons while non metals have 3,5,6 or 7 valence electrons
0%
Metals have 1 or 3 valence electrons while non metals have 2,5 valence electrons
0%
Metals have 2 or 3 valence electrons while non metals have 1,5,6 or 7 valence electrons
0%
Metals have 1,2 or 3 valence electrons while non metals have 4,5,6 or 7 valence electrons

Which element has an electronic configuration of 2, 8, 8 ?

0%
He
0%
Ne
0%
Ar
0%
Kr

Explanation

Elements in group VIII A (He, Ne, Ar, Kr, Xe and Rn) are called noble gases or rare gases.
Electronic configuration of

$Ar: 1s^2 2s^22p^6 3s^23p^6$

Hence, the correct option is (C).

The formula of the hydroxide of the element having electronic configuration 2, 8, 2 is :

0%
$MgOH$
0%
$Mg(OH)_3$
0%
$Mg_2(OH)_3$
0%
$Mg(OH)_2$

Explanation

Element having electronic configuration 2, 8, 2 is Mg and it forms hydroxide as

$Mg(OH)_2$ .

The character of the hydroxide of the element

$Al$ :

0%
acidic
0%
basic
0%
neutral
0%
amphoteric

Explanation

Period 3, group 13 element

$Al$ is a metal so it forms hydroxide as

$Al(OH)_3$ which has both acidic and basic properties.

Increase in nuclear charge of an atom decreases the tendency of the atom to loose electrons.

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True
0%
False

Arrange the following elements in order of their increasing ionization energies

$O, S, Se, Te, Po$ .

0%
$Se,Te, S, Po, O$
0%
$O,S,Se,Te, Po$
0%
$Po, Te,Se,S,O$
0%
$Te, O,S,Po,Se$

Explanation

Ionization energy decreases from top to bottom in a group due to increase in the size of atom.Due to increase in size of atom its ionization energy decreases from top to bottom , So increasing order will be

$Po < Te < Se < S <O$ .

Hence, option

$C$ is the correct answer.

Which of the following statements are correctly matched?

0%
$d$ -block element : electronic configuration is $n{s}^{1-2}(n-1){d}^{1-10}$
0%
$p$ -block element : electronic configuration is $n{s}^{1-2}n{d}^{1-6}$
0%
$s$ -block element : electronic configuration is $n{s}^{1-2}$
0%
$Ce$ : $f$ -block's first member

Explanation

The electronic configuration of

$d$ -block element is

$ns^{1-2}(n-1)d^{1-10}$

The electronic configuration of

$p$ -block element is

$ns^2np^{1-6}$

The electronic configuration of

$s$ block element is

$ns^{1-2}$

So, the correct options are

$A$ and

$C$

What is the electronic arrangement of G ?

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2, 2, 5
0%
2, 4, 3
0%
2, 7
0%
2, 5, 2

Ionization energy is influenced by:

0%
size of an atom
0%
charge on a nucleus
0%
electrons present in the inner shells
0%
None of the above

Explanation

$IE$ is influenced by:
(A) Size of an atom: The ionization energy depends upon the size of an atom because with the increase in size, the distance between nucleus and valence electrons increases and hence, the force of attraction between the nucleus and valence electron decreases.
(B) Charge on a nucleus: As the nuclear charge increases, the force of attraction between nucleus and valence electrons increases and hence, it is difficult to remove an electron from the valence shell. Thus, with the increase in nuclear charge, the ionization energy increases.
(C) Electrons present in inner shells: selections are closer to the nucleus than p-electrons, which are closer than d-electrons and these, in turn, are closer than f-electrons of the same principal energy level. Hence, selections experience more attraction from the nucleus than p, d and f electrons. Thus, ionization energy to remove an electron from a given energy level decrease in order

$s > p > d > f$ .

Electronic configuration of element T is 2, 8, 7. Under which category this element belongs to?

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Alkali metal
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Alkaline metal
0%
Noble gases
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Non-metal

Match column I with column II and select the correct answer using the codes given below:
Column I Column II
(A)

$_{ 6 }^{ 14 }{ C }$ and

$_{ 6 }^{ 13 }{ C }$ (p) Different number of neutrons
(B)

$_{ 20 }^{ 40 }{ C a}$ and

$_{ 18 }^{ 40 }{ Ar }$ (q) Different number of protons and neutrons
(C)

$K \left(19\right)$ (r) Non-metal
(D) Chlorine

$\left(17\right)$ (s) Metal

0%
A - (p); B - (q); C - (r); D - (s)
0%
A - (q); B - (p); C - (s); D - (r)
0%
A - (p); B - (q); C - (s); D - (r)
0%
A - (p); B - (s); C - (q); D - (r)

Explanation

a) Isotopes have different numbers of neutrons.
b) Isobars differ in the number of protons and neutrons but mass number is same.
c) Potassium is a metal.
d) Chlorine is a non-metal.

Hence, it is

$A \rightarrow (p); B \rightarrow (q); C \rightarrow (s); D \rightarrow (r)$

In the Periodic Table, the Ionization potential in a group _____ from top to bottom.

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increases
0%
decreases
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does not change
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can not be predicted

The electronic configuration of chlorine atom is :

0%
$1s^22s^2p^63s^3p^5$
0%
$1s^22s^3p^63s^3p^4$
0%
$1s^22s^2p^63s^2p^5$
0%
$1s^22s^2p^63s^2p^6$

Explanation

$Cl : 1s^22s^2p^63s^2p^5$

$Cl^-: 1s^22s^2p^63s^2p^6$

The electronic configurations of three elements X, Y and Z are

$2, 8$ ;

$2, 8, 7$ ; and

$2, 8, 2$ respectively. Which of the following is correct?

0%
$X$ is a metal
0%
$Y$ is a metal
0%
$Z$ is a non-metal
0%
$Y$ is a non-metal and $Z$ is a metal

Explanation

$X=2,8=10$ : It is the electronic configuration of neon.

$Y=2,8,7=17$ : It is the configuration of the halogen family of chlorine atom which is non-metal

$Z=2,8,2=12$ : It is the configuration of alkaline earth metals of magnesium atom which is metal.

Therefore, X is a noble gas, Y is a non-metal, and Z is a metal.

Option D is correct.

Select the correct I.E. order for the following species.

0%
$Cl\,>\,Br$
0%
$Br^-\,>\,Cl^-$
0%
$Cl\,>\,Cl^-$
0%
$Br\,>\,Br^-$

Explanation

(A)

$Cl\,>\, Br$ is correct because Size of

$Cl$ is smaller than

$Br$ so it requires more energy for the removal of electron hence more will be the ionization potential.

(B)

$Br^-\,>\, Cl^-$ is the wrong statement because as the size of the

$Cl^{-}$ is less

$Cl^{-}$ should have more IP value
(C).

$Cl\,>\, Cl^-$ correct because I.E. of a neutral atom is more than anion
(D).

$Br\,>\, Br^-$ correct because I.E. of a neutral atom is more than anion
Hence options

$A,\ C$ and

$D$ are correct.

Ionization energy depends upon :

0%
principal quantum number
0%
azimuthal quantum number
0%
magnetic quantum number
0%
spin quantum number

For which set of elements, "diagonal relationship" exists?

0%
$B, Si$
0%
$Li, Mg$
0%
$B, Mg$
0%
$Be, Al$

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 . These pairs (lithium

$(Li)$ and megnesium

$(Mg)$ , berilium

$(Be)$ and aluminium

$(Al)$ , boron

$(B)$ and silicon

$(Si)$ etc.) exhibit similar properties; for example, boron and silicon are both semiconductors, forming halides that are hydrolysed in water and have acidic oxides.
Hence options A,B & D are correct.

In group Ia of alkali metals, the ionization potential decreases down the group. Therefore, lithium is a:

0%
good reducing agent
0%
poor reducing agent
0%
good oxidising agent
0%
poor oxidising agent

Explanation

Alkali metals are very good reducing agents because of their great tendency to lose electrons. The reducing character increases from Na to Cs.

However, Li is a stronger reducing agent than Na due to greater hydration energy.

Lithium being small in size has high ionization enthalpy. On the other hand, because of the small size, it is extensively hydrated and has very high hydration enthalpy. This high hydration enthalpy compensates the high energy needed to remove an electron. Thus Li has a greater tendency to lose electrons in solution than other alkali metals. Thus, Li is the strongest reducing agent.

Hence, the correct option is

$\text{A}$

The chemistry of lithium is very similar to that of magnesium even though they are placed in different groups. Its reason is that:

0%
both are found together in nature
0%
both have nearly the same size
0%
both have similar electronic configurations
0%
the ratio of their charge to size is nearly the same

Which of the following has the lowest ionization enthalpy?

0%
$4s^1$
0%
$3d^2$
0%
$3p^6$
0%
$2p^6$

Explanation

Consider the options:-
A)

$4s^1$ - Since it has only one electron, it can be easily removed. Due to shielding effect, the energy is less.
B)

$3d^2$ - The

$I.E$ is considerable. The energy will be greater than that of

$4s^1$ because of lesser shielding effect.
C)

$3p^6$ - Fully filled orbital, high value of

$I.E$ .
D)

$2p^6$ - Fully filled orbital, high value of

$I.E$ .
Hence, option A is the right answer.

Aluminium is diagonally related to :

0%
$Li$
0%
$Si$
0%
$Be$
0%
$B$

Identify the least stable ion amongst the following :

0%
$Li^-$
0%
$Be^-$
0%
$B^-$
0%
$C^-$

Explanation

Hint: Half filled $\&$ fully filled configurations are more stable due to exchange energy.

Explanation

· The configuration of $Li$ is $1s^{2}2s^{1}$ . Adding one $e^{-}$ will fill its $2s$ subshell providing it extra stability.

· The configuration of $Be$ is $1s^{2}2s^{2}$ . Adding one $e^{-}$ will disturb its stable fully filled configuration. So, it is the least stable among all.

· The configuration of $B$ is $1s^{2}2s^{2}2p^{1}$ . It does not have half-filled or fully filled configuration so adding one $e^{-}$ will not disturb its stable configuration.

· The configuration of $C$ is $1s^{2}2s^{2}2p^{2}$ . It does not have half-filled or fully filled configuration so adding one $e^{-}$ its configuration becoes $1s^2 2s^2 2p^3$ that it is a half-filled configuration and becomes stable.

Correct Option : $B$ .

Compound of a metal

$'M'$ is

$M_{2}O_{3}$ .The formula of its nitride will be -

0%
$M_{3}N$
0%
$MN$
0%
$M_{3}N_{2}$
0%
$M_{2}N_{3}$

Explanation

Compound of a metal M formed

$M_{2}O_{3}$ . It represent that M have 3 valency electrons. So the nitride formed will be MN.

I.P. of sodium is 5.14 eV then I.P. of potassium will be :

0%
equal to sodium
0%
5.68 eV
0%
4.34 eV
0%
10.28 eV

A large difference between the third and fourth ionization energies indicates the presence of :

0%
4 valence electrons in an atom.
0%
5 valence electrons in an atom
0%
3 valence electrons in an atom
0%
8 valence electrons in an atom.

Explanation

c.) For possible

$ns^2 np^1$ configuration, the removal of fourth electron will be possibly from an inert gas electron configuration. So there will be high jump in the fourth ionisation enthalpy than the third ionisation enthalpy which will take place from

$ns^1$ electron configuration.

Which of the following decreases in going down the halogen group?

0%
Ionic radius
0%
Atomic radius
0%
Ionisation potential
0%
Boiling point

As we go down in the electro-chemical series of metals, the reactivity ______ .

0%
decreases and then increases
0%
increases and then decreases
0%
decreases
0%
increases

Which is most acidic oxide ?

0%
$Cl_2O$
0%
$Cl_2O_3$
0%
$Cl_2O_6$
0%
$Cl_2O_7$

Explanation

Hint :As oxidation state of central atom increases acidic nature increases.

Correct Answer :

$D$

Explanation of Correct Option:

The acidic nature increases with increase of percentage of oxygen and oxidation number of central atom. Therefore, the increasing order of the acidic character will be

$Cl_2O < Cl_2O_3 < Cl_2O_6 < Cl_2O_7$ .

Oxidation number of

$Cl:$

$+1$

$+3$

$+6$

$+7$

More the positive charge on chlorine atom greater is the acidic nature of oxide, therefore,

$\mathrm{Cl}_{2} \mathrm{O}_{7}$ is most acidic oxide.

Explanation of Incorrect Option:

In other oxides the oxidation state of chlorine less than

$+7$ hence less acidic than

$\mathrm{Cl}_{2} \mathrm{O}_{7}$

The correct order of ionization energies of

$\displaystyle F^{-},Cl^{-}F$ and

$Cl$ is:

0%
$\displaystyle Cl< F< Cl^{-}< F^{-}$
0%
$\displaystyle Cl^{-}< F^{-}< Cl< F$
0%
$\displaystyle F^{-}< Cl^{-}< Cl< F$
0%
$\displaystyle Cl^{-}< Cl< F^{-}< F$

Explanation

The correct order of ionization energies of

$\displaystyle F^{-},Cl^{-}, F$ and

$Cl$ is

$\displaystyle F^{-}< Cl^{-}< Cl< F$ .

$Cl^->F^-$ because the size of F is less than

$Cl^{-}$ , so the ionization potential will be more for F than

$Cl^{-}$ .

$F>Cl$ because the size of F is less than Cl,so the ionization potential will be more for F than Cl.

In a group, the ionization energy decreases from top to bottom.
The size of the anion is greater than the size of the neutral atom. Due to this, the attraction of the nucleus for the valence electrons is larger in case of neutral atom. Hence, the ionization energy of the neutral atom is larger than that of anion. In case of halogens, the addition of an electron leads to breaking of the noble gas configuration. The ionization energy of fluoride ion is lowest due to electron electron repulsion.

Fluorine is the most reactive among all the halogens, because of its:

0%
small size
0%
low dissociation energy of F - F bond
0%
large size
0%
high dissociation energy of F - F bond

Explanation

Fluorine is the most reactive among all the halogens, because of it's low dissociation energy of

$F - F$ bond. The dissociation energies (kcal/mol) in

$F_2, Cl_2, Br_2$ and

$I_2$ are 38, 57, 45.5 and 35.6 respectively.

In the case of fluorine:

(a) short bond length and larger repulsion between non-bonding electrons lead to low dissociation energy.

(b) absence of multiple bonding leads to low dissociation energy.

Which of the following is mismatched with reference to third period?

0%
Largest size ___ Argon
0%
Strongest oxidant ____ chlorine
0%
IP of phosphorus ____ greater than nitrogen
0%
IP of phosphorus ______ greater than sulphur

Explanation

IP of phosphorus is not greater than nitrogen because half-filled orbital of nitrogen and atomic size of

$P > N$ so IP is

$N > P.$

IP is influenced by :

0%
size of atom
0%
effective nuclear charge
0%
electrons present in inner shell
0%
all

With reference to ionisation potential which one of the following set is correct?

0%
$Li > K > B$
0%
$B > Li > K$
0%
$Cs > Li > K$
0%
$Cs < Li < K$

Explanation

The decreasing order of the ionization potentials is

$\displaystyle B > Li > K$ .
Ionization potential increases across a period.
In a group, the ionization potential decreases from top to bottom.

Which of the following statement is correct?

0%
Ionization energy may be negative for some elements.
0%
Second electron gain enthalpy always remains positive for all the elements.
0%
Negative value of electron gain enthalpy of Fluorine is minimum in its group.
0%
Ionization energy of Ga is slightly lower than Al.

As one moves down the group from top to bottom, which one among the following will not be observed?

0%
Increase in Ionization energy
0%
Decrease in electron affinity
0%
Decrease in electronegativity
0%
Increase in atomic radii

Which of the following is the increasing order of electron affinity of halogens ?

0%

$I < Br < Cl < F$
0%
$I < Br < F < Cl$
0%
$F < Cl < Br < I$
0%
$Br < F< I < Cl$

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 5 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 5 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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