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Test: Ionic Bonds & Fajans' Rule - NEET MCQ


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Test: Ionic Bonds & Fajans' Rule - Question 1

Direction (Q. Nos. 1-16) This section contains 16 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q. The correct statement for the molecule Csl-3 is

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 1



Cs is electropositive element hence, it forms Cs+ ion. Hence, no covalent bonding.

Cs+ has stable inert gas configuration thus, Cs3+ is not formed.

Test: Ionic Bonds & Fajans' Rule - Question 2

Mg2+, 02- is formed when

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 2


Smaller the IE of Mg, easier the formation of Mg2+.
Greater the EA of O, easier the formation of O2-.

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Test: Ionic Bonds & Fajans' Rule - Question 3

Solubility of KCI is maximum in

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 3

KCI is an ionic compound. Force of attraction (F) between K+


Greater the value of D, smaller the value of F, hence larger the solubility.
Note D for CH3OCH3 is least out of given solvents, thus solubility is minimum.

Test: Ionic Bonds & Fajans' Rule - Question 4

Lewis structure of MgCI2 and Al2O3 are

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 4

Mg(1s2 2s2 2p63s2) is an electropositive element and Cl(1s2 2s2 2p6 3s2 3p5)is an electronegative element.

Mg2+ attains stable inert gas configuration.


Cl- also attains stable inert gas configuration.

Al3+ attains stable inert gas configuration. 

O2- attains stable inert gas configuration.

Test: Ionic Bonds & Fajans' Rule - Question 5

Lewis structure of the elements M and Z are shown below.

Compound formed is

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 5


(Three electrons extra than octet, it can donate three electrons)

(Three electrons short of octet, it can gain three electrons) 
Thus, M3+ Z3-
Compound is MZ,

Test: Ionic Bonds & Fajans' Rule - Question 6

Different kinds of bonds and interaction present within CuSO4 • 5H2O. They can be

I. σ-bond
II. π-bond
III. coordinate bond
IV. electrostatic force of attraction
V. H-bond due to dipole-dipole interaction
VI. H-bond due to ion-dipole interaction 

Select the correct types of bonds/interactions.

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 6





IV. Electrostatic force of attraction between  and
V. H2O molecules joined together by dipole-dipole interaction.
VI. Outer H2O molecule joined to by ion-dipole interaction.

Test: Ionic Bonds & Fajans' Rule - Question 7

Consider the following compounds

I. K4[Fe(CN)6]
II. NH4Cl
III. H2SO4
IV. [Ni (CO)4]

Q.
Ionic, covalent and coordinate bonds are present in 

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 7




Test: Ionic Bonds & Fajans' Rule - Question 8

Out of the following, maximum covalent nature is in

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 8

By Fajans rule,
Smaller the size of cation, larger the size of anion. Larger the charge on cation or anion, larger the polarising power and thus, greater the covalent nature.

Test: Ionic Bonds & Fajans' Rule - Question 9

Arrange NaCI, MgCI2, AICI3, SiCI4 in increasing solubility in ether (non-polar solvent).

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 9

Like dissolves like. Polar solvents (as H2O) dissolve polar compounds (ionic compounds) and non-polar solvents (as ether) dissolve non-polar compounds (covalent compounds).
By Fajans’ rule smaller the size of cation, larger the size of anion. Larger the charge on cation or anion then larger the polarising power and thus, larger the covalent nature and thus, solubility in non-polar solvents.


Test: Ionic Bonds & Fajans' Rule - Question 10

Covalent nature of NaF, Na2O and Na3N in the increasing order is

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 10

By Fajans' rule,
Smaller the size of cation, larger the size of anion. Larger the charge on cation or anion, larger the polarising power and thus, greater the covalent nature.
NaF, Na2O, Na3N, Na+ is same.


Test: Ionic Bonds & Fajans' Rule - Question 11

Solubility of CuS(l), ZnS (II) and Na2S (III) in water is in the order

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 11

By Fajans' rule,
Smaller the size of cation, larger the size of anion.
Larger the charge ord-electrons present, then larger the polarising power of cation and larger the covalent nature and smaller the solubility in water.

Cu2+ due to shielding effect causes more screening than Zn2+.
Thus, polarising power of Na+ <Zn2+ < Cu2+ hence, ionic nature and solubility in water is in order
CuS < ZnS < Na2S

Test: Ionic Bonds & Fajans' Rule - Question 12

Which pair is not in the correct order of lattice energy?

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 12

Based on Fajans' rule,
Smaller the size of cation, larger the size of anion, larger the charge,
d-f-Aelectrons present then larger the polarising power, larger the covalent nature and larger the lattice energy.


Test: Ionic Bonds & Fajans' Rule - Question 13

Most stable compound is

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 13

By Fajans’ rule,
Smaller the size of cation, larger the size of anion, larger the charge, then larger the polarising power, larger the covalent nature, means low melting point.
Smaller the ionic nature, means low melting point 


Hence, most stable is LiF.

Test: Ionic Bonds & Fajans' Rule - Question 14

Out of the following ions, which pair wil make the compound most covalent?

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 14

By Fajans’ rule, smaller the size of cation, larger the size of anion, larger the charge on cation or anion,
d- and f-electrons present on the ion then greater the polarising power of the ion to polarise the other ion, thus larger the covalent nature.


Thus, com pound with Si4+ (smallest cation) and C4- (largest anion) will be most covalent.

Test: Ionic Bonds & Fajans' Rule - Question 15

Aqueous solution of a mixture contains LiCI, CuCI, NaCI and AICI3. This is shaken with ether. What is/are left in water?

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 15

Polarising power of Li+ > Na+ and AI3+ > Na+
Thus, LiCI and AICI3 have covalent nature and extracted into ether.
CuCI due to d-electrons has covalent nature. (Polarising power of Cu+ > Na+).
Thus, only NaCI (ionic) remains in water.

Test: Ionic Bonds & Fajans' Rule - Question 16

Select the correct statement(s).

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 16

Cu+ = [Ar] 3d10, Cu+ has inner 18-electrons shell. Shielding effect is decreased. Thus, electronegativity is increased. This results in greater polarising power of Cu+ .
Hence, CuCI is covalent.

*Multiple options can be correct
Test: Ionic Bonds & Fajans' Rule - Question 17

Direction (Q. Nos. 17-20) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.

Q. Which of the following can be calculated based on Born-Haber cycle of formation of a lattice A+B- (s) from A(s) and 6 (g) ?

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 17

Born Haber cycle involves



*Multiple options can be correct
Test: Ionic Bonds & Fajans' Rule - Question 18

Which of the following Lewis acid-Lewis base interactions are associated with the expansion of octet?

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 18

(a) In SbF5 (expansion of octet from 5 to 10 and further to 12).
(b) NH3, BF3 both have complete octet. NH3 is electron rich, donates lone pair to BF3.
(c) AICI3 is electron deficient, accepts lone pair from Cl-.
(d) In SF4, expansion of octet from 8 to 10 and further to 12.

*Multiple options can be correct
Test: Ionic Bonds & Fajans' Rule - Question 19

An ionic compound A+B- is most like to be formed from A and B when 

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 19

A → A+ + e- is favourable since, IE of A is low.
B + e- → B- is favourable since, EA of B is high.
Thus, A+B- is favoured by low IE of A and high EA of B.

Test: Ionic Bonds & Fajans' Rule - Question 20

Consider following types of bonds.

I. Ionic bond
II. Covalent σ - bond
III. Coordinate bond
IV. Covalent π - bond  

Q.
All these bonds are present in 

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 20

Test: Ionic Bonds & Fajans' Rule - Question 21

Direction (Q. Nos. 21-22) This section contains  a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d)

A+B- (ionic) is formed in the following steps from gaseous atoms



Q. 
How much energy is required to transfer one electron from A to B? 

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 21



Test: Ionic Bonds & Fajans' Rule - Question 22

A+B- (ionic) is formed in the following steps from gaseous atoms

Q. What is the maximum size of Athat would lend itself for the formation of an energetically stable bond?

Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 22



*Answer can only contain numeric values
Test: Ionic Bonds & Fajans' Rule - Question 23

Direction (Q. Nos. 23) This section contains 1 question. when worked out will result in an integerfrom  to 9 (both inclusive)

Q.
An ionic bond is established between a positive ion A+ and negative ion B- . How many times strength of the ionic bond is affected by doubling the charge on A+ and making the radius halved?


Detailed Solution for Test: Ionic Bonds & Fajans' Rule - Question 23



Thus, ionic strength becomes 8 times.

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