Revisal Problems (Past 13 Years) JEE Advanced (Chemical Bonding) - JEE MCQ

CN^- and CO are strong ligands. Thus, unpaired electrons in Ni or Ni²⁺ are paired. H₂O is a week ligand, thus unpaired electrons are not affected.
Ni [Ar]



Note There is no unpaired electron in [Ni(CO)₄]
— diamagnetic
Also, [Ni (CN)₄]^2- - diamagnetic but [Ni(H₂O)4] has unpaired electrons. Thus, it is paramagnetic.

Based on Band-theory of metallic bonding, as energy increases, conducting nature decreases and element changes from conductor to insulator.

(p - Ip) and (Ip - bp) repulsions are m inim ised by taking lone-pairs on equatorial position.

In PF₅, there are two types of angles thus has two types of bond lengths.
In SF₆, there is only one type of bond angle thus, only one type of bond.
There is no lone pair on P in PF₅.
Thus, statement I is correct and statement II is incorrect.

Lone pairs on Cl are at the equatorial positions to minimise repulsion.

Two F-atoms are at the vertices thus linear structure.
Two lone pairs on Cl—sp³ hybridisation thus bent.
Thus, statement I is correct but statement II is incorrect.

C,Si-—group 14. Thus, Statement I is correct larger size of Si results in poor sideway overlap of s-orbitals to form π-bond.
Thus, both statements I and II are correct and statement II is the correct explanation of statement I.

Maleic acid (cis-isomer) is more polar than its frans-isomer. Hence, maleic acid is more ionised than fumaric acid.
(b) Maleate ion (I) is stable due to intramolecular H-bonding. Hence, of maleic acid is smaller than that of fumaric acid in which intramolecular H-bonding is absent (II).

= First ionisation constant
= Second ionisation constant 


Nickel dimethyl glyoximate complex is stable due to intramolecular H-bonding.

By the coulombic force of attraction (F) between cation and anion is

z₁ = z₂ = (+ 1 or -1) charges on cation or anion.
r = internuclear distance
D = Dielectric constant (same for H₂O solvent).
If F is small, ions goes into solution and thus makes salt soluble in water.

Larger (r) means smaller (F).
(a) r increases hence solubility increases true.
(b) as in (a) true.
(c) If one of the ions is very big (as Cs), the solubility decreases with increasing size of the other ion.
F < C I < Br < I,
size
Thus, Csl < CsBr < ,CsCI < CsF
Thus, true, solubility
(d) NO₃^- ion is larger.
size of Li⁺ < Na⁺ < K⁺ < Cs⁺
Hence, solubility decreases with increase in size of M⁺
CsNO₃ < KNO₃ < NaNO₃ < LiNO₃^- true.

Based on MO theory
(a) C₂ (12 electrons) - all paired electrons
(b) N₂ (14 electrons) - all paired electrons
(c) O₂ (16 electrons) - two electrons unpaired
(d) S₂ (32 electrons) - two electrons unpaired
Thus, C₂ or N₂ are diamagnetic.

N = Number of unpaired electrons


Bond energy as bond order magnetic moment  

(a) l₃^- sp³d
(b) PCl₃ sp³
(c) XeF₂ sp³d
(d) SF₄ sp³d

(a) Smaller the energy gap, larger the conduction.
(b) Unpaired electrons in graphite is responsible for conduction used in bonding in hexagonal layer.


Bonding electrons = 4
Antibonding electrons = 2

Thus, Li₂ is stable.
(d) Diamond is non-conducting graphite is conducting.

due to one lone pair


It has three (C—H) α-bonds and singly occupied p-orbital. Thus, sp².

sp² (one lone pair is a part of hybrid orbital).


(one unpaired electron is a part of hybrid orbtial).


 (one lone pair is a part of hybrid orbital)

(a) 
(B—H) bond has only one electron thus B is electron deficient.
(b) Octet of B is complete.
(c) Octet of B is complete.

Octet of B is incomplete.

In IF₇, there are five equatorial bonds of equal bond length. 

P-atoms are sp³d -hybridised.

A, B and C are inclined at 120° with one each other form equatorial bonds. D and E at 90° with (ABC) plane. D is above and E is below this plane forming axial bonds. Axial bonds suffer more repulsive interaction from equatorial bonds, hence axial bonds are longer and sligtly weaker than the equatorial bonds.

sp²-hybridised carbon bond angle =120°

sp³-hybridised carbon bond angle < 109°28'

Unpaired electron is in unhybridised p₂-orbital.

Thus sp²-hybridised bond angle = 120°.

Thus (b), (c)

sp²-hybridised carbon bond angle =120°

sp³-hybridised carbon bond angle < 109°28'

Unpaired electron is in unhybridised p₂-orbital.

Thus sp²-hybridised bond angle = 120°.

​Thus (b), (c)

Each N-atom has one lone pair and three α-bonds —sp³ hybridised.

(Ip - bp) repulsion makes it. Trigonal pyramidal.
Y : N[Si(CH₃)₃]₃ Lone pair on N-atom is transferred to the empty d-orbitals of silicon (pπ - dπ overlapping) and its Lewis base character is lost.
Thus, y has trigonal planar structure.

as in (X) trigonal pyramidal.

Each N-atom has one lone pair and three α-bonds —sp³ hybridised.

(Ip - bp) repulsion makes it. Trigonal pyramidal.
Y : N[Si(CH₃)₃]₃ Lone pair on N-atom is transferred to the empty d-orbitals of silicon (pπ - dπ overlapping) and its Lewis base character is lost.
Thus, y has trigonal planar structure.

as in (X) trigonal pyramidal.

O₂^- undergoes disproportionation.

Cr-metal ion is bridged and has tetrahedral structure, (ii) → (r)

MnO₄^- is reduced.
NO₃^- is oxidised thus a redox reaction.
NO₃^- has triangular planar structure.

Fe²⁺ is oxidised and NO₃^- is reduced thus redox reaction.
Thus, (iv) → 4 (p)

(i) B₂ (10 electrons)

Two unpaired electrons-paramagnetic.
Bonding electrons = 6
Anit-bonding electrons = 4


(B— B) bond is formed by mixing of s and p-orbitals of valence shell.

Thus, (i) → 4 (p, q, r, t)
(ii) N₂ (14 electrons)

No unpaired electron - diamagnetic
Bonding electrons = 10
Anti-bonding electrons = 4



One unpaired electrons thus paramagnetic Bonding electrons = 10
Anti-bonding electrons = 7



(iv) O₂ (16 electrons)
There are two electrons in anti-bonding
Thus, it is paramagnetic.
Bond order = 2

m (HF) of single unit = 20 gL^-1
Thus, number of HF molecules = 78/20 = 4

Distance between A and X = 4 units
Thus, distance between two X -atoms = 4 x 2 = 8 units.

Na (11):1s²2s²2p⁶3s¹
Electron in (3s) orbital is a conduction band , Each orbital has one electron, thus one energy level.


= 7.0 x 10²⁰ atoms
= 7.0 x 10²⁰ conduction bands. 
Thus,

n = 7