Chemical Bonding & Molecular Structure — Page 7 | JEE Chemistry

Q61

The bond order and the magnetic characteristics of CN- are :

A 3, paramagnetic

B

2{1 \over 2}

, paramagnetic

C 3, diamagnetic

D

2{1 \over 2}

, diamagnetic

Correct Answer

Option C

Solution

CN- has 14 electrons. Moleculer orbital configuration of CN- is

{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}

Here is no unpaired electron so it is diamagnetic. Nb = 10 Na = 4

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 4} \right]

= 3

Q62

Arrange the following bonds according to their average bond energies in descending order : C–Cl, C–Br, C–F, C–I

A C–Br > C–I > C–Cl > C–F

B C–Cl > C–Br > C–I > C–F

C C–I > C–Br > C–Cl > C–F

D C–F > C–Cl > C–Br > C–I

Correct Answer

Option D

Solution

Bond length order in carbon halogen bonds are in the order of C – F

<

C – Cl

<

C – Br

<

C – I Hence, Bond energy order C – F

>

C – Cl

>

C – Br

>

C – I

Q63

If AB4 molecule is a polar molecule, a possible geometry of AB4 is

A Tetrahedral

B see-saw

C Square pyramidal

D Square planar

Correct Answer

Option B

Solution

Compound Electronic Geometry Moleculer Geometry or Moleculer Shape Lone Pair Polarity AB4 Tetrahedral Tetrahedral 0 Nonpolar AB4 Trigonal Bipyramidal see-saw 1 Polar AB4 Octahedral (Square Bipyramidal) Square Planar 2 Nonpolar Note : The difference between Electronic Geometry and Moleculer Geometry(Moleculer Shape) is that in Electronic Geometry we include lone pair but in Moleculer Geometry(Moleculer Shape) we do not include lone pair.

Here in this question it is not clear which geometry they are talking about, correct answer should be either Trigonal Bipyramidal or see-saw.

Q64

Match with : (Species) (Hybrid Orbitals)

List - I		List - II
(a)	$S{F_4}$	(i)	$s{p^3}{d^2}$
(b)	$I{F_5}$	(ii)	${d^2}s{p^3}$
(c)	$NO_2^ +$	(iii)	$s{p^3}d$
(d)	$NH_4^ +$	(iv)	$s{p^3}$
()		(v)	$sp$

A (a)-(i), (b)-(ii), (c)-(v) and (d)-(iii)

B (a)-(ii), (b)-(i), (c)-(iv) and (d)-(v)

C (a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)

D (a)-(iv), (b)-(iii), (c)-(ii) and (d)-(v)

Correct Answer

Option C

Solution

(a) SF4 - sp3d hybridization (b) IF5 - sp3d2 hybridization (c) NO

_2^ +

- sp hybridization (d) NH

_4^ +

- sp3 hybridization

Q65

The number of sp2 hybrid orbitals in a molecule of benzene is :

A 24

B 12

C 6

D 18

Correct Answer

Option D

Solution

Each carbon atom is sp2 hybrid $\therefore$ 3 sp2 hybrid orbitals are formed by each carbon atom Total sp2 orbitals = 6 × 3 = 18

Q66

The correct set from the following in which both pairs are in correct order of melting point is :

A LiCl > LiF ; NaCl > MgO

B LiCl > LiF ; MgO > NaCl

C LiF > LiCl ; NaCl > MgO

D LiF > LiCl ; MgO > NaCl

Correct Answer

Option D

Solution

Correct option is i.e.

LiF > LiCl; MgO > NaCl.

Melting point is directly proportional to lattice energy.

Lattice energy is the energy required to separate a mole of an ionic solid into gaseous ions.

It depends upon charge of ions and size of ions.

M.P. \propto L.E. \propto {{Charge} \over {Size}}

\begin{array}{ll}{Li \to + 1} & {Li \to + 1} \\ {F = - 1} & {Cl \to - 1} \end{array}

Both LiF and LiCl having same charge, so melting point will depend on size.

Larger the size of anion, lesser the lattice energy and hence, melting point order is LiF > LiCl.

Similarly,

\begin{array}{ll}{MgO} & {NaCl} \\ {Mg \to 2 + } & {Na \to 1 + } \\ {O \to 2 - } & {Cl \to 1 - } \end{array}

MgO having + 2 charge which is greater than NaCl (+ 1) charge.

So, greater the charge on the ions greater will be lattice energy and hence, melting point order is MgO > NaCl.

Q67

The correct shape and

I - I - I

bond angles respectively in

I_3^ -

ion are :

A Linear; 180

^\circ

B T-shaped; 180

^\circ

and 90

^\circ

C Trigonal planar; 120

^\circ

D Distorted trigonal planar; 135

^\circ

and 90

^\circ

Correct Answer

Option A

Solution

Hybridisation of central

I

in

I_3^ -

is sp3d with 3 lone pair and 2 bond pair. Shape Linear Lone pair 3 lone pair Bond angle 180

^\circ

(for linear molecule)

Q68

According to molecular orbital theory, the species among the following that does not exist is :

A

{O_2}^{2 - }

B

B{e_2}

C

H{e_2}^ -

D

H{e_2}^ +

Correct Answer

Option B

Solution

Note : According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.

We know, Bond order

= {1 \over 2}

[Nb $-$ Na] Nb = No of electrons in bonding molecular orbital Na $=$ No of electrons in anti bonding molecular orbital (4)

\,\,\,\,

upto 14 electrons, molecular orbital configuration is Here Na = Anti bonding electron $=$ 4 and Nb = 10 (5)

\,\,\,\,

After 14 electrons to 20 electrons molecular orbital configuration is - - - Here Na = 10 and Nb = 10 Molecular orbital configuration of O

_2^{2 - }

(18 electrons) is

{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ *

\therefore\,\,\,\,

Nb = 10 Na = 8

\therefore\,\,\,\,

BO =

{1 \over 2}

[ 10 $-$ 8] = 1

\,\,\,

Configuration of

He_2^ +

(3 electrons) is =

{\sigma _{1{s^2}}}\,\,\sigma _{1{s^1}}^ *

\therefore\,\,\,

Bond order =

{1 \over 2}

(2 $-$ 1) = 0.5

\,\,\,

Configuration of

He_2^ -

(5 electrons) is =

{\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * {\sigma _{2{s^1}}}\,

\therefore\,\,\,

Bond order =

{1 \over 2}

(3 $-$ 2) = 0.5

\,\,\,

Configuration of

Be_2

(4 electrons) is =

{\sigma _{1{s^2}}}

\sigma _{1{s^2}}^ *

$\therefore$ Bond order =

{1 \over 2}

(2 $-$ 2) = 0

Q69

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Dipole-dipole interactions are the only non-covalent interactions, resulting in hydrogen bond formation. Reason R : Fluorine is the most electronegative element and hydrogen bonds in HF are symmetrical. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both A and R are true and R is the correct explanation of A

B A is true but R is false

C A is false but R is true

D Both A and R are true but R is NOT the correct explanation of A

Correct Answer

Option C

Solution

Dipole - Dipole are not only the interaction responsible for hydrogen bond formation.

Ion-dipole can also be responsible for hydrogen bond formation.

F is most electronegative element and anhydrous HF in solid phase has symmetrical hydrogen bonding.

Q70

Given below are two statements : one is labeled as Assertion A and the other is labelled as Reason R : Assertion A : The H

-

O

-

H bond angle in water molecule is 104.5

^\circ

. Reason R : The lone pair - lone pair repulsion of electrons is higher than the bond pair - bond pair repulsion. In the light of the above statements, choose the correct answer from the options given below :

A Both A and R are true, and R is the correct explanation of A

B A is true but R is false

C A is false but R is true

D Both A and R are true, but R is not the correct explanation of A

Correct Answer

Option A

Solution

due to lp - lp repulsion bond angle decrease.