Chemical Bonding & Molecular Structure — Page 15 | JEE Chemistry

Q141

Which of the following is paramagnetic ?

A NO+

B CO

C

O_2^{2 - }

D B2

Correct Answer

Option D

Solution

Those species which have unpaired electrons are called paramagnetic species. (a) NO+ has 14 electrons.

Moleculer orbital configuration of NO+ 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}}

Here is no unpaired electron, So it is Diamagnetic. (b) CO has 14 electrons. Moleculer orbital configuration of CO 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. (c)

O_2^{2 - }

has 18 electrons. Moleculer orbital configuration of

O_2^{2 - }

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}^ *

Here also no unpaired electron present, so it is diamagnetic. (d) B2 has 10 electrons.

Molecular orbital configuration of B2 is

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

Here two unpaired electrons present. So it is paramagnetic.

Q142

Among the following species, the diamagnetic molecule is

A CO

B B2

C O2

D NO

Correct Answer

Option A

Solution

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species. (a) CO has 14 electrons.

Moleculer orbital configuration of CO 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. (b) B2 has 10 electrons. Molecular orbital configuration of B2 is

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

Here two unpaired electrons present. So it is paramagnetic. (c)

O_2

has 16 electrons. Moleculer orbital configuration of

O_2

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^1}^ * = \,\,\pi _{2p_y^1}^ *

Here 2 unpaired electron present, so it is paramagnetic. (d) NO has 15 electrons.

Moleculer orbital configuration of NO 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^1}^ * = \,\,\pi _{2p_y^0}^ *

Here is 1 unpaired electron, So it is Paramagnetic.

Q143

Among the following molecules / ions,

C_2^{2 - },N_2^{2 - },O_2^{2 - },{O_2}

which one is diamagnetic and has the shortest bond length?

A

{O_2}

B

C_2^{2 - }

C

N_2^{2 - }

D

O_2^{2 - }

Correct Answer

Option B

Solution

Note : (1)

\,\,\,\,

Bond strength $\propto$ Bond order (2)

\,\,\,\,

Bond length $\propto$

{1 \over {Bond\,\,order}}

(3)

\,

Bond order

= {1 \over 2}

[Nb $-$ Na] Nb = No of electrons in bending 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 (A) In O atom 8 electrons present, so in O2, 8 $\times$ 2 = 16 electrons present.

Then in

O_2^{2 - }

no of electrons = 18

\therefore\,\,\,\,

Molecular orbital configuration of O2 (16 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^1}^ * \,\, = \pi _{2p_y^1}^ *

\therefore\,\,\,\,

Na = 6 Nb = 10

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 6} \right] = 2

Here 2 unpaired electrons are present so it is paramagnetic. (D) 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 Here no unpaired electrons are present so it is diamagnetic. (B)

C_2^{2 - }

has 14 electrons. Moleculer orbital configuration of

C_2^{2 - }

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}}

\therefore\,\,\,\,

Na = 4 Nb = 10

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 4} \right] = 3

Here no unpaired electron present, so it is diamagnetic. (C)

N_2^{2 - }

has 16 electrons. Moleculer orbital configuration of

N_2^{2 - }

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

\therefore\,\,\,\,

Na = 6 Nb = 10

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 6} \right] = 2

Here 2 unpaired electron present, so it is paramagnetic. As Bond length $\propto$

{1 \over {Bond\,\,order}}

so among two diamagnetic ions

C_2^{2 - }

and

O_2^{2 - }

, bond order of

C_2^{2 - }

is more so it will have shorter bond length.

Q144

According to MO theory the bond orders for

\mathrm{O}

_2^{2 - }

,

\mathrm{CO}

and

\mathrm{NO^+}

respectively, are

A 1, 3 and 3

B 2, 3 and 3

C 1, 2 and 3

D 1, 3 and 2

Correct Answer

Option A

Solution

.tg .tg Species B.O. O

_2^{2 - }

1 CO 3 NO

^ \oplus

3 Note : (1)

\,

Bond order

= {1 \over 2}

[Nb $-$ Na] Nb = No of electrons in bending 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 (A) In O atom 8 electrons present, so in O2, 8 $\times$ 2 = 16 electrons present.

Then in

O_2^{2 - }

no of electrons = 18 $\therefore$ 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 (B) CO has 14 electrons. Moleculer orbital configuration of CO 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}}

$\therefore$ Nb = 10 Na = 4

\therefore\,\,\,\,

BO =

{1 \over 2}

[ 10 $-$ 4] = 3 (C) NO+ has 14 electrons. Moleculer orbital configuration of NO+ 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}}

$\therefore$ Nb = 10 Na = 4

\therefore\,\,\,\,

BO =

{1 \over 2}

[ 10 $-$ 4] = 3

Q145

In which of the following process, the bond order has increased and paramagnetic character has charged to diamagnetic ?

A NO

\to

NO+

B N2

\to

N2+

C O2

\to

O2+

D O2

\to

O22

-

Correct Answer

Option A

Solution

Molecular orbital configuration of NO (15 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^1}^ * \, = \,\pi _{2p_y^o}^ *

\therefore\,\,\,\,

Nb = 10 Na = 5

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 5} \right]

= 2.5 And in NO one unpaired electron is present , so it is paramagnetic.

Similarly Molecular orbital configuration of NO+ (14 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}}\,

\therefore\,\,\,\,

Nb = 10 Na = 4

\therefore\,\,\,\,

BO =

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

= 3 And in NO+ no unpaired electron is present , so it is diamagnetic.

Q146

Match items of with those of : (Property) (Example)

List - I		List - II
(a)	Diamagnetism	(i)	MnO
(b)	Ferrimagnetism	(ii)	${O_2}$
(c)	Paramagnetism	(iii)	NaCl
(d)	Antiferromagnetism	(iv)	$F{e_3}{O_4}$ Choose the most appropriate answer from the options given below :

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

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

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

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

Correct Answer

Option C

Solution

A.

NaCl is diamagnetic because all electrons are paired in Na+ and in Cl− .

So, it shows diamagnetism.

B.

Fe3O4 is ferrimagnetic because of the presence of the unequal alignment of magnetic moment in opposite direction.

C.

O2 molecule has two unpaired electrons.

So, it is paramagnetic.

D.

In MnO, the orientation of the electrons of Mn and O is such that they cancel their effects, hence it is antiferromagnetic.

Q147

Identify the incorrect statement for PCl5 from the following.

A In this molecule, orbitals of phosphorous are assumed to undergo sp3d hybridization.

B The geometry of PCl5 is trigonal bipyramidal.

C PCl5 has two axial bonds stronger than three equatorial bonds.

D The three equatorial bonds of PCl5 lie in a plane.

Correct Answer

Option C

Solution

PCl5 All three equatorial bonds in a plane sp3d hybridization Trigonal bipyramidal Axial bonds are weaker than equatorial bonds.

Q148

The correct order of increasing intermolecular hydrogen bond strength is :

A HCN 2O 3

B HCN 4 3

C CH4 3

D CH4 3 < HCN

Correct Answer

Option C

Solution

Due to the high difference in electronegativity of H and N the H-bond strength of NH3 is highest.

There is no H-bond in CH4.

CH4 3

Q149

Which of the following compounds contain(s) no covalent bond(s)? KCl, PH3, O2, B2H6, H2SO4

A KCl, B2H6

B KCl, B2H6, PH3

C KCl, H2SO4

D KCl

Correct Answer

Option D

Solution

Here only KCl is ionic compound all other compound has covalent bond.

Q150

Match with . (Molecule) (Bond order)

List - I		List - II
(a)	$N{e_2}$	(i)	1
(b)	${N_2}$	(ii)	2
(c)	${F_2}$	(iii)	0
(d)	${O_2}$	(iv)	3

A (a)

\to

(i), (b)

\to

(ii), (c)

\to

(iii), (d)

\to

(iv)

B (a)

\to

(iv), (b)

\to

(iii), (c)

\to

(ii), (d)

\to

(i)

C (a)

\to

(iii), (b)

\to

(iv), (c)

\to

(i), (d)

\to

(ii)

D (a)

\to

(ii), (b)

\to

(i), (c)

\to

(iv), (d)

\to

(iii)

Correct Answer

Option C

Solution

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 (a) Molecular orbital configuration of Ne2 (20 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}^ *

\sigma _{2p_z^2}^*

\therefore\,\,\,\,

Na = 10 Nb = 10

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 10} \right] = 0

(Note : All inert gases has BO = 0 and it does not exist as molecule.

Here Ne is also an inert gas.) (b) Moleculer orbital configuration of

N_2

(14 electrons) 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}}

\therefore\,\,\,\,

Na = 4 Nb = 10

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 4} \right] = 3

(d) Molecular orbital configuration of F2 (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\,\,\,\,

Na = 8 Nb = 10

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 8} \right] = 1

(d) Molecular orbital configuration of O2 (16 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^1}^ * \,\, = \pi _{2p_y^1}^ *

\therefore\,\,\,\,

Na = 6 Nb = 10

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 6} \right] = 2