Chemical Bonding & Molecular Structure — Page 4 | JEE Chemistry

Q31

Which one of the following species is diamagnetic in nature?

A

He_2^+

B H2

C

H_2^+

D

H_2^-

Correct Answer

Option B

Solution

TIPS/Formulae : A diamagnetic substance contains no unpaired electron.

{H_2}

is diamagnetic as it contains all paired electrons

\mathop {{H_2} = \sigma _b^2}\limits_{\left( {diamagnetic} \right)} \,\,,\,\,\mathop {H_2^ + = \sigma _b^1,}\limits_{\left( {paramagnetic} \right)} \,\,\mathop {H_2^ - = \sigma _b^2,}\limits_{\left( {paramagnetic} \right)}

\mathop {\sigma _a^{ * 1};He_2^ + }\limits_{\left( {paramagnetic} \right)} = \mathop {\sigma _b^2,\sigma _a^{ \ne 1}}\limits_{\left( {paramagnetic} \right)}

Q32

Lattice energy of an ionic compounds depends upon

A Charge on the ion only

B Size of the ion only

C Packing of ions only

D Charge on the ion and size of the ion

Correct Answer

Option D

Solution

The lattice energy of an ionic compound is a measure of the strength of the bonds in that ionic compound.

Specifically, it is the energy required to separate one mole of an ionic solid into its constituent ions in the gaseous state.

The lattice energy depends on several factors, primarily the charge on the ions and the size (or radius) of the ions.

1.

Charge on the ions: The lattice energy is directly proportional to the product of the charges of the cation and the anion.

Higher charges result in stronger electrostatic attraction between the ions, thereby increasing the lattice energy.

This relationship can be derived from Coulomb's law, which states that the force of attraction between two charged particles is directly proportional to the product of their charges.

Mathematically:

U \propto \frac{q_1 \cdot q_2}{r}

where $U$ is the lattice energy, $q_1$ and $q_2$ are the charges on the ions, and $r$ is the distance between the centers of the ions.

2.

Size of the ions: The lattice energy is inversely proportional to the distance between the ions, which depends on the sum of their ionic radii.

Smaller ions can come closer together, increasing the electrostatic attractions between them, thereby increasing the lattice energy.

Larger ions, being further apart, result in a lower lattice energy.

Considering these two factors, the correct choice is: Option D: Charge on the ion and size of the ion

Q33

Which of the following molecules/ions does not contain unpaired electrons?

A

O_2^{2−}

B B2

C

N_2^+

D O2

Correct Answer

Option A

Solution

(A) 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}^ *

So O

_2^{2 - }

has no unpaired electrons. (B) Molecular orbital configuration of B2 (10 electrons) 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 in B2, 2 unpaired electrons present. (C) Moleculer orbital configuration of

N_2^{ + }

(13 electrons) =

{\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^1}}

Here in

N_2^{ + }

, 1 unpaired electron present. (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}^ *

So O

_2

has 2 unpaired electrons.

Q34

The decreasing values of bond angles from NH3 (106o) to SbH3 (91o ) down group-15 of the periodic table is due to

A increasing bp-bp repulsion

B increasing p-orbital character in sp3

C decreasing lp-bp repulsion

D decreasing electronegativity

Correct Answer

Option C

Solution

Nitrogen(N) atom is smaller in size so it's lone pair is relatively unstable.

That is why repulsion between lone pairs present on nitrogen atom and bonded pairs of electrons is relatively high.

But size of Sb is more so its lone pair is relatively stable.

That is why it does not want to perticipate in bond pair lone pair repulsion and decreases lp-bp repulsion.

Q35

The maximum number of 90° angles between bond pair of electrons is observed in

A dsp3 hybridization

B sp3d2 hybridization

C dsp2 hybridization

D sp3d hybridization

Correct Answer

Option B

Solution

Here eight 90o angles between bond pair and bond pair.

Those angles are $\angle$ 1M2, $\angle$ 2M3, $\angle$ 3M4, $\angle$ 4M1, $\angle$ 5M1, $\angle$ 5M2, $\angle$ 5M3, $\angle$ 5M4.

Here twelve 90o angles between bond pair and bond pair.

Those angles are $\angle$ 1M2, $\angle$ 2M3, $\angle$ 3M4, $\angle$ 4M1, $\angle$ 5M1, $\angle$ 5M2, $\angle$ 5M3, $\angle$ 5M4, $\angle$ 6M1, $\angle$ 6M2, $\angle$ 6M3, $\angle$ 6M4.

Here four 90o angles between bond pair and bond pair.

Those angles are $\angle$ 1M2, $\angle$ 2M3, $\angle$ 3M4, $\angle$ 4M1.

Here six 90o angles between bond pair and bond pair.

Those angles are $\angle$ 1M3, $\angle$ 1M4, $\angle$ 1M5, $\angle$ 2M3, $\angle$ 2M4, $\angle$ 2M5.

Q36

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizinig order of the polarizing power of the cationic species, K+, Ca2+, Mg2+, Be2+?

A Mg2+ < Be2+ < K+ < Ca2+

B K+ < Ca2+ < Mg2+ < Be2+

C Be2+ < K+ < Ca2+ < Mg2+

D Ca2+ < Mg2+ < Be2+ < K+

Correct Answer

Option B

Solution

As charge/size ratio of a cation determines its polarizing power so high charge and small size of the cations increases polarisation.

As the size of the given cations decreases as K+ > Ca2+ > Mg2+ > Be2+ Hence, polarising power decreases as K+ < Ca2+ < Mg2+ < Be2+

Q37

The bond dissociation energy of B - F in BF3 is 646 kJ mol-1 whereas that of C - F in CF4 is 515 kj mol-1. The correct reason for higher B - F bond dissociation energy as compared to that of C - F is

A stronger

\sigma

bond between B and F in BF3 as compared to that between C and F in CF4

B significant

p\pi - p\pi

interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4

C lower degree of

p\pi - p\pi

interaction between B and F in BF3 than that between C and F in CF4

D smaller size of B - atom as compared to that of C- atom.

Correct Answer

Option B

Solution

In BF3, B is sp2 hybridised and by Back Bonding method strong p $\pi$ -p $\pi$ bond is created between filled p-orbital of F and vacant p-orbital of B which leads to shortening of B–F bond length which results in higher bond dissociation energy of the B–F bond.

However in CF4, C does not have any vacant p-orbitals to undergo $\pi$ -bonding.

Q38

Using MO theory, predict which of the following species has the shortest bond length?

A

O_2^+

B

O_2^-

C

O_2^{2-}

D

O_2^{2+}

Correct Answer

Option D

Solution

Note : (1)

\,\,\,\,

Bond length $\propto$

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

(2)

\,

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 In O atom 8 electrons present, so in O2, 8 $\times$ 2 = 16 electrons present.

Then in

O_2^ +

no of electrons = 15 in

O_2^ -

no of electrons = 17 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

Molecular orbital configuration of O

_2^ +

(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 Molecular orbital configuration of O

_2^ {2+ }

(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}}\,\pi _{2p_x^0}^ * \, = \,\pi _{2p_y^o}^ *

\therefore\,\,\,\,

Nb = 10 Na = 4

\therefore\,\,\,\,

BO =

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

= 3 Molecular orbital configuration of

O_2^ -

(17 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^1}^ *

\therefore\,\,\,\,

Nb = 10 Na = 7

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 7} \right]

= 1.5 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 As Bond length $\propto$

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

So

O_2^{2+}

has the shortest bond length.

Q39

In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed?

A

C_2 \to C_2^+

B

N_2 \to N_2^+

C

NO \to NO^+

D

O_2 \to O_2^+

Correct Answer

Option C

Solution

(A) Moleculer orbital configuration of

C_2

(12 electrons) =

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

\therefore\,\,\,\,

Na = 4 Nb = 8

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {8 - 4} \right] = 2

Here no unpaired electron present, so it is diamagnetic. Moleculer orbital configuration of

C_2^+

(11 electrons) =

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

\therefore\,\,\,\,

Na = 4 Nb = 7

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {7 - 4} \right] = 1.5

Here 1 unpaired electron present, so it is paramagnetic. (B) Moleculer orbital configuration of

N_2

(14 electrons) =

{\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. Moleculer orbital configuration of

N_2^+

(13 electrons) =

{\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^1}}

\therefore\,\,\,\,

Na = 4 Nb = 9

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {9 - 4} \right] = 2.5

Here 1 unpaired electron present, so it is paramagnetic. (C) Moleculer 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^0}^ *

\therefore\,\,\,\,

Na = 5 Nb = 10

\therefore\,\,\,\,

BO =

{1 \over 2}\left[ {10 - 5} \right] = 2.5

Here is 1 unpaired electron, So it is paramagnetic. Moleculer 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}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ *

\therefore\,\,\,\,

Na = 4 Nb = 10

\therefore\,\,\,\,

BO =

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

Here is no unpaired electron, So it is diamagnetic. (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

Here 2 unpaired electrons present, so it is paramagnetic. Molecular orbital configuration of O

_2^ +

(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 Here 1 unpaired electrons present, so it is also paramagnetic.

Q40

Which of the following hydrogen bonds is the strongest?

A O−H…….N

B F−H…….F

C O−H…….O

D O−H…….F

Correct Answer

Option B

Solution

Among F, O and N, F is most electronegative so F pulls bond pair of electron in F - H towards itself and develops highly positive charge on H atom.

This highly positive charged H atom creates stongest hydrogen bonding by taking lone pair of electron form electronegative atom F/N/O.

Hence among the given options F – H ......

F is the strongest bond.

Note : F – H ......

N has strongest hydogen bonding among F – H ......

F, F – H ......

O and F – H ......

N because N is least electronegative among F, O and N and can easily donate lone pair of electron.