Chemical Bonding & Molecular Structure — Page 9 | NEET Chemistry

Q81

In a regular octahedral molecule, MX 6 the number of X

-

M

-

X bonds at 180 o is

A three

B two

C six

D four

Correct Answer

Option A

Solution

The MX 6 molecule with regular octahedral geometry is as follows : Thus, there are three X—M—X bonds with bond angle 180 o .

Q82

Which one of the following statements is not correct for sigma- and pi- bonds formed between two carbon atoms?

A sigma-bond is stronger than a pi-bond.

B Bond energies of sigma- and pi-bonds are of the order of 264 kJ/mol and 347 kJ/mol, respectively

C Free rotation of atoms about a sigma-bond is allowed but not in case of a pi-bond.

D Sigma-bond determines the direction between carbon atoms but a pi-bond has no primary effect in this regard.

Correct Answer

Option B

Solution

As sigma bond is stronger than the $\pi$ (pi) bond, so it must be having higher bond energy than $\pi$ (pi) bond.

Q83

In NO 3

-

ion number of bond pair and lone pair of electrons on nitrogen atom are

A 2, 2

B 3, 1

C 1, 3

D 4, 0

Correct Answer

Option D

Solution

{O^ - }\hbox{---}\mathop N\limits^{\mathop \uparrow \limits^O } = O

In NO 3 - ion, nitrogen has 4 bond pair of electrons and no lone pair of electrons.

Q84

Which of the following has p

\pi

-

d

\pi

bonding?

A NO 3

-

B SO 3 2

-

C BO 3 3

-

D CO 3 2

-

Correct Answer

Option B

Solution

Electronic structure of S atom in excited state In 'S' unhybride d- orbital is present, which will involved in bond formation with oxygen atom.

In oxygen two unpaired p- orbital is present in these one is involved in $\sigma$ bond formation while other is used in $\pi$ bond formation.

So, in SO 3 2 $-$ , p $\pi$ $-$ d $\pi$ bonding present.

Q85

Which of the following is isoelectronic?

A CO 2 , NO 2

B NO 2

-

, CO 2

C CN

-

, CO

D SO 2 , CO 2

Correct Answer

Option C

Solution

In CO, the number of electrons $\Rightarrow$ 6 + 8 = 14 [ Z of C = 6 and O = 8] Electronic configuration of molecular orbital of CO :

{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}

CN - also get (6 + 7 + 1) 14 electrons and the configuration is similar to that CO. So, CN and CO is isoelectronic.

Q86

In X - H --- Y, X and Y both are electronegative elements. Then

A electron density on X will increase and on H will decrease

B in both electron density will increase

C in both electron density will decrease

D on X electron density will decrease and on H increases.

Correct Answer

Option A

Solution

As X and Y both are electronegative elements thus both attracts the electron density from H thus, electron density on H decreases and on X it increases.

Q87

In which of the following bond angle is maximum ?

A NH 3

B NH 4 +

C PCl 3

D SCl 2

Correct Answer

Option B

Solution

Bond angle is maximum in NH 4 + tetrahedral molecule with bond angle 109°.

Q88

Which of the following two are isostructural?

A XeF 2 , IF 2

-

B NH 3 , BF 3

C CO 3 2

-

, SO 3 2

-

D PCl 5 , ICl 5

Correct Answer

Option A

Solution

Compounds with same shape and same hybridisation are known as isostructural.

XeF 2 , IF 2 - $\to$ both are sp 3 hybridised linear molecules.

Q89

Nitrogen forms N 2 , but phosphorus does not form P 2 , however, it converts P 4 , reason is

A triple bond present between phosphorus atom

B p

\pi

-

p

\pi

bonding is weak

C p

\pi

-

p

\pi

bonding is strong

D multiple bonds form easily.

Correct Answer

Option B

Solution

For strong $\pi$ -bonding, p $\pi$ - p $\pi$ bonding should be strong.

In case of P, due to larger size as compared to N-atom, p $\pi$ - p $\pi$ bonding is not so strong.

Q90

Right order of dissociation energy N 2 and N 2 + is

A N 2 > N 2 +

B N 2 = N 2 +

C N 2 + > N 2

D none

Correct Answer

Option A

Solution

N_2

has 14 electrons. Moleculer orbital configuration of

N_2

=

{\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\,\,\,\,

N b = 10 N a = 4 $\therefore$ BO =

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

= 3 N 2 + has 13 electrons. Moleculer orbital configuration of N 2 + =

{\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\,\,\,\,

N b = 9 N a = 4 $\therefore$ BO =

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

= 2.5 As the bond order in N 2 is more than N 2 + so the dissociation energy of N 2 is higher than N 2 + .