Chemical Bonding & Molecular Structure — Page 2 | NEET Chemistry

Q11

Amongst the following, the total number of species NOT having eight electrons around central atom in its outer most shell, is

\mathrm{NH}_{3}, \mathrm{AlCl}_{3}, \mathrm{BeCl}_{2}, \mathrm{CCl}_{4}, \mathrm{PCl}_{5}

:

A 2

B 4

C 1

D 3

Correct Answer

Option D

Solution

Total number of species = 3

Q12

The correct order of energies of molecular orbitals of

\mathrm{N}_{2}

molecule, is

A

\sigma 1 \mathrm{~s} < \sigma^{*} 1 \mathrm{~s} < \sigma 2 \mathrm{~s} < \sigma^{*} 2 \mathrm{~s} < \sigma 2 \mathrm{p}_{\mathrm{z}} < \left(\pi 2 \mathrm{p}_{\mathrm{x}}=\pi 2 \mathrm{p}_{\mathrm{y}}\right) < \left(\pi^{*} 2 \mathrm{p}_{\mathrm{x}}=\pi^{*} 2 \mathrm{p}_{\mathrm{y}}\right) < \sigma^{*} 2 \mathrm{p}_{\mathrm{z}}

B

\sigma 1 \mathrm{~s} <\sigma^{*} 1 \mathrm{~s} <\sigma 2 \mathrm{~s} < \sigma^{*} 2 \mathrm{~s} < \sigma 2 \mathrm{p}_{\mathrm{z}} < \sigma^{*} 2 \mathrm{p}_{\mathrm{z}} < \left(\pi 2 \mathrm{p}_{\mathrm{x}}=\pi 2 \mathrm{p}_{\mathrm{y}}\right) < \left(\pi^{*} 2 \mathrm{p}_{\mathrm{x}}=\pi^{*} 2 \mathrm{p}_{\mathrm{y}}\right)

C

\sigma 1 \mathrm{~s} <\sigma^{*} 1 \mathrm{~s} <\sigma 2 \mathrm{~s} < \sigma^{*} 2 \mathrm{~s} < \left(\pi 2 \mathrm{p}_{\mathrm{x}}=\pi 2 \mathrm{p}_{\mathrm{y}}\right) < \left(\pi^{*} 2 \mathrm{p}_{\mathrm{x}}=\pi^{*} 2 \mathrm{p}_{\mathrm{y}}\right) < \sigma 2 \mathrm{p}_{\mathrm{z}} < \sigma^{*} 2 \mathrm{p}_{\mathrm{z}}

D

\sigma 1 \mathrm{~s} <\sigma^{*} 1 \mathrm{~s} <\sigma 2 \mathrm{~s} < \sigma^{*} 2 \mathrm{~s} < \left(\pi 2 \mathrm{p}_{\mathrm{x}}=\pi 2 \mathrm{p}_{\mathrm{y}}\right) < \sigma 2 \mathrm{p}_{\mathrm{z}} < \left(\pi^{*} 2 \mathrm{p}_{\mathrm{x}}=\pi^{*} 2 \mathrm{p}_{\mathrm{y}}\right) < \sigma^{*} 2 \mathrm{p}_{\mathrm{z}}

Correct Answer

Option D

Solution

Molecular orbital (energy) diagram / sequence of N

_2

Q13

Taking stability as the factor, which one of the following represents correct relationship?

A InI

_3

> lnI

B AlCl > AlCl

_3

C TlI > TlI

_3

D TlCl

_3

> TlCl

Correct Answer

Option C

Solution

T

l^+

& I

^-

> T

l^{+3}

& 3I

^-

due to inert pair effect T

l^{+}

is more stable than T

l^{+3}

.

Q14

The correct order of bond angles in the following compounds/species is :

A CO 2 3 2 O <

\mathop {\mathrm{N{H_4}}}\limits^ +

B H 2 O 3 2

C H 2 O 3 2

D H 2 O 3 2

Correct Answer

Option B

Solution

CO 2 $\Rightarrow$ sp 2 hybridisation, bond angle = 180

^\circ

NH

_4^ +

$\Rightarrow$ sp 3 hybridisation, bond angle = 109

^\circ

28' NH 3 $\Rightarrow$ sp 3 hybridisation with one lone pair on central atom, bond angle

\simeq

107

^\circ

H 2 O $\Rightarrow$ sp 3 hybridisation with two lone pairs on central atom, bond angle

\simeq

104.5

^\circ

Q15

Amongst the following which one will have maximum 'lone pair - lone pair' electron repulsions?

A CIF 3

B IF 5

C SF 4

D XeF 2

Correct Answer

Option D

Solution

XeF 2 having maximum lone pairs, so, it has maximum 'lone pair-lone pair' electron repulsions.

Q16

Which amongst the following is incorrect statement?

A The bond orders of O

_2^ +

, O 2 , O

_2^ -

and O

_2^ {2-}

are 2.5, 2, 1.5 and 1, respectively

B C 2 molecule has four electrons in its two degenerate

\pi

molecular orbitals

C H

_2^ +

ion has one electron

D O

_2^ +

ion is diamagnetic

Correct Answer

Option D

Solution

Due to one unpaired electron in $\pi$ * 2p molecular orbital, O

_2^ +

is a paramagnetic ion.

Q17

BF 3 is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are :

A sp 2 and 8

B sp 3 and 4

C sp 3 and 6

D sp 2 and 6

Correct Answer

Option D

Solution

$\bullet$ Number of electrons around boron atom is 6.

$\bullet$ Hybridization of B is sp 2 .

$\bullet$ Shape is trigonal planar.

Q18

The correct sequence of bond enthalpy of 'C-X' bond is :

A CH 3 - Cl > CH 3 - F > CH 3 - Br > CH 3 - l

B CH 3 - F < CH 3 - Cl < CH 3 - Br < CH 3 - l

C CH 3 - F > CH 3 - Cl > CH 3 - Br > CH 3 - l

D CH 3 - F < CH 3 - Cl > CH 3 - Br > CH 3 - l

Correct Answer

Option C

Solution

On moving down the group from F to I, the size of atom increases.

Order of the size of halogen atoms is I > Br > Cl > F.

So, the bond length of C—X bond also increases from F to I and hence, the bond enthalpy decreases from F to I.

Correct order of bond length of C—X bond is H 3 C - I > H 3 C - Br > H 3 C - Cl > H 3 C - F.

Correct order of bond enthalpy is H 3 C - F > H 3 C - Cl > CH 3 - Br > H 3 C - I.

Q19

Which of the following molecules is non-polar in nature?

A NO 2

B POCl 3

C CH 2 O

D SbCl 5

Correct Answer

Option D

Solution

SbCl 5 : Net vector summation of bond moments will be zero so SbCl 5 is a non-polar molecule.

$\bullet$ NO 2 : $\bullet$ POCl 3 : $\bullet$ CH 2 O :

Q20

Which of the following set of molecules will have zero dipole moment?

A Boron trifluoride, hydrogen, fluoride, carbon dioxide, 1, 3-dichlorobenzene

B Nitrogen trifluoride, beryllium difluoride, water, 1, 3-dichlorobenzene

C Boron trifluoride, beryllium difluoride, carbon dioxide, 1, 4-dichlorobenzene

D Ammonia, beryllium difluoride, water, 1, 4-dichlorobenzene

Correct Answer

Option C

Solution

Due to their symmetrical structure, BF 3 , BeF 2 , CO 2 and 1, 4-dichloro benzene molecules have a zero dipole moment.