Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (More than One Correct Answer)

According to Molecular Orbital Theory, which of the following statements is(are) correct?

A

$$C_2^{2-}$$ is expected to be diamagnetic

B

$$O_2^{2+}$$ expected to have a longer bond length than O_{2}

C

$$N_2^+$$ and $$N_2^-$$ have the same bond order

D

$$He_2^+$$ has the same energy as two isolated He atoms

Option (A) : According to molecular orbital theory, the arrangement of the electrons in the molecular orbitals is as follows :

For $$C_2^{2 - }$$, the total number of electrons is 14. The molecular orbital configuration is

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

There is no unpaired electron and thus it is diamagnetic.

Option (B) : For $$O_2^{2 + }$$, the total number of electrons is 14. The molecular orbital configuration is

$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^0}^ * \,\, = \pi _{2p_y^0}^ * $$

Thus the bond order = (10 $$-$$ 4)/2 = 3;

Molecular orbital configuration of O_{2} (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. for O_{2}, bond order = 2.

As bond order is inversely proportional to bond length, thus the bond length of $$O_2^{2 + }$$ is less than the bond length of O_{2}.

Option (C) : For $$N_2^ + $$, the total number of electrons is 13. The molecular orbital configuration is

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

Bond order of $$N_2^ + $$ = (9 $$-$$ 4)/2 = 2.5

$$N_2^{ - }$$ has 15 electrons.

Moleculer orbital configuration of $$N_2^{ - }$$ is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^0}^ *$$

$$\therefore\,\,\,\,$$N_{a} = 5

N_{b} = 10

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right] = 2.5$$

Thus, the bond orders are the same.

Option (D) : As some energy is released during the formation of $$He_2^ + $$ from two isolated He atoms, thus it has lesser energy as compared to two isolated He atoms.

For $$C_2^{2 - }$$, the total number of electrons is 14. The molecular orbital configuration is

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

There is no unpaired electron and thus it is diamagnetic.

Option (B) : For $$O_2^{2 + }$$, the total number of electrons is 14. The molecular orbital configuration is

$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^0}^ * \,\, = \pi _{2p_y^0}^ * $$

Thus the bond order = (10 $$-$$ 4)/2 = 3;

Molecular orbital configuration of O

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

So O$$_2$$ has 2 unpaired electrons. for O

As bond order is inversely proportional to bond length, thus the bond length of $$O_2^{2 + }$$ is less than the bond length of O

Option (C) : For $$N_2^ + $$, the total number of electrons is 13. The molecular orbital configuration is

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

Bond order of $$N_2^ + $$ = (9 $$-$$ 4)/2 = 2.5

$$N_2^{ - }$$ has 15 electrons.

Moleculer orbital configuration of $$N_2^{ - }$$ is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^0}^ *$$

$$\therefore\,\,\,\,$$N

N

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right] = 2.5$$

Thus, the bond orders are the same.

Option (D) : As some energy is released during the formation of $$He_2^ + $$ from two isolated He atoms, thus it has lesser energy as compared to two isolated He atoms.

2

MCQ (More than One Correct Answer)

The molecules that will have dipole moment are

A

2, 2-dimethylpropane

B

trans-2-pentene

C

cis-3-hexene

D

2,2,3,3-tetramethylbutane

3

MCQ (More than One Correct Answer)

Which of the following have identical bond order?

A

CN^{-}

B

$$O_2^-$$

C

NO^{+}

D

CN^{+}

4

MCQ (More than One Correct Answer)

The linear structure is assumed by

A

SnCl_{2}

B

NCO^{-}

C

CS_{2}

D

$$NO_2^+$$

On those following papers in MCQ (Multiple Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

JEE Advanced 2020 Paper 1 Offline (1) *keyboard_arrow_right*

JEE Advanced 2019 Paper 1 Offline (1) *keyboard_arrow_right*

JEE Advanced 2016 Paper 2 Offline (1) *keyboard_arrow_right*

JEE Advanced 2016 Paper 1 Offline (1) *keyboard_arrow_right*

IIT-JEE 1992 (2) *keyboard_arrow_right*

IIT-JEE 1991 (1) *keyboard_arrow_right*

IIT-JEE 1986 (1) *keyboard_arrow_right*

Some Basic Concepts of Chemistry *keyboard_arrow_right*

Structure of Atom *keyboard_arrow_right*

States of Matter *keyboard_arrow_right*

Thermodynamics *keyboard_arrow_right*

Equilibrium *keyboard_arrow_right*

Solid State & Surface Chemistry *keyboard_arrow_right*

Solutions *keyboard_arrow_right*

Electrochemistry *keyboard_arrow_right*

Chemical Kinetics and Nuclear Chemistry *keyboard_arrow_right*

Redox Reactions *keyboard_arrow_right*

Periodic Table & Periodicity *keyboard_arrow_right*

Chemical Bonding & Molecular Structure *keyboard_arrow_right*

Isolation of Elements *keyboard_arrow_right*

Hydrogen *keyboard_arrow_right*

p-Block Elements *keyboard_arrow_right*

d and f Block Elements *keyboard_arrow_right*

Coordination Compounds *keyboard_arrow_right*

Basics of Organic Chemistry *keyboard_arrow_right*

Hydrocarbons *keyboard_arrow_right*

Haloalkanes and Haloarenes *keyboard_arrow_right*

Alcohols, Phenols and Ethers *keyboard_arrow_right*

Aldehydes, Ketones and Carboxylic Acids *keyboard_arrow_right*

Compounds Containing Nitrogen *keyboard_arrow_right*

Biomolecules *keyboard_arrow_right*