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1

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)
Which of the following pairs of ions are isoelectronic and isostructural ?
A
SO32$$-$$,  NO3$$-$$
B
ClO3$$-$$,  SO32$$-$$
C
CO32$$-$$, SO32$$-$$
D
ClO3$$-$$, CO32$$-$$,

Explanation

Species Hybridisation Shape No. of e-
$$SO_3^{2-}$$ sp3 Pyramidal 42
$$ClO_3^{-}$$ sp3 Pyramidal 42
$$CO_3^{2-}$$ sp2 Triangular planar 32
$$NO_3^{-}$$ sp2 Triangular planar 32
2

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)
The correct bond order in the following species is
A
O2+ < O2$$-$$ < O22+
B
O2$$-$$ < O2+ < O22+
C
O22+ < O2+ < O2$$-$$
D
O22+ < O2$$-$$ < O2+

Explanation

$$\,$$ Bond order $$ = {1 \over 2}$$ [Nb $$-$$ Na]

Nb = No of electrons in bonding molecular orbital

Na $$=$$ No of electrons in anti bonding molecular orbital

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 = 14

$$\therefore\,\,\,\,$$ 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 $$O_2^ - $$ (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 $$_2^{2 + }$$ (14 electrons) is

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

$$\therefore\,\,\,\,$$ Nb = 10

Na = 4

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 4] = 3

$$O_2^-$$ $$O_2$$ $$O_2^+$$ $$O_2^{2+}$$
B.O. : 1.5 2.0 2.5 3.0
3

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)
Maximum bond angle at nitrogen is present in which of the following ?
A
NO2+
B
NO3$$-$$
C
NO2
D
NO2$$-$$

Explanation

Species $$NO_3^-$$ $$NO_2$$ $$NO_2^-$$ $$NO_2^+$$
Hybridisation sp2 sp2 sp2 sp(linear)
Bond Angle 120o 134o 115o 180o
4

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)
Which of the following options represents the correct bond order ?
A
O2$$-$$  > O2 < O2+
B
O2$$-$$  <  O2  >  O2+
C
O2$$-$$  >  O2  >  O2+
D
O2$$-$$  <  O2  <  O2+

Explanation

$$\,$$ Bond order $$ = {1 \over 2}$$ [Nb $$-$$ Na]

Nb = No of electrons in bonding molecular orbital

Na $$=$$ No of electrons in anti bonding molecular orbital

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

$$\therefore\,\,\,\,$$ 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 $$O_2^ - $$ (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

$$O_2^-$$ $$O_2$$ $$O_2^+$$
B.O. : 1.5 2.0 2.5

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