1
MCQ (Single Correct Answer)

### JEE Main 2016 (Online) 10th April Morning Slot

The bond angle H - X - H is the greatest in the compound :
A
CH4
B
NH3
C
H2O
D
PH3

## Explanation

CH4 - sp3 hybridisation - Bond angle (109o28')

NH3 - sp3 hybridisation - Bond angle ( 107.8o)

H2O - sp3 hybridisation - Bond angle ( 104.5o)

PH3 - No hybridisation - Bond angle ( 90o)

Note : According to DRAGO RULE, lone pair of electrons are in pure s- orbital of the atom P. That is why PH3 has no hybridisation.
2
MCQ (Single Correct Answer)

### JEE Main 2017 (Offline)

Which of the following species is not paramagnetic?
A
CO
B
O2
C
B2
D
NO

## Explanation

Those species which have unpaired electrons are called paramagnetic species.

(a)    CO has 14 electrons.

Moleculer orbital configuration of CO is

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

Here is no unpaired electron so it is diamagnetic.

(b)   $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ is

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

Here 2 unpaired electron present, so it is paramagnetic.

(c)   B2 has 10 electrons.

Molecular orbital configuration of B2 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 two unpaired electrons present. So it is paramagnetic.

(d)   NO has 15 electrons.

Moleculer orbital configuration of NO is

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

Here is 1 unpaired electron, So it is Paramagnetic.
3
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 8th April Morning Slot

Which of the following is paramagnetic ?
A
NO+
B
CO
C
$O_2^{2 - }$
D
B2

## Explanation

Those species which have unpaired electrons are called paramagnetic species.

(a)   NO+ has 14 electrons.

Moleculer orbital configuration of NO+ is

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

Here is no unpaired electron, So it is Diamagnetic.

(b)    CO has 14 electrons.

Moleculer orbital configuration of CO is

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

Here is no unpaired electron so it is diamagnetic.

(c)   $O_2^{2 - }$ has 18 electrons.

Moleculer orbital configuration of $O_2^{2 - }$ is

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

Here also no unpaired electron present, so it is diamagnetic.

(d)   B2 has 10 electrons.

Molecular orbital configuration of B2 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 two unpaired electrons present.

So it is paramagnetic.
4
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 8th April Morning Slot

sp3d2 hybridization is not displayed by :
A
BrF5
B
SF6
C
[CrF6]3$-$
D
PF5

## Explanation

Hybridization (X) = ${1 \over 2}$ [ VE + MA – c + a ]

where, VE = No. of valence electrons of central atom
MA = No. of monovalent atoms/groups surrounding the central atom,
c = Charge on the cation,
a = Charge on the anion

(A) [BrF5] : X = ${1 \over 2}$[ 7 + 5 - 0 + 0] = 6 = sp3d2 hybridized

(B) [SF6] : X = ${1 \over 2}$[ 6 + 6 - 0 + 0] = 6 = sp3d2 hybridized

(C) [CrF6]3$-$ : Cr+3 = [Ar]3d3

(D) [PF5] : X = ${1 \over 2}$[ 5 + 5 - 0 + 0] = 5 = sp3d hybridized

Note : [CrF6]3$-$ shows d2sp3 hybridization not sp3d2. So option (C) should also be the correct answer.

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