1

### 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.
2

### 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.
3

### JEE Main 2017 (Online) 9th April Morning Slot

The group having triangular planar structures is :
A
BF3, NF3, $CO_3^{2 - }$
B
$CO_3^{2 - }$, $NO_3^ -$, SO3
C
NH3, SO3, $CO_3^{2 - }$
D
NCl3, BCl3, SO3

## Explanation

(A) BF3 : sp2 Hybridization : Triangular Planar

NF3 : sp3 Hybridization : Tetrahedral

$CO_3^{2 - }$ : sp2 Hybridization : Triangular Planar

(B) $CO_3^{2 - }$ : sp2 Hybridization : Triangular Planar

$NO_3^ -$ : sp2 Hybridization : Triangular Planar

SO3 : sp2 Hybridization : Triangular Planar

(C) NH3 : sp3 Hybridization : Tetrahedral

SO3 : sp2 Hybridization : Triangular Planar

$CO_3^{2 - }$ : sp2 Hybridization : Triangular Planar

(D) NCl3 : sp3 Hybridization : Tetrahedral

BCl3 : sp2 Hybridization : Triangular Planar

SO3 : sp2 Hybridization : Triangular Planar
4

### JEE Main 2018 (Offline)

According to molecular orbital theory, which of the following will not be a viable molecule?
A
${\rm H}e_2^{2 + }$
B
${\rm H}e_2^{ + }$
C
${\rm H}_2^{- }$
D
${\rm H}_2^{2 - }$

## Explanation

Note :

According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.

(a)$\,\,\,$ Configuration of $He_2^{2 + }$ (2 electrons) is = ${\sigma _{1{s^2}}}$

$\therefore\,\,\,$ Bond order = ${1 \over 2}$ (2 $-$0) = 1

(b)$\,\,\,$ Configuration of $He_2^ +$ (3 electrons) is = $\sigma 1{s^2}$ $\sigma _{1{s^1}}^ *$

$\therefore\,\,\,$ Bond order = ${1 \over 2}$ (2 $-$1) = 0.5

(c) $\,\,\,$ Configuration of $H_2^ -$ (3 electrons) is = ${\sigma _{1{s^2}}}$ $\sigma _{1{s^1}}^ *$

$\therefore$ Bond order = ${1 \over 2}$ (2 $-$ 1) = 0.5

(d) $\therefore\,\,\,$ Configuration of $He_2^{2 - }$ is = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ *$

$\therefore\,\,\,$ Bond order = $= {1 \over 2}$ (2 $-$ 2) = 0

$\therefore\,\,\,$ $H_2^{2 - }$ is not a viable molecule.