### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2007

Which of the following species exhibits the diamagnetic behaviour?
A
$O_2^{2−}$
B
NO
C
$O_2^+$
D
O2

## Explanation

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(a)    $O_2^{2−}$ 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 is no unpaired electron so it is diamagnetic.

(b)   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.

(c)   $O_2$ 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.

(d)   $O_2^{+}$ has 15 electrons.

Moleculer orbital configuration of $O_2^{+}$ is

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

Here 1 unpaired electron present, so it is paramagnetic.
2

### AIEEE 2007

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizinig order of the polarizing power of the cationic species, K+, Ca2+, Mg2+, Be2+?
A
Mg2+ < Be2+ < K+ < Ca2+
B
K+ < Ca2+ < Mg2+ < Be2+
C
Be2+ < K+ < Ca2+ < Mg2+
D
Ca2+ < Mg2+ < Be2+ < K+

## Explanation

As charge/size ratio of a cation determines its polarizing power so high charge and small size of the cations increases polarisation.

As the size of the given cations decreases as

K+ > Ca2+ > Mg2+ > Be2+

Hence, polarising power decreases as

K+ < Ca2+ < Mg2+ < Be2+
3

### AIEEE 2006

The decreasing values of bond angles from NH3 (106o) to SbH3 (91o ) down group-15 of the periodic table is due to
A
increasing bp-bp repulsion
B
increasing p-orbital character in sp3
C
decreasing lp-bp repulsion
D
decreasing electronegativity

## Explanation

Nitrogen(N) atom is smaller in size so it's lone pair is relatively unstable. That is why repulsion between lone pairs present on nitrogen atom and bonded pairs of electrons is relatively high.

But size of Sb is more so its lone pair is relatively stable. That is why it does not want to perticipate in bond pair lone pair repulsion and decreases lp-bp repulsion.
4

### AIEEE 2006

In which of the following molecules/ions are all the bonds not equal?
A
XeF4
B
$BF_4^−$
C
SF4
D
SiF4

## Explanation

(a)   XeF4  is sp3d2 hybridised with 4 bond pairs and 1 lone pair and structure is square planar. Here all the bond lengths are equal.

(b) $\,\,\,\,$ BF$_4^ -$, 4 bond pair present so angle is 109o 28' and sp3 hybridised. So structure is regular tetrahedral. Here all the bond lengths are equal.

(c)   SF4 is sp3d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or see-saw with the bond angles equal to < 120o and 179o instead of the expected angles of 120o and 180o respectively. Here axial and equitorial both bonds are presents. And we know axial bonds are longer and weaker.

(d) SiF4 is sp3 hybridisation and regular tetrahedral geometry. Here all the bond lengths are equal.