### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

Which one of the following has the regular tetrahedral structure?
(Atomic nos : B = 5, S = 16, Ni = 28, Xe = 54)
A
XeF4
B
[Ni(CN)4]2-
C
$BF_4^-$
D
SF4

## Explanation

Regular Tetrahedral structure is possible in sp3 hybridization where central atom has 4 bond pair and no lone pair.

(a)   XeF4  is sp3d2 hybridised and structure is square planar.

(b)   [Ni(CN)4]$-$2 is coordinate compound and oxidation number of Ni is +2.

Electronic configuration of Ni+2 is $=$ [Ar]3d8

But because of CN$-$ ion which is a strong field ligand , it can perform pairing of electron.

And the structure of dsp2 hybridization is square planar.

(C) $\,\,\,\,$ BF$_4^ -$, 4 bond pair present so angle is 109o 28' and sp3 hybridised. So structure is regular tetrahedral.

(d)

SF4 is sp3d hybridised and structure is see-saw.
2

### AIEEE 2004

The states of hybridization of boron and oxygen atoms in boric acid (H3BO3) are respectively
A
sp2 and sp2
B
sp3 and sp3
C
sp3 and sp2
D
sp2 and sp3

## Explanation

Structure of  H3BO3  is

Here Boron has 3 sigma bond. So it is sp2 hybridised.

And oxygen has two sigma bond and two lone pair. So it sp3 hybridised.
3

### AIEEE 2004

The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements is true for these two species?
A
Bond length in NO+ is greater than in NO
B
Bond length is unpredictable
C
Bond length in NO+ in equal to that in NO
D
Bond length in NO is greater than in NO+

## Explanation

Molecular orbital configuration of NO (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

Similarly Molecular orbital configuration of NO+ (14 electrons) is

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

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

Na = 4

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

Note :
(2) $\,\,\,\,$ Bond length $\propto$ ${1 \over {Bond\,\,order}}$

So, bond length in NO > NO+
4

### AIEEE 2004

The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is
A
H2S < SiH4 < NH3 < BF3
B
H2S < NH3 < BF3 < SiH4
C
H2S < NH3 < SiH4 < BF3
D
NH3 < H2S < SiH4 < BF3

## Explanation

1. In H2S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H2S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(S) atom and size of S$-$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, px, py, pz and it look like this :

As we know the angle between and py is 90o, so bond angle is 90o.

2. In NH3, 3 Bond pair(BP) + 1 lone pair (LP) present so angle between bond 107o.

3. $\,\,\,\,$ BF3 has sp2 hybridization. So bond angle is 120o.

4.
Here Si is sp3 hybridised and shape is regular tetrahedral and bond angle 109o 28'