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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

MCQ (Single Correct Answer)
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+
2

AIEEE 2004

MCQ (Single Correct Answer)
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'
3

AIEEE 2003

MCQ (Single Correct Answer)
Which one of the following compounds has the smallest bond angle in its molecule?
A
OH2
B
SH2
C
NH3
D
SO2

Explanation

(a)   H2O is sp3 hybridized, and oxygen atom has 2 bond pair and 2 lone pair. So the angle between two O $$-$$ H bond is 104.5o

(b)   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.

(c)   NH3 has sp3 hybridization and N atom has 3 bond pair and one lone pair so bond angle is 107o

(d)   SO2 is sp2 hybridized and bond angle is 117o.

4

AIEEE 2003

MCQ (Single Correct Answer)
The pair of species having identical shapes for molecules of both species is
A
XeF2, CO2
B
BF3, PCl3
C
PF5, IF5
D
CF4, SF4

Explanation




Questions Asked from Chemical Bonding & Molecular Structure

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