JEE MAIN - Chemistry (2021 - 20th July Evening Shift - No. 10)
The hybridisations of the atomic orbitals of nitrogen in NO$$_2^ - $$, NO$$_2^ + $$ and NH$$_4^ + $$ respectively are.
sp3, sp2 and sp
sp, sp2 and sp3
sp3, sp and sp2
sp2, sp and sp3
Explanation
Hybridisation = Number of sigma bonds + number of lone pair of electrons + number of coordinate bond.
For $$NO_2^ - $$ hybridisation = 2 sigma NO bonds + 1 lone pair on N = 3; so $$NO_2^ - $$ is sp2-hybridised.
_20th_July_Evening_Shift_en_10_1.png)
For $$NO_2^ + $$ hybridisation = 2 sigma NO bonds = 2; so $$NO_2^ + $$ sp-hybridised.
_20th_July_Evening_Shift_en_10_2.png)
For $$NO_4^ + $$ Hybridisation = 4 sigma NH bonds = 4; so $$NO_4^ + $$ is sp3-hybridised.
_20th_July_Evening_Shift_en_10_3.png)
Therefore, hybridisation of N in $$NO_2^ - $$, $$NO_2^ + $$, $$NO_4^ + $$ are sp2, sp, sp3 respectively.
For $$NO_2^ - $$ hybridisation = 2 sigma NO bonds + 1 lone pair on N = 3; so $$NO_2^ - $$ is sp2-hybridised.
For $$NO_2^ + $$ hybridisation = 2 sigma NO bonds = 2; so $$NO_2^ + $$ sp-hybridised.
For $$NO_4^ + $$ Hybridisation = 4 sigma NH bonds = 4; so $$NO_4^ + $$ is sp3-hybridised.
Therefore, hybridisation of N in $$NO_2^ - $$, $$NO_2^ + $$, $$NO_4^ + $$ are sp2, sp, sp3 respectively.
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