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NEET CHEMISTRYMedium

The hybridisations of atomic orbitals of nitrogen in NO+^+, NO3−_3^- and NH3_3 respectively are:

A

sp, sp3^3 and sp2^2

B

sp2^2, sp3^3 and sp

C

sp, sp2^2 and sp3^3

D

sp2^2, sp and sp3^3

Step-by-Step Solution

  1. Formula for Hybridisation: The steric number (H) or number of hybrid orbitals can be calculated using the formula: H=12[V+M−C+A]H = \frac{1}{2} [V + M - C + A], where V = valence electrons of central atom, M = number of monovalent atoms, C = positive charge, A = negative charge.
  2. Analysis of NO+^+ (Nitrosonium ion):
  • Central atom N (V=5V=5). No monovalent atoms (M=0M=0). Cationic charge (C=1C=1).
  • H=12[5+0−1+0]=2H = \frac{1}{2} [5 + 0 - 1 + 0] = 2.
  • H=2H=2 corresponds to sp hybridisation .
  1. Analysis of NO3−_3^- (Nitrate ion):
  • Central atom N (V=5V=5). No monovalent atoms (M=0M=0). Anionic charge (A=1A=1).
  • H=12[5+0−0+1]=3H = \frac{1}{2} [5 + 0 - 0 + 1] = 3.
  • H=3H=3 corresponds to sp2^2 hybridisation .
  1. Analysis of NH3_3 (Ammonia):
  • Central atom N (V=5V=5). Three monovalent H atoms (M=3M=3).
  • H=12[5+3−0+0]=4H = \frac{1}{2} [5 + 3 - 0 + 0] = 4.
  • H=4H=4 corresponds to sp3^3 hybridisation .
  1. Conclusion: The correct order is sp, sp2^2, and sp3^3.
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