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NEET CHEMISTRYMedium

Which of the two ions from the list given below have the geometry that is explained by the same hybridisation of orbitals, NO2−NO_2^-, NO3−NO_3^-, NH2−NH_2^-, NH4+NH_4^+, SCN−SCN^-?

A

NH4+NH_4^+ and NO3−NO_3^-

B

SCN−SCN^- and NH2−NH_2^-

C

NO2−NO_2^- and NH2−NH_2^-

D

NO2−NO_2^- and NO3−NO_3^-

Step-by-Step Solution

  1. Determine Hybridisation (HH) using the formula H=12(V+M−C+A)H = \frac{1}{2}(V + M - C + A), where VV is valence electrons, MM is monovalent atoms, CC is cationic charge, and AA is anionic charge.
  2. Analyze NO2−NO_2^- (Nitrite Ion):
  • N is central (V=5V=5). Oxygen is divalent (M=0M=0). Charge (A=1A=1).
  • H=12(5+0−0+1)=3H = \frac{1}{2}(5 + 0 - 0 + 1) = 3. Hybridisation is sp2sp^2.
  1. Analyze NO3−NO_3^- (Nitrate Ion):
  • N is central (V=5V=5). M=0M=0. A=1A=1.
  • H=12(5+0−0+1)=3H = \frac{1}{2}(5 + 0 - 0 + 1) = 3. Hybridisation is sp2sp^2.
  1. Analyze NH4+NH_4^+ (Ammonium Ion):
  • N is central (V=5V=5). H is monovalent (M=4M=4). Charge (C=1C=1).
  • H=12(5+4−1+0)=4H = \frac{1}{2}(5 + 4 - 1 + 0) = 4. Hybridisation is sp3sp^3.
  1. Analyze NH2−NH_2^- (Amide Ion):
  • N is central (V=5V=5). H is monovalent (M=2M=2). Charge (A=1A=1).
  • H=12(5+2−0+1)=4H = \frac{1}{2}(5 + 2 - 0 + 1) = 4. Hybridisation is sp3sp^3.
  1. Analyze SCN−SCN^- (Thiocyanate Ion):
  • C is central (V=4V=4). Forms 2 sigma bonds (one with S, one with N).
  • Hybridisation is spsp.
  1. Conclusion: Both NO2−NO_2^- and NO3−NO_3^- exhibit sp2sp^2 hybridisation.
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