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NEET CHEMISTRYMedium

In which of the following pairs of molecules/ions, the central atoms have sp2sp^2 hybridization?

A

NO2−NO_2^- and NH3NH_3

B

BF3BF_3 and NO2−NO_2^-

C

NH2−NH_2^- and H2OH_2O

D

BF3BF_3 and NH2−NH_2^-

Step-by-Step Solution

  1. Method: Determine the steric number (SNSN) using the formula: SN=12(V+M−C+A)SN = \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. SN=3SN=3 corresponds to sp2sp^2 hybridization .
  2. Analyze BF3BF_3:
  • Boron (Group 13) has V=3V=3. Fluorine is monovalent (M=3M=3).
  • SN=12(3+3−0+0)=3SN = \frac{1}{2}(3 + 3 - 0 + 0) = 3. Hybridization is sp2sp^2.
  1. Analyze NO2−NO_2^- (Nitrite Ion):
  • Nitrogen (Group 15) has V=5V=5. Oxygen is divalent (M=0M=0). Charge (A=1A=1).
  • SN=12(5+0−0+1)=3SN = \frac{1}{2}(5 + 0 - 0 + 1) = 3. Hybridization is sp2sp^2.
  1. Analyze other species:
  • NH3NH_3: SN=12(5+3)=4→sp3SN = \frac{1}{2}(5 + 3) = 4 \rightarrow sp^3.
  • NH2−NH_2^-: SN=12(5+2+1)=4→sp3SN = \frac{1}{2}(5 + 2 + 1) = 4 \rightarrow sp^3.
  • H2OH_2O: SN=12(6+2)=4→sp3SN = \frac{1}{2}(6 + 2) = 4 \rightarrow sp^3.
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