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

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Accepted Answer

1. **Method:** Determine the steric number ($SN$) using the formula: $SN = \frac{1}{2}(V + M - C + A)$, where $V$ is valence electrons, $M$ is monovalent atoms, $C$ is cationic charge, and $A$ is anionic charge. $SN=3$ corresponds to $sp^2$ hybridization .
2. **Analyze $BF_3$:**
   - Boron (Group 13) has $V=3$. Fluorine is monovalent ($M=3$).
   - $SN = \frac{1}{2}(3 + 3 - 0 + 0) = 3$. Hybridization is **$sp^2$**.
3. **Analyze $NO_2^-$ (Nitrite Ion):**
   - Nitrogen (Group 15) has $V=5$. Oxygen is divalent ($M=0$). Charge ($A=1$).
   - $SN = \frac{1}{2}(5 + 0 - 0 + 1) = 3$. Hybridization is **$sp^2$**.
4. **Analyze other species:**
   - $NH_3$: $SN = \frac{1}{2}(5 + 3) = 4 \rightarrow sp^3$.
   - $NH_2^-$: $SN = \frac{1}{2}(5 + 2 + 1) = 4 \rightarrow sp^3$.
   - $H_2O$: $SN = \frac{1}{2}(6 + 2) = 4 \rightarrow sp^3$.

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