Which of the two ions from the list given below have the geometry that is explained by the same hybridisation

Question

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

Accepted Answer

1. **Determine Hybridisation ($H$)** using the formula $H = \frac{1}{2}(V + M - C + A)$, where $V$ is valence electrons, $M$ is monovalent atoms, $C$ is cationic charge, and $A$ is anionic charge.
2. **Analyze $NO_2^-$ (Nitrite Ion):**
   - N is central ($V=5$). Oxygen is divalent ($M=0$). Charge ($A=1$).
   - $H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$. Hybridisation is **$sp^2$**.
3. **Analyze $NO_3^-$ (Nitrate Ion):**
   - N is central ($V=5$). $M=0$. $A=1$.
   - $H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$. Hybridisation is **$sp^2$**.
4. **Analyze $NH_4^+$ (Ammonium Ion):**
   - N is central ($V=5$). H is monovalent ($M=4$). Charge ($C=1$).
   - $H = \frac{1}{2}(5 + 4 - 1 + 0) = 4$. Hybridisation is **$sp^3$**.
5. **Analyze $NH_2^-$ (Amide Ion):**
   - N is central ($V=5$). H is monovalent ($M=2$). Charge ($A=1$).
   - $H = \frac{1}{2}(5 + 2 - 0 + 1) = 4$. Hybridisation is **$sp^3$**.
6. **Analyze $SCN^-$ (Thiocyanate Ion):**
   - C is central ($V=4$). Forms 2 sigma bonds (one with S, one with N).
   - Hybridisation is **$sp$**.
7. **Conclusion:** Both $NO_2^-$ and $NO_3^-$ exhibit **$sp^2$** hybridisation.

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