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, $NO_2^-$, $NO_3^-$, $NH_2^-$, $NH_4^+$, $SCN^-$?

TopperSquare Expert · Accepted Answer

The correct answer is Option (D): $NO_2^-$ and $NO_3^-$. 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.

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

Step-by-Step Solution

Exam Context & Concepts Covered

Everything you need to crack NEET

Track Your Progress

Platform Features

15,000+ MCQs

AI Analytics

Detailed Solutions

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^-NO2−​, NO3−NO_3^-NO3−​, NH2−NH_2^-NH2−​, NH4+NH_4^+NH4+​, SCN−SCN^-SCN−?