Q19 / 180 · COMEDK 2021 · Chemistry
Which of the following pairs of ions in iso-electronic and iso-structural?
- Correct answerA$\mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}$
- B$\mathrm{SO}_{3}^{2-}, \mathrm{NO}_{3}^{-}$
- C$\mathrm{ClO}_{3}^{-}, \mathrm{CO}_{3}^{2-}$
- D$\mathrm{SO}_{3}^{2-}, \mathrm{CO}_{3}^{2-}$
Solution
Iso-electronic species have the same number of electrons, and iso-structural species have the same geometry and hybridization.
For $\mathrm{CO}_{3}^{2-}$:
Total electrons = $6 + 3 \times 8 + 2 = 32$.
Hybridization: $\dfrac{1}{2}(4 + 0 + 2) = 3$ (sp$^2$ hybridization). Geometry: Trigonal planar.
For $\mathrm{NO}_{3}^{-}$:
Total electrons = $7 + 3 \times 8 + 1 = 32$.
Hybridization: $\dfrac{1}{2}(5 + 0 + 1) = 3$ (sp$^2$ hybridization). Geometry: Trigonal planar.
Since both $\mathrm{CO}_{3}^{2-}$ and $\mathrm{NO}_{3}^{-}$ have 32 electrons and sp$^2$ hybridized trigonal planar structures, they are both iso-electronic and iso-structural.
Answer: $\mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}$
For $\mathrm{CO}_{3}^{2-}$:
Total electrons = $6 + 3 \times 8 + 2 = 32$.
Hybridization: $\dfrac{1}{2}(4 + 0 + 2) = 3$ (sp$^2$ hybridization). Geometry: Trigonal planar.
For $\mathrm{NO}_{3}^{-}$:
Total electrons = $7 + 3 \times 8 + 1 = 32$.
Hybridization: $\dfrac{1}{2}(5 + 0 + 1) = 3$ (sp$^2$ hybridization). Geometry: Trigonal planar.
Since both $\mathrm{CO}_{3}^{2-}$ and $\mathrm{NO}_{3}^{-}$ have 32 electrons and sp$^2$ hybridized trigonal planar structures, they are both iso-electronic and iso-structural.
Answer: $\mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}$
Hence, the correct answer is option: 'A'.