Question

What is the hybridization of ${{\mathrm{SO}}_4}^{2-}$ion?

Answer 1

Step 1: Formula used for calculation:

• $\mathrm{Hybridization}=\frac{1}{2}\left[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}\right]$
• Where $\mathrm{V}$ is the number of valence electrons of the central atom, $\mathrm{M}$ is the monovalent atoms attached to the central atom, $\mathrm{C}$ is the positive charge, and $\mathrm{A}$represents the negative charge.

Step 2: Calculation for Hybridization of Sulfate ion ${{\mathrm{SO}}_4}^{2-}$:

• The central atom is sulfur$\left(\mathrm{S}\right)$ and its atomic number is $16$ so its electronic configuration is $\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^4$.
• So, $\mathrm{S}$ has $6$ valence electrons.
• $\mathrm{Hybridization}\mathrm{of}{{\mathrm{SO}}_4}^{2-}=\frac{1}{2}\left[6+0-0+2\right]=\frac{1}{2}\times 8=4$
• The hybridization number is $4$, thus the hybridization of ${{\mathrm{SO}}_4}^{2-}$ion is ${\mathrm{sp}}^3$.