The ion that has \(sp^3d^2\) hybridization for the central atom, is :
Show Hint
- \(\mathrm{[ICl_2]^-}\)
- \(\mathrm{[ICl_4]^-}\)
- \(\mathrm{[IF_6]^-}\)
- \(\mathrm{[BrF_2]^-}\)
The Correct Option is B
Solution and Explanation
\(\mathrm{[ICl_2]^-}\): \(\frac{7 + 2 + 1}{2} = 5\), \(sp^3d\).
\(\mathrm{[ICl_4]^-}\): \(\frac{7 + 4 + 1}{2} = 6\), \(sp^3d^2\) (square planar).
\(\mathrm{[IF_6]^-}\): \(\frac{7 + 6 + 1}{2} = 7\), \(sp^3d^3\).
\(\mathrm{[BrF_2]^-}\): \(\frac{7 + 2 + 1}{2} = 5\), \(sp^3d\).
Only \(\mathrm{[ICl_4]^-}\) has 6 electron pairs, i.e. \(sp^3d^2\).
Answer: (2) \(\mathrm{[ICl_4]^-}\)
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