The correct statement about \(\mathrm{ICl_5}\) and \(\mathrm{ICl_4^-}\) is
Show Hint
- \(\mathrm{ICl_5}\) is square pyramidal and \(\mathrm{ICl_4^-}\) is square planar.
- \(\mathrm{ICl_5}\) is trigonal bipyramidal and \(\mathrm{ICl_4^-}\) is tetrahedral.
- \(\mathrm{ICl_5}\) is square pyramidal and \(\mathrm{ICl_4^-}\) is tetrahedral
- Both are isostructural.
The Correct Option is A
Solution and Explanation
\(\mathrm{ICl_4^-}\): \(7 + 4 + 1 = 12\) electrons around I = 6 pairs: 4 bond pairs + 2 lone pairs, \(sp^3d^2\); the two lone pairs sit trans, so the shape is square planar.
Hence \(\mathrm{ICl_5}\) is square pyramidal and \(\mathrm{ICl_4^-}\) is square planar; the two are not isostructural.
Answer: (1) \(\mathrm{ICl_5}\) is square pyramidal and \(\mathrm{ICl_4^-}\) is square planar.
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