The ion which is not tetrahedral in shape is:

{B F_{4}}^{-}

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N {H_{4}}^{+}

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X e O_{4}

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{I C l_{4}}^{-}

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Solution

The correct option is D ${I C l_{4}}^{-}$
The ion $I C l_{4}^{-}$ contains $4$ bond pairs of electrons and $2$ lone pairs of electrons. The central iodine atom undergoes $s p^{3} d^{2}$ hybridization . The expected geometry is octahedral and the actual geometry is square planar.
${B F_{4}}^{-}, N H_{4}^{+}$ and $X e O_{4}$ are tetrahedral in shape.

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X e O_{4}

s p^{3}

Which of the following conversions involve change in both shape and hybridisation?

a) CH4 -> C2H6

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c) BF3 -> BF4-

d) H2O ->H3O+

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