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Question

The shape of $$XeF_4$$ is:


A
tetrahedral
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B
octahedral
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C
square planar
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D
pyramidal
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Solution

The correct option is A square planar
The shape of $$XeF_4$$ is square planar. The central $$Xe$$ atom has 4 bond pairs of electrons and two lone pairs of electrons. It undergoes $$sp^3d^2$$ hybridization which results in octahedral electron geometry and square planar molecular geometry. The two lone pairs are at opposite corners of an octahedron. This minimizes repulsion between them.

Chemistry

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