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Question

The correct order of increasing s character (in perencetage) in the hybrid orbitals in below molecules/ions is (assume all hybrid orbitals are exactly equivalent):

CO23I;XeF4II; I3III; NCl3IV; BeCl2(g)V

A
II< III< IV< I< V
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B
II< IV< III< V< I
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C
III< II< I< V< IV
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D
II< IV< III< I< V
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Solution

The correct option is A II< III< IV< I< V
More is the percentage of s-character in a compound more will be the electronegativity of central metal atom. The hybridization of given compound will be as follows.
SpeciesHybridisationCO23sp2XeF4sp3d2I3sp3dNCl3sp3BeCl2sp

Lesser is the number of p-orbitals more will be the percentage of s-character.

So the correct order of S-character in hybrid orbitals is sp>sp2>sp3>sp3d>sp3d2 . So the correct option is II<III<IV<I<V.
Hence option A is correct.


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