JEE MAIN - Chemistry (2019 - 10th January Morning Slot - No. 12)
The type of hybridisation and number of lone pair (s) of electrons of Xe in XeOF4, respectively, are:
sp3d and 2
sp3d2 and 2
sp3d and 1
sp3d2 and 1
Explanation
H = $${1 \over 2}$$ (V + M - c + a)
$$ \therefore $$ H = $${1 \over 2}$$(8 + 4) = 6
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From structure, it is clear that it has five bond pairs and one lone pair.
$$ \therefore $$ H = $${1 \over 2}$$(8 + 4) = 6
From structure, it is clear that it has five bond pairs and one lone pair.
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