The hybridization of P exhibited in PF_{5} is | vedclass

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(A) The central atom $P$ in $PF_{5}$ has $5$ valence electrons and forms $5$ sigma bonds with $5$ $F$ atoms.The steric number is calculated as: $	ext

Vedclass · Accepted Answer

$1$

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(A) The central atom $P$ in $PF_{5}$ has $5$ valence electrons and forms $5$ sigma bonds with $5$ $F$ atoms.The steric number is calculated as: $	ext{Steric Number} = \frac{1}{2} (V + M - C + A) = \frac{1}{2} (5 + 5 - 0 + 0) = 5$.$A$ steric number of $5$ corresponds to $sp^{3}d$ hybridization.Comparing $sp^{3}d$ with $sp^{x}d^{y}$,we get $x=3$ and $y=1$.Therefore,the value of $y$ is $1$.

The hybridization of $P$ exhibited in $PF_{5}$ is $sp^{x}d^{y}$. The value of $y$ is $...$

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