Determine the number of sigma bonds in P_4O_{ | vedclass

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Question

(D) The structure of $P_4O_{10}$ consists of a $P_4$ tetrahedral core where each $P$ atom is bonded to three other $P$ atoms via an oxygen bridge $(-P

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) The structure of $P_4O_{10}$ consists of a $P_4$ tetrahedral core where each $P$ atom is bonded to three other $P$ atoms via an oxygen bridge $(-P-O-P-)$. There are $6$ such $P-O-P$ bonds. Additionally,each of the $4$ phosphorus atoms is bonded to a terminal oxygen atom via a double bond $(P=O)$. This gives $4$ $P=O$ bonds. Each $P-O-P$ bond contains $1$ $\sigma$ bond,and each $P=O$ bond contains $1$ $\sigma$ bond and $1$ $\pi$ bond. Total $\sigma$ bonds = $6$ (from $P-O-P$) + $4$ (from $P=O$) + $6$ (from $P-O-P$ bridges) is incorrect; let us recount: There are $6$ $P-O-P$ linkages,each having $2$ $\sigma$ bonds ($P-O$ and $O-P$),totaling $12$ $\sigma$ bonds. There are $4$ $P=O$ bonds,each having $1$ $\sigma$ bond,totaling $4$ $\sigma$ bonds. Total $\sigma$ bonds = $12 + 4 = 16$.

List-$I$	List-$II$
$A. PCl_5$	$I. \text{Tetrahedral}$
$B. BrF_5$	$II. \text{Square Planar}$
$C. BF_4^-$	$III. \text{Trigonal bipyramidal}$
$D. [Ni(CN)_4]^{2-}$	$IV. \text{Square pyramidal}$

Determine the number of $\sigma$ bonds in $P_4O_{10}$.

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