Define the order of d_{P-O} (single bond leng | vedclass

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(B) The reaction of $P_4$ with limited $O_2$ produces $P_4O_6$ (compound $X$).The reaction of $P_4$ with excess $O_2$ produces $P_4O_{10}$ (compound $

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$X > Y$

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(B) The reaction of $P_4$ with limited $O_2$ produces $P_4O_6$ (compound $X$).The reaction of $P_4$ with excess $O_2$ produces $P_4O_{10}$ (compound $Y$).In $P_4O_6$,each $P$ atom is bonded to three $O$ atoms via single bonds $(P-O)$.In $P_4O_{10}$,each $P$ atom is bonded to three $O$ atoms via single bonds $(P-O)$ and one $O$ atom via a double bond $(P=O)$.The presence of the $P=O$ double bond in $P_4O_{10}$ increases the oxidation state of $P$ and exerts a stronger inductive effect,which pulls electron density away from the $P-O$ single bonds.This results in a shortening of the $P-O$ single bond length in $P_4O_{10}$ compared to $P_4O_6$.Therefore,the bond length $d_{P-O}$ in $X$ $(P_4O_6)$ is greater than in $Y$ $(P_4O_{10})$,i.e.,$X > Y$.

Define the order of $d_{P-O}$ (single bond length) in the given compounds $X$ and $Y$:
$P_4 \xrightarrow{O_2} 'X'$
$P_4 \xrightarrow{O_2 \text{ (excess)}} 'Y'$

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Define the order of $d_{P-O}$ (single bond length) in the given compounds $X$ and $Y$:$P_4 \xrightarrow{O_2} 'X'$$P_4 \xrightarrow{O_2 \text{ (excess)}} 'Y'$