The correct order of equatorial FSF bond angl | vedclass

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(C) According to Bent's rule,the bond angle is inversely proportional to the $p$-character of the hybrid orbitals forming the bonds.$(I) \ SF_4$: The

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$(I) > (II) > (III)$

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(C) According to Bent's rule,the bond angle is inversely proportional to the $p$-character of the hybrid orbitals forming the bonds.$(I) \ SF_4$: The lone pair occupies an equatorial position in the trigonal bipyramidal geometry. The lone pair-bond pair repulsion compresses the equatorial $FSF$ bond angle to approximately $101^{\circ}$.$(II) \ OSF_4$: The highly electronegative oxygen atom is double-bonded to the sulfur atom. According to Bent's rule,the double bond prefers an orbital with more $s$-character,leaving more $p$-character for the $S-F$ bonds. This results in a smaller bond angle compared to $SF_4$.$(III) \ H_2CSF_4$: The $CH_2$ group is attached via a double bond. The presence of the $CH_2$ group further influences the hybridization,leading to an equatorial $FSF$ bond angle of approximately $97^{\circ}$.Comparing the effects,the bond angle decreases as the $p$-character of the $S-F$ bonds increases due to the presence of more electronegative or double-bonded substituents. The correct order is $(I) > (II) > (III)$.

Molecule	$P-X$ (axial) bond length / $P-X$ (equatorial) bond length
$PF_5$	$a / b$
$PF_4CH_3$	$c / d$
$PF_3(CH_3)_2$	$e / f$
$PCl_5$	$g / h$

The correct order of equatorial $FSF$ bond angle in the following compounds:
$(I) \ SF_4$ $(II) \ OSF_4$ $(III) \ H_2CSF_4$

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Molecule $P-X$ (axial) bond length / $P-X$ (equatorial) bond length

$PF_5$ $a / b$

$PF_4CH_3$ $c / d$

$PF_3(CH_3)_2$ $e / f$

$PCl_5$ $g / h$

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The correct order of equatorial $FSF$ bond angle in the following compounds:$(I) \ SF_4$ $(II) \ OSF_4$ $(III) \ H_2CSF_4$