The correct order of the bond angles of the c | vedclass

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(D) $BeCl_2$ has a linear structure,so the bond angle is $180^{\circ}$.$BF_3$ has a trigonal planar structure,so the bond angle is $120^{\circ}$.$SiCl

Vedclass · Accepted Answer

$BeCl_2 > BF_3 > SiCl_4 > SF_6$

Vedclass · Answer

(D) $BeCl_2$ has a linear structure,so the bond angle is $180^{\circ}$.$BF_3$ has a trigonal planar structure,so the bond angle is $120^{\circ}$.$SiCl_4$ has a tetrahedral structure,so the bond angle is $109.5^{\circ}$.$SF_6$ has an octahedral structure,so the bond angle is $90^{\circ}$.Therefore,the correct order of bond angles is $BeCl_2 (180^{\circ}) > BF_3 (120^{\circ}) > SiCl_4 (109.5^{\circ}) > SF_6 (90^{\circ})$.

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 the bond angles of the compounds $SiCl_4, BF_3, BeCl_2$ and $SF_6$ is

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