Which has a larger bond angle between PH_3 an | vedclass

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(B) The bond angle in $PH_4^+$ is larger than in $PH_3$.In $PH_4^+$,the phosphorus atom is $sp^3$ hybridized with a tetrahedral geometry,resulting in

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$PH_4^+$

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(B) The bond angle in $PH_4^+$ is larger than in $PH_3$.In $PH_4^+$,the phosphorus atom is $sp^3$ hybridized with a tetrahedral geometry,resulting in a bond angle of approximately $109.5^{\circ}$.In $PH_3$,the phosphorus atom is also $sp^3$ hybridized,but it contains one lone pair of electrons. According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion,which compresses the bond angle to approximately $93.6^{\circ}$.

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

Which has a larger bond angle between $PH_3$ and $PH_4^+$? Why?

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

$PF_5$ $a / b$

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

$PCl_5$ $g / h$

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According to $VSEPR$ theory,in which species do all the atoms lie in the same plane?
$1.$ $CH^{+}_3$
$2.$ $CH^{-}_3$

The structure of $NH_3$ is:

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