Bond angles of ${\rm{N}}{{\rm{H}}_3},{\rm{P}}{{\rm{H}}_3},{\rm{As}}{{\rm{H}}_3}$ and ${\rm{Sb}}{{\rm{H}}_3}$ is in the order

${\rm{N}}{{\rm{H}}_3} > P{{\rm{H}}_3} > As{{\rm{H}}_3} > Sb{{\rm{H}}_3}$

${\rm{P}}{{\rm{H}}_3} > As{{\rm{H}}_3} > Sb{{\rm{H}}_3} > N{{\rm{H}}_3}$

${\rm{Sb}}{{\rm{H}}_3} > As{{\rm{H}}_3} > P{{\rm{H}}_3} > N{{\rm{H}}_3}$

${\rm{Sb}}{{\rm{H}}_3} > As{{\rm{H}}_3} > N{{\rm{H}}_3} > P{{\rm{H}}_3}$

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JEE CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

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Understanding +4 | -1

Bond angles of

\mathrm{N}{\mathrm{H}_3},\mathrm{P}{\mathrm{H}_3},\mathrm{As}{\mathrm{H}_3}

and

\mathrm{Sb}{\mathrm{H}_3}

is in the order

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{H}_3} > P{\mathrm{H}_3} > As{\mathrm{H}_3} > Sb{\mathrm{H}_3}

Option D (

\mathrm{N}{\mathrm{H}_3} > P{\mathrm{H}_3} > As{\mathrm{H}_3} > Sb{\mathrm{H}_3}

) is correct because as the size of the central atom increases down the group, the bond angle decreases due to decreasing electronegativity and increasing bond length.

JEE CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

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