The decreasing values of bond angles from ${\rm{N}}{{\rm{H}}_3}$($106{\;^0}$) to ${\rm{Sb}}{{\rm{H}}_3}$ ($101{\;^0}$) down group-$15$ of the periodic table is due to:

Increasing $bp$ - $bp$ repulsion

Increasing $p$-orbital character in $s{p^3}$

Decreasing $lp$ - $bp$ repulsion

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NEET UG CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

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The decreasing values of bond angles from

\mathrm{N}{\mathrm{H}_3}

(

106{\;^0}

) to

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

(

101{\;^0}

) down group-

15

of the periodic table is due to:

View answer and explanation

Correct option: D

Decreasing electronegativity

Option D (Decreasing electronegativity) is correct because as electronegativity decreases down the group, the bond pair electrons are further away from the central atom, reducing bond angle.

NEET UG CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

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