The decreasing values of bond angles from $\mathrm{NH}_{3}(106^{\circ})$ to $\mathrm{SbH}_{3}(101^{\circ})$ down group-15 of the periodic table is due to

(D) decreasing electronegativity

(B) increasing $p$-orbital character in $sp^{3}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Chemical Bonding and Molecular Structure

The decreasing values of bond angles from

NH_{3} (10 6^{\circ})

SbH_{3} (10 1^{\circ})

down group-15 of the periodic table is due to

(A)increasing bp-bp repulsion
(B)increasing $p$ -orbital character in $s p^{3}$
(C)decreasing lp-bp repulsion
(D)decreasing electronegativity✓ Correct

Answer: (D)

Solution

Down group 15 from

N H_{3}

Sb H_{3}

the central atom grows larger and distinctly less electronegative. As its electronegativity falls, the bonding pairs are held more loosely and lie further from the central atom, so the repulsion between adjacent bond pairs weakens and the bonds close up; the hydride angle accordingly shrinks toward the pure

p

-orbital value near

9 0^{\circ}

. The steady fall in bond angle is therefore caused by the decreasing electronegativity of the central atom.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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JEE Main · 2006 · 30 Apr 2006 · Offline · Chemical Bonding and Molecular Structure

Solution

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