Question

Which of the following orders is/are correct?
$1.Sb{H}_3>N{H}_3>As{H}_3>P{H}_3$ (Boiling point)
$2.N{H}_3>P{H}_3>As{H}_3>Sb{H}_3$ (Thermal stability)
$3.N{H}_3>P{H}_3>As{H}_3>Sb{H}_3$ (Basic character)
$4.N{H}_3>P{H}_3>As{H}_3>Sb{H}_3$ (Bond angle)

• A. $1,2$ and $3$
• B. $2,3$ and $4$
• C. $1,3$ and $4$
• D. $1,2,3$ and $4$

Answer 1

The correct option is D $1,2,3$ and $4$

$1.$ $N{H}_3$ has a higher boiling point than $As{H}_3$ and $P{H}_3$ on account of H-Bonding. $Sb{H}_3$ has the highest boiling point as it has the highest molecular weight and $As{H}_3$ has a higher boiling point than $P{H}_3$ due to its higher molecular weight.

$2.$ As the size of an element increases down the group, the strength of the $M-H$ bond decreases and thus, thermal stability decreases.

$3.$ Basicity decreases down the group because the size of the central atom increases down the group.

$4.$ The bond angles are $N{H}_3={107.8}^{\circ };P{H}_3={93.6}^{\circ };As{H}_3={91.8}^{\circ };Sb{H}_3={91.33}^{\circ }$
As we go down the group, the size of the central atom increases and hence, bond pair-bond pair repulsion decreases down the group.