Which of the following statement is true for $H_{2} S$ ?

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The sulphur atom is ${sp}^{3}$ hybridized and bonding take place in excited state.

(b) and (c) both

Its molecular geometry is non-linear and $H - S - H$ bond angle is smaller than $H - 0 - H$

bond angle in $H_{2} O$

$H_{2} S$ and ${BF}_{3}$ form adduct readily through coordinate bond between boron and sulphur

answer is D.

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Detailed Solution

The sulphur atom is ${sp}^{3}$ hybridized and bonding takes place in ground state. Two filled orbitals $3 s^{2}$ and $3 p_{x}^{2}$

$H_{2} S \to {BF}_{3}$ has coordinated bond, S from $H_{2} S$ gives lone pair and B receives it.
Bond angle difference leads to form hydrogen bonds in water.
so that two lone pairs of oxygen in water engaged in hydrogen bond.
Formation of hydrogen bonds reduce the lone pair - bond pair repulsion. That's why bond angle is greater in water when compare to hydrogen sulfide. In $H_{2} S$ the ${sp}^{3}$ hybrid, orbitals are bigger than $H_{2} O$ and so the bond pair-bond pair repulsion decreases.