Question

The correct order of bond angles (smaller first) in \[{{\rm{H}}_{\rm{2}}}{\rm{S,}}\,{\rm{N}}{{\rm{H}}_{\rm{3}}}{\rm{,B}}{{\rm{F}}_{\rm{3}}}{\rm{,Si}}{{\rm{H}}_{\rm{4}}}\]is,
A) \[{{\rm{H}}_{\rm{2}}}{\rm{S < N}}{{\rm{H}}_{\rm{3}}}{\rm{ < Si}}{{\rm{H}}_{\rm{4}}} < {\rm{B}}{{\rm{F}}_{\rm{3}}}\]
B) \[{\rm{N}}{{\rm{H}}_{\rm{3}}}{\rm{ < }}{{\rm{H}}_{\rm{2}}}{\rm{S}} < {\rm{Si}}{{\rm{H}}_{\rm{4}}} < {\rm{B}}{{\rm{F}}_{\rm{3}}}\]
C) \[{{\rm{H}}_{\rm{2}}}{\rm{S}} < {\rm{N}}{{\rm{H}}_{\rm{3}}} < {\rm{Si}}{{\rm{H}}_{\rm{4}}} < {\rm{B}}{{\rm{F}}_{\rm{3}}}\]
D) \[{{\rm{H}}_{\rm{2}}}{\rm{S}} < {\rm{N}}{{\rm{H}}_{\rm{3}}} < {\rm{B}}{{\rm{F}}_{\rm{3}}} < {\rm{Si}}{{\rm{H}}_{\rm{4}}}\]

Answer 1

Hint: To check the bond angle, we have to check the hybridization first. If the hybridizations are \[sp,s{p^2},s{p^3}\] , then the bond angles will be \[180^\circ ,120^\circ ,109^\circ 28'\] respectively.

Complete step by step solution:Let's check the hybridization of the compounds.
In \[{{\rm{H}}_{\rm{2}}}{\rm{S}}\] , two atoms of hydrogen are bonded to the sulfur atom and the sulfur atom has two lone pairs on it. So, the total count of groups is 4, therefore, its hybridization is \[s{p^3}\].

Let's check the hybridization of \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] . In this compound, three atoms of hydrogen are bonded to nitrogen atoms. And, one lone pair is on the nitrogen atom. So, here also, the count of groups is 4, therefore, its hybridization is \[s{p^3}\].

Now, we will find the hybridization of \[{\rm{Si}}{{\rm{H}}_{\rm{4}}}\] . Here, four atoms of hydrogen form a bond with the silicon atom. And there is no pair present on the N atom. So, its hybridization is \[s{p^3}\].
In \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\] , boron is bonded to three fluorine atoms. And there is no lone pair on the Boron atom. So, the total count of groups is 3. Therefore, its hybridization is \[s{p^2}\] .
As we know, the bond angle of an \[s{p^2}\]hybridized atom is larger than \[s{p^3}\]hybridized atom. So, the bond angle of \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\] is highest.
Now, we have to compare the bond angle of all \[s{p^3}\]hybridized compounds. If the hybridization of a set of compounds is the same, then we have to count the count of lone pairs on this compound. The more the count of lone pairs, the less the bond angle be.
Here, in \[{{\rm{H}}_{\rm{2}}}{\rm{S}}\], count of lone pairs present=2
In \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\], count of lone pairs=1
In \[{\rm{Si}}{{\rm{H}}_{\rm{4}}}\], count of lone pairs=0
So, the order of bond angle of these compounds is,

\[{{\rm{H}}_{\rm{2}}}{\rm{S}} < {\rm{N}}{{\rm{H}}_3} < {\rm{Si}}{{\rm{H}}_{\rm{4}}} < {\rm{B}}{{\rm{F}}_{\rm{3}}}\]

Option ‘C’ is correct

Note: The VSEPR theory guides us to understand the bond angle of many compounds. This theory gives the order of repulsion of the bond pair (bp) and lone pair (lp) is bp-bp