Question

Which of the following compounds is not linear?
A) \[{\rm{SnC}}{{\rm{l}}_{\rm{2}}}\]
B) \[{\rm{HCl}}\]
C) \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]
D) \[{\rm{HgC}}{{\rm{l}}_{\rm{2}}}\]

Answer 1

Hint: VSEPR theory helps to predict the molecular shapes considering the lone pairs and bond pairs around the centrally placed atom. Here, first we have to find out hybridization of these molecules.

Formula Used:\[H = \dfrac{{V + X - C + A}}{2}\]
Here, V is for the valence electrons, X is for the count of monovalent atoms and C denotes the cationic charge and A denotes the anionic charge.

Complete step by step solution:A linear shape is mostly acquired by the compounds which are \[sp\] hybridized. If the value of H is 2, then the molecule is sp hybridized.
Let’s find out the hybridization of all the compounds one by one.
Option A is \[{\rm{SnC}}{{\rm{l}}_{\rm{2}}}\]. For \[{\rm{SnC}}{{\rm{l}}_{\rm{2}}}\], count of valence electrons=2, count of monovalent atoms=2
 \[H = \dfrac{{4 + 2}}{2} = 3\]
So, \[{\rm{SnC}}{{\rm{l}}_{\rm{2}}}\]is \[s{p^2}\] hybridized and its structure is trigonal planar.
Option B is \[{\rm{HCl}}\]. As there is only one bond and electronegativity of chlorine atoms is more than hydrogen atom. So, its shape is linear.
Option C is \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]. count of valence electrons=4, count of monovalent atoms=0
\[H = \dfrac{4}{2} = 2\]
As H is 2, so, it is \[sp\] hybridized and its shape is linear.
Option D is \[{\rm{HgC}}{{\rm{l}}_{\rm{2}}}\]. count of valence electrons=2, count of monovalent atoms=2
\[H = \dfrac{4}{2} = 2\]
As H is 2, so, it is also \[sp\] hybridized and its shape is also linear.

Therefore, option A is right.

Note: Always remember that, the linear shape is also obtained by \[s{p^3}d\] hybridized molecules on one condition. If the count of lone pairs is three and the bond pair is two, then the \[s{p^3}d\] hybridized molecule acquires a linear shape.