Question

Which molecule is linear?
A) \[{\rm{N}}{{\rm{O}}_{\rm{2}}}\]
B) \[{\rm{Cl}}{{\rm{O}}_{\rm{2}}}\]
C) \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]
D) \[{{\rm{H}}_{\rm{2}}}{\rm{S}}\]

Answer 1

Hint: Here, first we will calculate the hybridization of the central atom of all the compounds. If the value of hybridization of a molecule is two, then the molecule is sp hybridised and the shape of the molecule is linear.

Formula used: \[H = \dfrac{{V + X - C + A}}{2}\]; Here, H stands for hybridization, V is the count of valence electrons, X is the count of monovalent groups bonded to the central atom, C stands for cationic charge and A stands for the charge of anion.

Complete Step by Step Answer:
Let’s find out the hybridization of each compound.
For \[{\rm{N}}{{\rm{O}}_{\rm{2}}}\], Count of valence electrons=5, count of monovalent atoms=0
 \[H = \dfrac{{5 + 0}}{2} = 2.5\]
So, the value of H is 2.5, this indicates that the hybridization N atom is not sp. So, this molecule does not have a linear shape.

Now, for \[{\rm{Cl}}{{\rm{O}}_{\rm{2}}}\], Count of valence electrons=7, count of monovalent atoms = 0. Therefore, \[H = \dfrac{{7 + 0}}{2} = 3.5\]. The value of H is 3.5, this indicates that the hybridization of the H atom is not sp. So, this molecule has no linear shape. For\[{\rm{C}}{{\rm{O}}_{\rm{2}}}\], Count of valence electrons=4, count of monovalent atoms = 0. Therefore, \[H = \dfrac{{4 + 0}}{2} = 2\]. The value of H is 2, this indicates that the hybridization of the C atom is sp. So, this molecule has a linear shape.

For \[{{\rm{H}}_{\rm{2}}}{\rm{S}}\], the valence electrons of S atom=6, Count of monovalent atoms = 2. Therefore, \[H = \dfrac{{6 + 2}}{2} = 4\]. The value of H is 4, this indicates that the hybridization of the C atom is \[s{p^3}\]. So, this molecule does not have a linear shape but is angular.
Therefore, option C is right.

Note: It is to be noted that, if the hybridization is \[s{p^3}\], the electron geometry of the molecule is tetrahedral. And if the molecule has\[s{p^2}\] hybridization, then the molecule's electron geometry is trigonal planar.