
The molecule of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] has \[180^\circ \] bond angle. It can be explained on the basis of
A) \[s{p^3}\] hybridization
B) \[s{p^2}\] hybridization
C) \[sp\] hybridization
D) \[{d^2}s{p^3}\] hybridization
Answer
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Hint: We know that, in a carbon dioxide molecule, two atoms of oxygen form bonds with the carbon atom. To know the bond angle of a compound, we have to first identify the molecular hybridization of it.
Complete step by step solution:Let's understand how to identify hybridization of a molecule. For that, we have to count the groups that surround the central atom. The surrounding groups are atoms that form covalent bonds with the central atom and the lone pairs of the central atom.
If the count of groups is four, then the hybridization is \[s{p^3}\] . If the count of groups is three, then the molecular hybridization is \[s{p^2}\] and if the count is two, then the hybridization is \[sp\] .
Let's find the hybridization in \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]. The structure of the carbon dioxide molecule is \[{\rm{O}} = {\rm{C}} = {\rm{O}}\] . So, the carbon atom is surrounded by two groups and a non lone pair is present on the atom of carbon. Therefore, \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]molecule is \[sp\]hybridized.
And we know, \[s{p^3}\] hybridization means bond angle is \[109^\circ 28'\] , \[s{p^2}\] hybridization means the bond angle is \[120^\circ \] and the \[sp\] hybridization means the bond angle is \[180^\circ \] .
Therefore, \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] molecule has a bond angle of \[180^\circ \].
Hence, option C is right.
Note: It is to be noted that the bond angle gets affected due to the presence of lone pairs on the central atom. The more the number of lone pairs on the central atom, the less the bond angle be. For example, the bond angle of \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] is less than \[{\rm{C}}{{\rm{H}}_{\rm{4}}}\] , although both are \[s{p^3}\] hybridized.
Complete step by step solution:Let's understand how to identify hybridization of a molecule. For that, we have to count the groups that surround the central atom. The surrounding groups are atoms that form covalent bonds with the central atom and the lone pairs of the central atom.
If the count of groups is four, then the hybridization is \[s{p^3}\] . If the count of groups is three, then the molecular hybridization is \[s{p^2}\] and if the count is two, then the hybridization is \[sp\] .
Let's find the hybridization in \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]. The structure of the carbon dioxide molecule is \[{\rm{O}} = {\rm{C}} = {\rm{O}}\] . So, the carbon atom is surrounded by two groups and a non lone pair is present on the atom of carbon. Therefore, \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]molecule is \[sp\]hybridized.
And we know, \[s{p^3}\] hybridization means bond angle is \[109^\circ 28'\] , \[s{p^2}\] hybridization means the bond angle is \[120^\circ \] and the \[sp\] hybridization means the bond angle is \[180^\circ \] .
Therefore, \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] molecule has a bond angle of \[180^\circ \].
Hence, option C is right.
Note: It is to be noted that the bond angle gets affected due to the presence of lone pairs on the central atom. The more the number of lone pairs on the central atom, the less the bond angle be. For example, the bond angle of \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] is less than \[{\rm{C}}{{\rm{H}}_{\rm{4}}}\] , although both are \[s{p^3}\] hybridized.
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