question_answer

The molecule of \[C{{O}_{2}}\] has 180° bond angle. It can be explanid on the basis of                                    [AFMC 2004]

A)                 \[s{{p}^{3}}\] hybridisation         

B)                 \[s{{p}^{2}}\] hybridisation

C)                 \[sp\] hybridisation        

D)                 \[{{d}^{2}}s{{p}^{3}}\] hybridization

Correct Answer: C

Solution :

           \[{{C}_{\text{ground state}}}=2{{s}^{2}},2{{p}_{x}}^{1}{{p}_{y}}^{1}\]; \[{{C}_{\text{excited state }}}=2{{s}^{1}},2{{p}_{x}}^{1}{{p}_{y}}^{1}{{p}_{z}}^{1}\]                    \[{{O}_{\text{ground state }}}=2{{s}^{2}},2{{p}_{x}}^{2}{{p}_{y}}^{1}{{p}_{z}}^{1}\]                    In the formation of \[C{{O}_{2}}\] molecule, hybridization of orbitals of carbon occur only to a limited extent involving only one s and one p orbitals there is thus sp hybridisation of valence shell orbitals of the carbon atom resulting in the formation of two sp hybrid orbitals.