question_answer

The molecule of \[\text{C}{{\text{O}}_{\text{2}}}\]has angle \[\text{18}{{\text{0}}^{\text{o}}}.\] It can be    explained on the basis of:

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

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

C) \[sp\]hybridisation         

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

Correct Answer: C

Solution :

Carbon (At .No = 6 ) \[EC=2,\,4(1{{s}^{2}}2{{s}^{2}}2{{p}^{2}})\] in excited state sp hybrids atom orbital which from n bond does not take part in hybridisation. Carbon has no lone pair so, according to VSEPR theory the sp hybrid orbitals stay as far away as possible. In this case, the bond angle being \[\text{18}{{\text{0}}^{\text{o}}}.\] \[s{{p}^{3}}\] hybrid molecules are tetrahedral with bond angle \[{{109}^{o}}28',\,\,s{{p}^{2}}\]hybrid orbitals are triangular planar with bond angle \[\text{12}{{\text{0}}^{\text{o}}}\text{.}\] \[\overset{s{{p}^{2}}}{\mathop{{{H}_{2}}C}}\,=\overset{s{{p}^{2}}}{\mathop{CH}}\,-\overset{sp}{\mathop{C}}\,\equiv \overset{sp}{\mathop{C}}\,-\overset{s{{p}^{3}}}{\mathop{C}}\,{{H}_{3}}\]