question_answer

In the following molecule: \[C{{H}_{3}}-C\equiv \overset{1}{\mathop{C}}\,-\overset{2}{\mathop{C}}\,{{H}_{2}}-\overset{\begin{smallmatrix}  O \\  3|| \end{smallmatrix}}{\mathop{C}}\,-OH\] arrange the carbon atoms numbered\[C-1,\,C-2\]  and\[C-3\] in the decreasing order of s-character.

Answer:

                Hybridisation:\[C-1=sp,\,C-2=s{{p}^{3}},\,C-3=s{{p}^{2}}\] . Hence, s-character will be in the order: \[\underset{(50%)}{\mathop{C-1}}\,\,\,\,>\,\,\,\underset{(33%)}{\mathop{C-3}}\,\,\,\,>\,\,\underset{(25%)}{\mathop{C-2}}\,\].