A) \[\operatorname{sp}-s{{p}^{2}}\]
B) \[{{\operatorname{sp}}^{3}}-s{{p}^{3}}\]
C) \[\operatorname{sp}-s{{p}^{3}}\]
D) \[{{\operatorname{sp}}^{2}}-s{{p}^{3}}\]
Correct Answer: D
Solution :
\[{{\operatorname{sp}}^{2}}-s{{p}^{3}}\] When both double and triple bonds are present in a molecule and if there is a choice, the, numbering is always done from that end which is nearer to the double bond. \[~\overset{1}{\mathop{C}}\,{{H}_{2}}=\overset{2}{\mathop{C}}\,H\overset{3}{\mathop{C}}\,{{H}_{2}}\overset{4}{\mathop{C}}\,{{H}_{2}}\overset{5}{\mathop{C}}\,\equiv \overset{6}{\mathop{C}}\,H\] \[\operatorname{Thus}, {{C}_{2}}-{{C}_{3}} bond is of type s{{p}^{2}}-s{{p}^{3}}.\]You need to login to perform this action.
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