Question

$\underset{6}{C{H}_3}-\underset{5}{CH}=\underset{4}{CH}-\underset{3}{C{H}_2}-\underset{2}{C}\equiv \underset{1}{CH}$

In the hydrocarbon the state of hybridization of carbons 1,3 and 5 are:

• A. $s{p}^3,s{p}^2,sp$
• B. $s{p}^2,sp,s{p}^3$
• C. $sp,s{p}^3,s{p}^2$
• D. $sp,s{p}^2,s{p}^3$

Answer 1

Answer: (C)

$sp,s{p}^3,s{p}^2$

$\underset{6}{C{H}_3}-\underset{5}{CH}=\underset{4}{CH}-\underset{3}{C{H}_2}-\underset{2}{C}\equiv \underset{1}{CH}$

Hybridization $1,3,5$

$1\Rightarrow SP$

$3=S{P}^3$

$5=S{P}^2$