Question

$\underset{6}{C}{H}_3-\underset{5}{C}H=\underset{4}{C}H-\underset{3}{C}{H}_2-\underset{2}{C}\equiv \underset{1}{C}H$
The hybridizations of the carbons numbered 1, 3 and 5 in the given hydrocarbon sequentially 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

The correct option is C $sp,s{p}^3,s{p}^2$

From their steric number, as the carbon atoms have no lone pairs and $C_1,C_3$ and $C_5$ are bonded to 2, 4 and 3 atoms respectively, their respective hybridizations are $sp,s{p}^3$ and $s{p}^2$.

Also,
$-C-C-⟹s{p}^3-C=C-⟹s{p}^2-C\equiv C-⟹sp=C=C=⟹sp$

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

$\stackrel{s{p}^3}{C}{H}_3-\stackrel{s{p}^2}{C}H=\stackrel{s{p}^2}{C}H-\stackrel{s{p}^3}{C}{H}_2-\stackrel{sp}{C}\equiv \stackrel{sp}{C}H$