Question

In the hydrocarbon,
6 5 4 3 2 1
$C{H}_3-CH=CH-C{H}_2-C\equiv CH$

The state of hybridization of carbons 1, 3 and 5 are in the following sequence:

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

Answer 1

Answer: (D)

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

$654321$

$C{H}_3-CH=CH-C{H}_3-C\equiv CH$

We have been asked to find out the hybridization of carbon as $1,3$ and $5$, just count the no. of sigma $\left(\sigma \right)$bonds to decide the state of hybridization of the Carbon

$1^{st}$ carbon: $\to 2\sigma$ bonds and $2\pi$bonds

$\therefore sp$hybridized

$3^{rd}$ carbon: $\to 4\sigma$ bonds and no $\pi$ bonds.

$\therefore s{p}^3$ hybridized

$5^{th}$ carbon :$\to 3\sigma$ bonds and $1\pi$ bond

$s{p}^2$ hybridized

$\therefore$ the correct answer is option D.$-sp$ ($1^{st}$carbon), $s{p}^3$ ($3^{rd}$carbon) and $s{p}^2$ ($5^{th}$carbon).