Question

Which of the following represents the given mode of hybridisation $s{p}^2-s{p}^2-sp-sp$ from left to right ?

• A. $H_{2}C=CH-C\equiv N$
• B. $HC\equiv C-C\equiv CH$
• C. $H_{2}C=C=C=C{H}_2$
• D. $H_{2}C=C-C=C{H}_2$

Answer 1

Answer: (A)

$H_{2}C=CH-C\equiv N$

For finding hybridisation, count the no. of sigma bonds. In option A, carbon $1$ from left side has $3$ sigma bonds, therefore $s{p}^2$, carbon $2$ also has $3$ sigma bonds so $s{p}^2$, carbon $3$ has $2$ sigma bonds therefore $sp$ and nitrogen also has $2$ sigma bonds therefore $sp$ hybridised.