In the following molecule,determine the hybri | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To determine the hybridisation of a carbon atom,we count the number of sigma $(\sigma)$ bonds and lone pairs (if any) attached to it.$1$. Carbon $

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) To determine the hybridisation of a carbon atom,we count the number of sigma $(\sigma)$ bonds and lone pairs (if any) attached to it.$1$. Carbon $a$ is part of a methyl group $(-CH_3)$. It is bonded to three hydrogen atoms and one carbon atom,forming four single bonds. Thus,it has four $\sigma$ bonds and is $sp^{3}$ hybridised.$2$. Carbon $b$ is part of a double bond $(C=C)$. It is bonded to one carbon atom (via a double bond,which counts as one $\sigma$ bond),one hydrogen atom,and one oxygen atom. It has three $\sigma$ bonds and is $sp^{2}$ hybridised.$3$. Carbon $c$ is part of the benzene ring. It is bonded to two other carbon atoms in the ring and one oxygen atom. It is involved in a double bond within the aromatic ring,giving it three $\sigma$ bonds. Thus,it is $sp^{2}$ hybridised.Therefore,the hybridisation of $a, b$ and $c$ are $sp^{3}, sp^{2}$ and $sp^{2}$ respectively.

In the following molecule,determine the hybridisation of carbon atoms $a, b$ and $c$ respectively:

Similar Questions

The hybridization of the central atom in $ICl^{+}_2$ is

Identify the pair of molecules in which the central atom has the same hybridization.

The ion that has $sp^3d^2$ hybridization for the central atom is

The hybridization of atomic orbitals of nitrogen in $NO_2^+$,$NO_3^-$,and $NH_4^+$ are:

In compound $X$,all the bond angles are exactly $109^o 28'$,$X$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module