Animated Solution for Chemistry - Organic Chemistry: In the following molecules, hybridisation of carbon $a,b$ and $c$ respectively are

Imagine you are a molecular detective, and your mission today is to uncover the hidden architecture of a fascinating organic molecule. We are given a structure and asked to find the hybridization of three specific carbon atoms, labeled a, b, and c.

At first glance, organic molecules can look like a complex web of lines and letters. But don't get intimidated! Every carbon atom follows a strict set of rules, and once you know the secret code, reading the molecule becomes as easy as reading a book.

Steric Number

To find the hybridization of any atom, we use a powerful and simple tool called the Steric Number.

The steric number is simply the sum of the number of sigma ($\sigma$) bonds and the number of lone pairs around the atom.

For carbon atoms in stable, neutral molecules, the number of lone pairs is almost always zero. This makes our job incredibly easy: we just need to count the sigma bonds!

Remember the golden rules of bonding: - A single bond contains exactly $1\sigma$ bond. - A double bond contains $1\sigma$ bond and $1\pi$ bond. - A triple bond contains $1\sigma$ bond and $2\pi$ bonds.

Based on the steric number, the hybridization is assigned as follows: - Steric Number = 4 $⟹s{p}^3$ - Steric Number = 3 $⟹s{p}^2$ - Steric Number = 2 $⟹sp$

Let's zoom in on carbon a. This carbon is part of a methyl group ($-C{H}_3$).

If we expand its structure, we see that it is bonded to three hydrogen atoms and one adjacent carbon atom. All of these connections are single bonds.

Therefore, carbon a forms exactly 4 $\sigma$ bonds.

Since there are no lone pairs, its steric number is 4. A steric number of 4 perfectly corresponds to $s{p}^3$ hybridization.

Now, let's shift our focus to carbon b. This carbon is part of an alkene chain, meaning it is involved in a double bond.

Let's count its connections carefully. It forms a single bond with a hydrogen atom, a single bond with the methyl carbon, and a double bond with the next carbon in the chain.

Remember our golden rule: a double bond only contributes one $\sigma$ bond (the other is a $\pi$ bond).

So, carbon b has a total of 3 $\sigma$ bonds. A steric number of 3 means that carbon b is $s{p}^2$ hybridized.

Finally, let's examine carbon c, which sits proudly inside the benzene ring.

Benzene rings are characterized by alternating single and double bonds (or a delocalized $\pi$ system). If we look at the specific localized structure drawn, carbon c is bonded to one hydrogen atom, and two other carbons in the ring—one via a single bond and one via a double bond.

Just like carbon b, this gives carbon c exactly 3 $\sigma$ bonds.

Thus, its steric number is also 3, making carbon c $s{p}^2$ hybridized.

We have successfully decoded the entire sequence!

The hybridization of carbons a, b, and c are $s{p}^3$, $s{p}^2$, and $s{p}^2$ respectively.

Comparing this with our options, we can confidently select Option (c) as the correct answer.

This is a classic, high-yield concept for JEE and NEET. Master the steric number rule, and you will never get a hybridization question wrong!