Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Underlined carbon is ${\text{sp}}^3$ hybridised in

Have you ever wondered how carbon, an element with only two unpaired electrons in its ground state, manages to form four perfectly symmetrical bonds in molecules like methane? The answer lies in a beautiful mathematical and physical concept called hybridization.

When atoms form bonds, their atomic orbitals (like $s$ and $p$ orbitals) mix together to form new, identical hybrid orbitals. This mixing allows the atom to form stronger, more stable bonds. For carbon, the most common hybridization states are $s{p}^3$, $s{p}^2$, and $sp$. But how do we quickly figure out which state a carbon atom is in?

To find the hybridization of any atom, we use a simple tool called the Steric Number (SN). The steric number tells us how many electron domains (regions of electron density) are surrounding the central atom.

The formula is beautifully simple:

For a neutral carbon atom in an organic molecule, life is even easier. Carbon almost never has lone pairs because it uses all four of its valence electrons to form bonds. Therefore, for carbon, the steric number is simply the number of $\sigma$ bonds it forms.

- If carbon forms 4 single bonds, it has $4\sigma$ bonds. $\text{SN}=4⟹s{p}^3$ hybridization. - If carbon forms 1 double bond and 2 single bonds, it has $3\sigma$ bonds and $1\pi$ bond. $\text{SN}=3⟹s{p}^2$ hybridization. - If carbon forms 1 triple bond and 1 single bond (or two double bonds), it has $2\sigma$ bonds and $2\pi$ bonds. $\text{SN}=2⟹sp$ hybridization.

Let's apply this powerful tool to the molecules given in our problem.

Option (a): Propene ($C{H}_3\underline{C}H=C{H}_2$) Let's zoom in on the underlined carbon. It is bonded to a methyl group ($C{H}_3$), a hydrogen atom ($H$), and it shares a double bond with a $C{H}_2$ group. A double bond consists of one $\sigma$ bond and one $\pi$ bond. Counting the $\sigma$ bonds around this carbon, we get $1+1+1=3$. Since the steric number is 3, this carbon is $s{p}^2$ hybridized.

Option (c): Acetamide ($C{H}_3\underline{C}ON{H}_2$) Here, the underlined carbon is part of a carbonyl group. It is single-bonded to a methyl group ($C{H}_3$), single-bonded to an amino group ($N{H}_2$), and double-bonded to an oxygen atom ($O$). Again, counting the $\sigma$ bonds gives us $1+1+1=3$. The steric number is 3, meaning this carbon is also $s{p}^2$ hybridized.

Option (d): Propanenitrile ($C{H}_{3}C{H}_2\underline{C}N$) In this molecule, the underlined carbon is part of a nitrile group. It is single-bonded to an ethyl group ($C{H}_{3}C{H}_2$) and triple-bonded to a nitrogen atom ($N$). A triple bond consists of one $\sigma$ bond and two $\pi$ bonds. Counting the $\sigma$ bonds, we get $1+1=2$. With a steric number of 2, this carbon is $sp$ hybridized.

Option (b): Ethylamine ($C{H}_3\underline{C}{H}_{2}N{H}_2$) Finally, let's look at ethylamine. The underlined carbon is bonded to a methyl group ($C{H}_3$), an amino group ($N{H}_2$), and two hydrogen atoms ($H$). All of these connections are single bonds. Counting the $\sigma$ bonds, we get $1+1+1+1=4$. A steric number of 4 perfectly corresponds to $s{p}^3$ hybridization.

By systematically drawing out the structures and counting the $\sigma$ bonds, we easily identified the hybridization state of each carbon. The only molecule containing an $s{p}^3$ hybridized underlined carbon is ethylamine. Therefore, the correct answer is (b).

Always remember, when in doubt, draw it out! Visualizing the bonds is the ultimate defense against silly mistakes in organic chemistry.