Q.Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecules.

Concept: Hybridisation and Molecular Geometry

Hybridisation is determined by counting the number of sigma bonds + lone pairs on the central atom. The geometry follows from the steric number.

(a) H₂C=O (Formaldehyde)

Carbon forms three sigma bonds (two C–H, one C=O) and has no lone pairs → steric number = 3 → $sp^2$ hybridisation.

The molecule is trigonal planar around carbon (all four atoms lie in one plane, with H–C–H ≈ 116°).

(b) CH₃F (Methyl fluoride)

Carbon forms four sigma bonds (three C–H, one C–F) and has no lone pairs → steric number = 4 → $sp^3$ hybridisation.

The molecule is tetrahedral (bond angles ≈ 109.5°). …

The hybridisation state of carbon tells us how its atomic orbitals mix to form new hybrid orbitals that accommodate bonds and lone pairs. The key insight is simple: count the number of electron domains (regions of electron density) around the carbon atom. Each single bond, double bond, triple bond, or lone pair counts as one domain. The number of domains dictates the hybridisation and geometry.

Now let's apply this systematically to each molecule.

(a) H₂C=O (Formaldehyde)

1. Draw the Lewis structure

Carbon forms a double bond with oxygen and single bonds with two hydrogen atoms:

2. Count electron domains around carbon

• One $\ce{C=O}$ double bond (counts as 1 domain)
• Two $\ce{C-H}$ single bonds (each counts as 1 domain)

Total: 3 electron domains.

3. Determine hybridisation

Three domains → $sp^2$ hybridisation.

The carbon uses three $sp^2$ hybrid orbitals (one for each $\ce{C-H}$ $\sigma$-bond and one for the $\ce{C=O}$ $\sigma$-bond). The unhybridised $p$ orbital on carbon overlaps sideways with oxygen's $p$ orbital to form the $\pi$-bond of the double bond.

4. Molecular shape

The three electron domains arrange themselves in a trigonal planar geometry to minimise repulsion. All four atoms (two H, one C, one O) lie in the same plane, with bond angles close to $120°$.

(b) CH₃F (Fluoromethane)

1. Draw the Lewis structure

Carbon forms single bonds with three hydrogen atoms and one fluorine atom:

2. Count electron domains around carbon

• Three $\ce{C-H}$ single bonds (each counts as 1 domain)
• One $\ce{C-F}$ single bond (counts as 1 domain)

Total: 4 electron domains.

3. Determine hybridisation

Four domains → $sp^3$ hybridisation.

Carbon mixes one $2s$ and three $2p$ orbitals to form four equivalent $sp^3$ hybrid orbitals, each forming a $\sigma$-bond with H or F.

4. Molecular shape

The four electron domains arrange in a tetrahedral geometry, with bond angles approximately $109.5°$. The slight electronegativity difference between F and H causes minor distortion, but the overall shape remains tetrahedral.

(c) HC≡N (Hydrogen cyanide)

1. Draw the Lewis structure

Carbon forms a single bond with hydrogen and a triple bond with nitrogen:

2. Count electron domains around carbon

• One $\ce{C-H}$ single bond (counts as 1 domain)
• One $\ce{C≡N}$ triple bond (counts as 1 domain)

Total: 2 electron domains. …