Q.Assertion (A): Formaldehyde is a planar molecule.
Reason (R): It contains an $sp^2$ hybridised carbon atom.

Concept: Hybridisation and molecular geometry — $sp^2$ hybridisation gives trigonal planar shape.

Reasoning:

1. In formaldehyde ($\mathrm{H_2C=O}$), the carbon is double-bonded to oxygen and single-bonded to two hydrogens — three sigma bonds and no lone pairs.
2. This requires $sp^2$ hybridisation of carbon, which has a trigonal planar arrangement with bond angles of about $120^\circ$.
3. Therefore, the entire molecule is planar. …

The key to this question lies in understanding how hybridisation determines molecular shape. When a carbon atom is $sp^2$ hybridised, it forms three sigma bonds using three equivalent hybrid orbitals that lie in a plane at $120^\circ$ angles. The remaining unhybridised $p$ orbital sticks out perpendicular to that plane, forming a pi bond. This arrangement forces the entire molecule to be planar — all atoms lie in the same flat plane.

Formaldehyde ($\mathrm{H_2C=O}$) has a central carbon that is double-bonded to oxygen and single-bonded to two hydrogens. That double bond is one sigma bond and one pi bond. The carbon therefore has three sigma bonds (two C–H and one C–O sigma), with no lone pairs. Three sigma bonds with no lone pairs means $sp^2$ hybridisation, which gives trigonal planar geometry — and that means the molecule is planar.

Let's walk through the reasoning step by step.

1. Identify the central atom's bonding. In $\mathrm{H_2C=O}$, the carbon is bonded to two hydrogens via single bonds and to oxygen via a double bond. A double bond counts as one sigma bond and one pi bond. So the carbon forms three sigma bonds in total.

2. Count sigma bonds and lone pairs on carbon. Carbon has no lone pairs here. The number of sigma bonds (3) plus the number of lone pairs (0) equals 3. This is the steric number.

3. Determine hybridisation from steric number. Steric number 3 corresponds to $sp^2$ hybridisation. The three $sp^2$ hybrid orbitals arrange themselves in a trigonal planar geometry with bond angles of approximately $120^\circ$.

4. Check planarity. In $sp^2$ hybridisation, all three hybrid orbitals lie in the same plane. The pi bond formed by the unhybridised $p$ orbital does not affect the positions of the atoms — it simply sits above and below the plane. Therefore, all atoms (C, O, and both H) lie in one plane. The molecule is planar. …

Method: Hybridisation and Molecular Geometry Analysis

Concept: The shape of a molecule is determined by the hybridisation of its central atom, which in turn depends on the number of sigma bonds and lone pairs around it.

Steps:

1. Identify the central atom

   In formaldehyde ($\text{HCHO}$), the central atom is carbon.

2. Count sigma bonds and lone pairs on carbon

   • Carbon forms:
     • Two $\sigma$ bonds with hydrogen atoms
     • One $\sigma$ bond with oxygen atom
   • Total sigma bonds = 3
   • Lone pairs on carbon = 0

3. Determine hybridisation

   • Hybridisation formula: $\text{Number of hybrid orbitals} = \text{Sigma bonds} + \text{Lone pairs}$
   • Here: $3 + 0 = 3$ → $sp^2$ hybridisation

4. Predict geometry

   • $sp^2$ hybridised atoms have trigonal planar geometry (bond angle $\approx 120^\circ$)

5. Check planarity …

Common Mistakes & How to Avoid Them

Mistake 1: Thinking $sp^2$ hybridisation always guarantees planarity

The error: Students assume that if a central atom is $sp^2$ hybridised, the entire molecule must be planar. While the $sp^2$ centre itself is trigonal planar, substituents can rotate or the molecule may have non-planar conformations due to steric hindrance or ring strain.

How to avoid:

• Remember: $sp^2$ hybridisation gives local planarity around that atom (bond angles ~120°).
• For global planarity, check if all atoms lie in the same plane. In formaldehyde ($\mathrm{H_2C=O}$), the molecule is small and rigid — the $\pi$ bond locks the atoms in one plane.
• Key test: If a molecule has an $sp^2$ carbon with no bulky groups and no ring constraints, it is usually planar.

Mistake 2: Confusing “planar” with “flat” or “linear”

The error: Some students think “planar” means the molecule is flat like a sheet, but they forget that a molecule can be planar even if atoms are not in a straight line.

How to avoid:

• Planar means all atoms lie in a single plane.
• Formaldehyde:
  • Carbon is $sp^2$ hybridised → trigonal planar geometry around C.
  • The $\mathrm{C=O}$ double bond and two $\mathrm{C-H}$ bonds all lie in the same plane.
  • Result: The entire molecule is planar.

Mistake 3: Misidentifying hybridisation of carbon in formaldehyde

The error: Students sometimes think carbon in $\mathrm{H_2C=O}$ is $sp^3$ (because they see four bonds: two to H, one double bond to O) or $sp$ (because of the double bond).

How to avoid:

• Count regions of electron density (bonds + lone pairs) around carbon:
  • 1 double bond = 1 region
  • 2 single bonds = 2 regions
  • Total = 3 regions → $sp^2$ hybridisation.
• Formula: $sp^2$ → 3 regions → trigonal planar → bond angles ~120°.

Mistake 4: Thinking the Reason is wrong because “planarity depends on more than hybridisation”

The error: Students argue that $sp^2$ hybridisation alone doesn’t guarantee planarity (which is true in general), so they mark the Reason as incorrect.

How to avoid:

• In this specific case, the Reason is correct — the $sp^2$ carbon is the reason formaldehyde is planar.
• The Assertion and Reason are both correct, and the Reason correctly explains the Assertion.
• Answer: Option (A). …