Skip to content
NCERT Exemplar · Q53

Q.Predict the shapes of the following molecules on the basis of hybridisation.
BCl3, CH4, CO2, NH3

Uttar Pradesh UpmspShort· 2mImportance★★★★★est
87% · 97/112 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. BCl₃ is trigonal planar, CH₄ is tetrahedral, CO₂ is linear, and NH₃ is pyramidal.

The key to predicting molecular shapes lies in understanding how atomic orbitals mix to form hybrid orbitals. This is the VSEPR (Valence Shell Electron Pair Repulsion) theory in action, but the foundation is hybridisation. The central atom's valence electrons arrange themselves in space to minimise repulsion, and the hybridisation tells us the geometry of that arrangement.

For each molecule, we first count the number of sigma bonds and lone pairs around the central atom. The total number of these "electron domains" gives the steric number, which directly tells us the hybridisation. Then, we consider the lone pairs — they occupy space but are not "seen" in the molecular shape, so they distort the geometry from the ideal hybrid orbital arrangement.

Let's work through each one.

  1. BCl₃ (Boron Trichloride)

    Boron is the central atom. It has 3 valence electrons. Each chlorine atom contributes one electron to form a single bond. Boron forms three sigma bonds with three chlorine atoms. There are no lone pairs on boron.

    Steric number = 3 (three sigma bonds, zero lone pairs). This corresponds to sp2sp^2 hybridisation. The three sp2sp^2 hybrid orbitals lie in a plane at 120∘120^\circ angles. Since there are no lone pairs, the molecular shape is exactly the same as the orbital arrangement: trigonal planar.

  2. CH₄ (Methane)

    Carbon is the central atom. Carbon has 4 valence electrons. Each hydrogen atom contributes one electron. Carbon forms four sigma bonds with four hydrogen atoms. No lone pairs exist on carbon.

    Steric number = 4. This gives sp3sp^3 hybridisation. The four sp3sp^3 hybrid orbitals point to the corners of a tetrahedron, with bond angles of 109.5∘109.5^\circ. Again, no lone pairs to distort the shape, so the molecule is tetrahedral.

  3. CO₂ (Carbon Dioxide)

    Carbon is the central atom. Carbon has 4 valence electrons. Each oxygen atom has 6 valence electrons. Carbon forms a double bond with each oxygen. A double bond consists of one sigma bond and one pi bond. For hybridisation, we only count the sigma bonds. So, carbon forms two sigma bonds (one to each oxygen). There are no lone pairs on carbon.

    Steric number = 2. This corresponds to spsp hybridisation. The two spsp hybrid orbitals are oriented linearly, 180∘180^\circ apart. The molecule is linear.

  4. NH₃ (Ammonia) …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.