Skip to content
NCERT Exemplar · Q7

Q.Which of the following species has tetrahedral geometry?

(i) BH4^-
(ii) NH2^-
(iii) CO3^2-
(iv) H3O^+
Dnh Dd CbseMCQ· 1mImportance★★★★★est
46% · 51/112 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 key idea is to count the steric number (bond pairs + lone pairs) around the central atom using the VSEPR theory. BH4^- has a steric number of 4 with no lone pairs, giving it a perfect tetrahedral geometry. The correct option is (A).

The question asks for a species with tetrahedral geometry. This is a classic VSEPR (Valence Shell Electron Pair Repulsion) problem. The shape of a molecule or ion is determined by the number of electron domains (bonding pairs and lone pairs) around the central atom, which repel each other to maximize separation.

For a tetrahedral geometry, the central atom must have four bonding pairs and zero lone pairs — a steric number of 4 with no lone pairs. Let's check each option.

  1. Option (A): BH4^-

    Boron is the central atom. It has 3 valence electrons. Each hydrogen contributes 1 electron, and the negative charge adds 1 more. Total valence electrons = 3+4×1+1=83 + 4 \times 1 + 1 = 8.

    Boron forms 4 single bonds with hydrogen, using all 8 electrons. There are 4 bond pairs and 0 lone pairs on boron. Steric number = 4.

    According to VSEPR, this gives a tetrahedral geometry. The bond angle is 109.5∘109.5^\circ.

  2. Option (B): NH2^-

    Nitrogen is central. Nitrogen has 5 valence electrons, each hydrogen gives 1, and the negative charge adds 1. Total = 5+2×1+1=85 + 2 \times 1 + 1 = 8.

    Nitrogen forms 2 single bonds with hydrogen (using 4 electrons), leaving 4 electrons as 2 lone pairs on nitrogen. Steric number = 2 (bonds) + 2 (lone pairs) = 4.

    With 2 lone pairs, the shape is bent (or V-shaped), not tetrahedral. The lone pairs repel more strongly, reducing the bond angle below 109.5∘109.5^\circ to about 104.5∘104.5^\circ.

  3. Option (C): CO3^2-

    Carbon is central. Carbon has 4 valence electrons, each oxygen contributes 6 (but we count only bonding electrons), and the 2- charge adds 2. Total = 4+3×6+2=244 + 3 \times 6 + 2 = 24.

    The Lewis structure shows carbon double-bonded to one oxygen and single-bonded to the other two (with resonance). Carbon has 3 bond pairs and 0 lone pairs. Steric number = 3.

    This gives a trigonal planar geometry with bond angles of 120∘120^\circ, not tetrahedral.

  4. Option (D): H3O^+ …

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.