Question 6 of 32intermediate🔍 AnalyzeLong Answer3 marks

Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Correct Answer

Both NH3NH_3 and H2OH_2O undergo sp3sp^3 hybridisation, ideally giving a tetrahedral geometry with a bond angle of 109.5∘109.5^\circ. However, lone pairs of electrons distort this geometry.

In NH3NH_3, nitrogen has three unpaired electrons and one lone pair. The force of repulsion between a lone pair and a bond pair is more than between two bond pairs. This distorts the molecule, reducing the bond angle to 107∘107^\circ and giving it a pyramidal geometry.

In H2OH_2O, oxygen has two unpaired electrons and two lone pairs. The presence of two lone pairs causes greater repulsion, further compressing the bond angle to 104.5∘104.5^\circ. Thus, the bond angle in water is less than in ammonia, resulting in a V-shape or angular geometry.

Exercise: EXERCISES | Q: 4.8 | (Page 34)
For More Understanding

Explanation

The textbook context explains that both NH3 and H2O have sp3 hybridisation. NH3 has one lone pair which exerts more repulsion than bond pairs, reducing the angle to 107 degrees. H2O has two lone pairs, causing even more repulsion and reducing the angle to 104.5 degrees.

Solution Steps

  1. Step 1: Identify that both molecules undergo sp3sp^3 hybridisation with ideal bond angles of 109.5∘109.5^\circ.

  2. Step 2: Explain that in NH3NH_3, one lone pair causes repulsion, reducing the bond angle to 107∘107^\circ.

  3. Step 3: Explain that in H2OH_2O, two lone pairs cause greater repulsion, further reducing the bond angle to 104.5∘104.5^\circ.

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