Chapter 5: Problem 12

Why is the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) bond angle in \(\mathrm{H}_{2} \mathrm{O}\) smaller than H-N-H bond angle in \(\mathrm{NH}_{3}\) ?

Short Answer

Expert verified

Answer: The H-O-H bond angle in H2O (104.5°) is smaller than the H-N-H bond angle in NH3 (107.5°) because of the greater electronic repulsion experienced between the two lone pairs of electrons on the oxygen atom as compared to the lone pair-bonding pair repulsion on the nitrogen atom.

Step by step solution

Determine the molecular shape around the central atom using the VSEPR theory

In H2O, there are 2 bonding pairs of electrons (O-H bonds) and 2 lone pairs on the oxygen atom. In NH3, there are 3 bonding pairs of electrons (N-H bonds) and 1 lone pair on the nitrogen atom. The valence shell electron pair repulsion (VSEPR) theory helps to understand the bond angle between these electrons.

Understand the electronic repulsion in both compounds based on VSEPR theory

According to the VSEPR theory, the electron pairs around a central atom arrange themselves to minimize repulsion. Lone pairs of electrons repel more than the bonding pairs because they are closer to the central atom and are less shielded by other atoms. In H2O, the two lone pairs on the oxygen atom are involved in greater electronic repulsion, which causes the H-O-H bond angle to be smaller compared to NH3.

Calculate the bond angles in H2O and NH3

In H2O, the molecular geometry around oxygen is bent, and the idealized bond angle is 109.5°, as predicted by the tetrahedral electron geometry. However, due to the electronic repulsion between the two lone pairs, the observed H-O-H bond angle is 104.5°. In NH3, the molecular geometry around nitrogen is trigonal pyramidal, with an idealized bond angle of 109.5°, assuming tetrahedral electron geometry. However, due to the electronic repulsion between the lone pair and the bonding pairs, the observed H-N-H bond angle is 107.5°.

Conclude why the H-O-H bond angle is smaller than the H-N-H bond angle

The H-O-H bond angle in H2O (104.5°) is smaller than the H-N-H bond angle in NH3 (107.5°) because of the greater electronic repulsion experienced between the two lone pairs of electrons on oxygen as compared to the lone pair-bonding pair repulsion on nitrogen.

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Why is the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) bond angle in \(\mathrm{H}_{2} \mathrm{O}\) smaller than H-N-H bond angle in \(\mathrm{NH}_{3}\) ?

Short Answer

Step by step solution

Determine the molecular shape around the central atom using the VSEPR theory

Understand the electronic repulsion in both compounds based on VSEPR theory

Calculate the bond angles in H2O and NH3

Conclude why the H-O-H bond angle is smaller than the H-N-H bond angle

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