Compare the bond angles in the following molecules: (a) \(\mathrm{CH}_{4}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
The bond angle decreases in the series \(\mathrm{CH}_{4}\) (109.5°) > \(\mathrm{NH}_{3}\) (107°) > \(\mathrm{H}_{2}\mathrm{O}\) (104.5°) due to increasing repulsion from lone pairs.

Step by step solution

01

Draw the Lewis structure

The Lewis structure for each molecule is as follows: \n(a) \(\mathrm{CH}_{4}\): Carbon forms four bonds and no lone pairs, making it a regular tetrahedron with Hydrogen. \n(b) \(\mathrm{NH}_{3}\): Nitrogen forms three bonds and one lone pair, forming a trigonal pyramidal shape with Hydrogen. \n(c) \(\mathrm{H}_{2} \mathrm{O}\): Oxygen forms two bonds and two lone pairs, making it a bent shape with Hydrogen.
02

Determine the Molecular Geometry

Each Lewis structure implies a specific molecular geometry: \n(a) \(\mathrm{CH}_{4}\): Tetrahedral \n(b) \(\mathrm{NH}_{3}\): Trigonal pyramidal \n(c) \(\mathrm{H}_{2} \mathrm{O}\): Bent
03

Infer the Bond Angle from the Shape

Using the molecular geometry, the bond angle for each compound is: \n(a) \(\mathrm{CH}_{4}\): 109.5° in a tetrahedral shape \n(b) \(\mathrm{NH}_{3}\): 107° in a trigonal pyramidal shape, less than \(\mathrm{CH}_{4}\) due to the presence of a lone pair causing more repulsion. \n(c) \(\mathrm{H}_{2} \mathrm{O}\): 104.5° in a bent shape, the smallest among the three due to the presence of two lone pairs causing more repulsion than the \(\mathrm{NH}_{3}\).

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