Question

Bond angle of \(109^{\circ} 28\) ' is found in (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{CH}_{3}\) (d) \(\mathrm{NH}_{4}^{+}\)

Short answer

The bond angle of \(109^{\circ} 28'\) is found in \(\mathrm{NH}_{4}^{+}\).

Step-by-step solution

Understanding Bond Angles

When we talk about bond angles in molecules, we refer to the angles formed between adjacent bonds. The bond angle of approximately \(109^{\circ} 28'\) is characteristic of a tetrahedral geometry. In a perfect tetrahedron, this angle is precisely \(109.5^{\circ}\). This means we should look at molecules or ions that can have a tetrahedral geometry.

Analyzing Given Options

We're provided with four options: \(\mathrm{NH}_{3}\), \(\mathrm{H}_{2} \mathrm{O}\), \(\mathrm{CH}_{3}\), and \(\mathrm{NH}_{4}^{+}\). We need to determine the shape and bond angles of each to see which matches a tetrahedral geometry.

Examining \(\mathrm{NH}_{3}\)

\(\mathrm{NH}_{3}\) (ammonia) has a trigonal pyramidal shape due to the lone pair on nitrogen. The bond angle here is approximately \(107^{\circ}\), not quite \(109.5^{\circ}\), thus it's not the answer.

Examining \(\mathrm{H}_{2} \mathrm{O}\)

Water \(\mathrm{H}_{2} \mathrm{O}\) has a bent molecular shape with two lone pairs affecting the angle, resulting in approximately \(104.5^{\circ}\). This is much less than \(109.5^{\circ}\), so it can't be the answer.

Examining \(\mathrm{CH}_{3}\)

The \(\mathrm{CH}_{3}\) radical is not typically isolated as a stable structure in common conditions, but it should have a planar geometry if it were stable. Therefore, it wouldn't have strictly defined bond angles like \(109.5^{\circ}\), disqualifying it.

Examining \(\mathrm{NH}_{4}^{+}\)

The ammonium ion \(\mathrm{NH}_{4}^{+}\) has a tetrahedral geometry due to the four hydrogen atoms symmetrically surrounding the nitrogen. This gives it bond angles of exactly \(109.5^{\circ}\), matching the given bond angle.

Key concepts

Bond Angles

When discussing bond angles in chemistry, we're talking about the angle between two adjacent bonds around a central atom. These angles are important because they help define the shape and stability of a molecule. In tetrahedral geometries, the bond angle is about \(109.5^{\circ}\). This is due to the repulsion between the bonds striving for maximum separation, a rule known as the "VSEPR Theory". For our problem, the ammonium ion \(\text{NH}_4^+\) exhibits this famous angle. This bond angle is representative of atoms that arrange themselves as far apart as possible while still connected to a central atom, similar to pointing to the corners of a regular tetrahedron.
Some key points about bond angles:

• They are intrinsic to molecular geometry.
• They influence physical and chemical properties of compounds.
• Strain in angles can change the reactivity of molecules.

Recognizing these angles in structures such as \(\text{NH}_4^+\) helps in understanding molecular interactions in a broader chemical context.

Molecular Shapes

Molecular shapes are determined by how atoms are arranged in space in a molecule. This arrangement is a direct consequence of the repulsion between electron pairs—both bonding lone pairs—around the central atom, fitting nicely under the VSEPR theory. Molecular shape greatly influences the molecule's properties, like polarity and reactivity. For example, the tetrahedral shape in molecules, such as \(\text{NH}_4^+\), comes from four areas of electron density repelling each other equally in three dimensions to minimize repulsion.
Important aspects of molecular shapes:

• Tetrahedral shape involves four bonds spaced evenly around a central atom.
• Shapes can be predicted using the VSEPR model.
• Shape affects molecular polarity, a crucial factor in interactions.

A molecule's three-dimensional structure can affect everything from its boiling point to solubility, showcasing the importance of understanding these shapes.

VSEPR Theory

The VSEPR Theory stands for "Valence Shell Electron Pair Repulsion" Theory. This theory is a cornerstone in predicting and understanding molecular geometry. According to VSEPR, the shape of a molecule is determined by electron pair repulsions. Whether those pairs are bonding pairs or lone pairs, they all repel each other. This pushes the bonds to organize in a manner that minimizes these repulsions, resulting in specific molecular geometries like linear, trigonal planar, or tetrahedral.
The connection of VSEPR to tetrahedral geometry:

• It predicts molecular shapes based on valence electron arrangements.
• Tetrahedral shapes occur when there are four atoms bonded to a central atom.
• The geometry minimizes repulsive forces, leading to the stable structure we observe in molecules like \(\text{NH}_4^+\).

By using VSEPR theory, chemists can predict molecular shapes and get experiential insights into the behavior of molecules in reactions and various states.