Question

Determine the geometry about each interior atom in each molecule and draw the molecule. (Skeletal structure is indicated in parentheses.) \begin{equation}\mathrm{ a. }{CH}_{3} \mathrm{NH}_{2}\left(\mathrm{H}_{3} \mathrm{CNH}_{2}\right)\end{equation} \begin{equation}\mathrm{ b. }{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}\left(\mathrm{H}_{3} \mathrm{CCOOCH}_{3} \text { both } \mathrm{O} \text { atoms attached to second } \mathrm{C}\right) \end{equation}\begin{equation} \mathrm{ c. }{NH}_{2} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{H}_{2} \mathrm{NCOOH} \text { both } \mathrm{O} \text { atoms attached to } \mathrm{C}\right) \end{equation}

Short answer

Molecule a: Carbon - tetrahedral; Nitrogen - trigonal pyramidal. Molecule b: First Carbon - tetrahedral; Second Carbon - trigonal planar; Oxygen - bent (for single-bonded O). Molecule c: Nitrogen - trigonal pyramidal; Carbon - trigonal planar; Oxygen - linear (for double-bonded O) and bent (for single-bonded O).

Step-by-step solution

Determine Geometry of Carbon and Nitrogen in Molecule a

For CH3NH2, the central carbon (C) atom has four single bonds, one to each of three hydrogen (H) atoms and one to the nitrogen (N) atom, giving it a tetrahedral geometry with approximate bond angles of 109.5 degrees. Nitrogen has three bonds, one to carbon and two to hydrogen atoms, and one lone pair of electrons. This gives the nitrogen atom in CH3NH2 a trigonal pyramidal geometry with bond angles slightly less than 109.5 degrees due to the lone pair's repulsion.

Draw Molecule a

Draw the carbon atom with four single bonds to three hydrogen atoms and one nitrogen atom. The nitrogen atom has two more single bonds, each to a hydrogen atom, and shows the lone pair.

Determine Geometry of Carbon and Oxygen in Molecule b

For CH3CO2CH3, there are two central carbon atoms. The first carbon (C1) is bonded to three hydrogen (H) atoms and one more carbon (C2), indicating a tetrahedral geometry for C1. The second carbon (C2) is double-bonded to an oxygen (O) and single-bonded to an oxygen (O) and a carbon (C1), thus for C2, the geometry is trigonal planar. Each oxygen is bonded to C2, one with a double bond and the other with a single bond and also bonded to a hydrogen.

Draw Molecule b

Draw C1 bonded to three H atoms and C2. Then, draw C2 bonded to C1, a double-bonded O, a single-bonded O, and attach an additional H to the single-bonded O. The carbon atoms are in a central position with O atoms appending C2.

Determine Geometry of Nitrogen, Carbon, and Oxygen in Molecule c

For NH2CO2H, nitrogen (N) has three bonds (two to hydrogen and one to carbon) and one lone pair, resulting in a trigonal pyramidal geometry. The central carbon (C) is double-bonded to one oxygen (O), single-bonded to the nitrogen (N) and single-bonded to another oxygen (O), which is also single-bonded to a hydrogen (H), resulting in a trigonal planar geometry for the carbon atom. Each oxygen has different bonding: one is double-bonded to C (linear geometry), and the other is single-bonded to C and to H (bent geometry).

Draw Molecule c

Draw the nitrogen with two single bonds to hydrogen atoms and one to the carbon atom, and show the lone pair on nitrogen. Next, draw the carbon atom double-bonded to one oxygen and single-bonded to the other, which in turn is bonded to hydrogen. The geometry around the central carbon atom should reflect the trigonal planar arrangement.

Key concepts

VSEPR theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used in chemistry to predict the geometry of individual molecules based on the number of electron pairs surrounding their central atoms. The foundational premise is that electron pairs, which can be bonding or nonbonding (lone pairs), repel each other and will, therefore, arrange themselves as far apart as possible to minimize repulsion.

This model helps us understand that the geometrical structure of molecules is significantly influenced by the number of lone pairs and bonded atoms around the central atom. This is because lone pairs are slightly more repulsive than bonding pairs, which can lead to variations in bond angles to accommodate these interactions. For instance, while the ideal bond angle in a tetrahedral molecule is 109.5 degrees, the presence of a lone pair in a trigonal pyramidal molecule reduces the bond angle slightly.

VSEPR theory is an invaluable tool for predicting and visualizing the shapes of molecules, providing an accessible means for students to grasp how molecules are structured in three dimensions.

Tetrahedral geometry

When discussing molecular geometries, 'tetrahedral' is one of the most fundamental shapes. It occurs when a central atom is surrounded by four other atoms, with none being a lone pair. The atoms are positioned at the corners of an imaginary tetrahedron with the central atom at its center.

In a perfect tetrahedral geometry, the bond angles are all approximately 109.5 degrees. An example where this geometry is observed is in a methane molecule (\( CH_4 \)), or in our exercise example, the carbon atom in \( CH_3NH_2 \) which is bonded to three hydrogen atoms and one nitrogen atom.

The tetrahedral geometry helps spread out the atoms and electron pairs evenly, minimizing repulsion and resulting in a stable structure. For students, visualizing a tetrahedron—a pyramid with a triangular base—can often aid in understanding this three-dimensional configuration.

Trigonal pyramidal geometry

Trigonal pyramidal geometry occurs in molecules where the central atom is bonded to three other atoms and has one lone pair of electrons. This arrangement causes the molecule to adopt a pyramidal shape with a triangular base.

The lone pair exerts a greater repulsive force than the bonded atoms which results in the bond angles being less than the ideal tetrahedral angle of 109.5 degrees. Ammonia (\( NH_3 \)) is a classic example of trigonal pyramidal geometry. In the context of the exercise, the nitrogen atom in \( CH_3NH_2 \) exhibits this geometry due to its three bonds and one lone pair.

It is essential for students to note that the lone pair's presence is responsible for the distortion from the perfect tetrahedral shape to the trigonal pyramidal geometry, highlighting the significance of electron pair repulsions in determining molecular shape.

Trigonal planar geometry

Molecules with trigonal planar geometry have a central atom bonded to three peripheral atoms with no lone pairs. This leads to a flat, triangular shape where all the bonded atoms and the central atom are in the same plane.

Perfect trigonal planar molecular geometries have bond angles of 120 degrees, equidistant from each other. An example of this geometry could be seen in boron trifluoride (\( BF_3 \)), and from the textbook solution, the carbon atom in the \( CH_3CO_2CH_3 \) molecule exhibits trigonal planar geometry at the second carbon bonded to two oxygens and one other carbon atom.

Recognizing the presence of double bonds is also significant in predicting trigonal planar geometry, as they often lead to such spatial arrangements due to their electronic structure. It's a neat exercise for students to identify central atoms with double bonds when inferring molecular geometries using the VSEPR theory.