Problem 20
Question
In which of the following molecules can you confidently predict the bond angles about the central atom, and for which would you be a bit uncertain? Explain in each case. (a) \(\mathrm{H}_{2} \mathrm{~S},\) (b) \(\mathrm{BCl}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{I}\) (d) \(\mathrm{CBr}_{4}\) (e) TeBr \(_{4}\)
Step-by-Step Solution
Verified Answer
We can confidently predict the bond angles for (b) BCl3, (c) CH3I, and (d) CBr4, as they have symmetric molecular geometries (trigonal planar and tetrahedral) with no lone pairs on the central atoms. The bond angles are 120° for BCl3, and 109.5° for both CH3I and CBr4. However, we are uncertain about the bond angles in (a) H2S and (e) TeBr4, as they both contain lone pairs on their central atoms (S and Te) affecting their bond angles.
1Step 1: 1. Determine the Lewis Structure
Draw the Lewis structure(electron-dot structure) for each molecule to understand the distribution of electrons around the central atom.
(a) H2S:
S is the central atom, surrounded by two H atoms and two lone pairs of electrons.
(b) BCl3:
B is the central atom, surrounded by three Cl atoms with no lone pairs.
(c) CH3I:
C is the central atom, surrounded by three H atoms and one I atom with no lone pairs.
(d) CBr4:
C is the central atom, surrounded by four Br atoms with no lone pairs.
(e) TeBr4:
Te is the central atom, surrounded by four Br atoms and one lone pair of electrons.
2Step 2: 2. Determine the Electronic Geometry and Molecular Geometry
Use the VSEPR theory to identify the electronic geometry and molecular geometry for each molecule. The notation AXmEn is used, where A stands for the central atom, X represents surrounding atoms, m represents the number of surrounding atoms, and n stands for the number of lone pairs on the central atom.
(a) H2S:
- Electronic geometry: AX2E2 (Tetrahedral)
- Molecular geometry: Bent
(b) BCl3:
- Electronic geometry: AX3 (Trigonal planar)
- Molecular geometry: Trigonal planar
(c) CH3I:
- Electronic geometry: AX4 (Tetrahedral)
- Molecular geometry: Tetrahedral
(d) CBr4:
- Electronic geometry: AX4 (Tetrahedral)
- Molecular geometry: Tetrahedral
(e) TeBr4:
- Electronic geometry: AX4E (Trigonal bipyramidal)
- Molecular geometry: See-saw
3Step 3: 3. Predict Bond Angles and Level of Confidence
Based on the molecular geometry, predict the bond angles, and then determine the level of confidence in the predictions.
(a) H2S (Bent):
- Bond angle: Approximately 104.5°
- Confidence: Uncertain, due to two lone pairs on the central atom (S) that can cause variations in bond angle.
(b) BCl3 (Trigonal Planar):
- Bond angle: 120°
- Confidence: Confident, as there are no lone pairs on the central atom (B) and the molecular geometry is symmetric.
(c) CH3I (Tetrahedral):
- Bond angle: 109.5°
- Confidence: Confident, as there are no lone pairs on the central atom (C) and the molecular geometry is symmetric.
(d) CBr4 (Tetrahedral):
- Bond angle: 109.5°
- Confidence: Confident, as there are no lone pairs on the central atom (C) and the molecular geometry is symmetric.
(e) TeBr4 (See-saw):
- Bond angle: <120° for Br-Te-Br (equatorial) and <90° for Br-Te-Br (axial)
- Confidence: Uncertain, due to the presence of a lone pair on the central atom (Te) that can cause variations in bond angle.
Key Concepts
Lewis StructureMolecular GeometryBond AnglesElectron Pair Repulsion
Lewis Structure
To fully understand molecules, we start by drawing the Lewis Structure. This representation shows how atoms are connected and illustrates the distribution of electrons, such as bonding pairs and lone pairs, around the central atom. For example, in the molecule \(\text{H}_2\text{S}\), sulfur acts as the central atom, which is bonded to two hydrogen atoms and has two lone pairs on it.
Similarly, \(\text{BCl}_3\) displays boron as the central atom bonded to three chlorine atoms without any lone pairs.
Similarly, \(\text{BCl}_3\) displays boron as the central atom bonded to three chlorine atoms without any lone pairs.
- \(\text{CH}_3\text{I}\) highlights carbon at the center surrounded by hydrogens and an iodine.
- In \(\text{CBr}_4\), carbon is again central with four bromine atoms around it.
- For \(\text{TeBr}_4\), tellurium is the centerpiece with four bromines and one lone pair.
Molecular Geometry
Once the Lewis Structure is known, we can determine the Molecular Geometry using the VSEPR (Valence Shell Electron Pair Repulsion) theory. This theory suggests that electron pairs will arrange themselves to minimize repulsion between them. This affects the shape of the molecule significantly.
For each molecule, the notation \(AX_mE_n\) is used to denote geometry. Here, "A" is the central atom, "X" is the number of surrounding atoms, and "E" is the number of lone pairs.
For each molecule, the notation \(AX_mE_n\) is used to denote geometry. Here, "A" is the central atom, "X" is the number of surrounding atoms, and "E" is the number of lone pairs.
- For \(\text{H}_2\text{S}\), the geometry is bent due to the two lone pairs (AX2E2).
- In \(\text{BCl}_3\), it forms a trigonal planar structure as it is simply \(AX3\), with no lone pairs affecting the shape.
- Both \(\text{CH}_3\text{I}\) and \(\text{CBr}_4\) display a tetrahedral arrangement (AX4).
- For \(\text{TeBr}_4\), the geometry is see-saw due to the lone pair, marked as AX4E.
Bond Angles
Bond angles are an essential feature of molecular geometry. They determine the angle formed between three atoms across at least two bonds. These angles can vary depending, mainly, upon the presence of lone electron pairs, which exert repulsion on bonding pairs, thereby altering the idealized angles.
Here are basic expectations for each molecule:
Here are basic expectations for each molecule:
- In \(\text{H}_2\text{S}\), the bending caused by lone pairs compresses the angle to about 104.5°.
- \(\text{BCl}_3\)’s planar symmetry allows for perfect 120° angles.
- Compounds like \(\text{CH}_3\text{I}\) and \(\text{CBr}_4\) both maintain angles of approximately 109.5° due to their tetrahedral shape.
- For \(\text{TeBr}_4\), angles vary due to its asymmetric see-saw shape, with Br-Te-Br typically less than 120° for equatorial and less than 90° for axial bonds.
Electron Pair Repulsion
According to VSEPR theory, Electron Pair Repulsion is a central concept that explains molecular shapes. This premise indicates that electron pairs surrounding a central atom repel each other.
Thus, to minimize repulsion and achieve stability, they space themselves as far apart as possible.
Thus, to minimize repulsion and achieve stability, they space themselves as far apart as possible.
- Take \(\text{H}_2\text{S}\); due to significant repulsion from the two lone pairs, the molecule bends.
- \(\text{BCl}_3\) lacks lone pairs, allowing for equal distribution and maintains a trigonal planar shape.
- The tetrahedral shape in both \(\text{CH}_3\text{I}\) and \(\text{CBr}_4\) arises due to the equal repulsion of four bonds.
- Lastly, the see-saw shape of \(\text{TeBr}_4\) is because of lone pair influence which distorts the relative symmetry.
Other exercises in this chapter
Problem 18
Would you expect the nonbonding electron-pair domain in \(\mathrm{NCl}_{3}\) to be greater or smaller in size than the corresponding one in \(\mathrm{PCl}_{3} ?
View solution Problem 19
In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape? (a) \(\mathrm{CO}_{2},(\mathbf{b}) \mat
View solution Problem 21
How many nonbonding electron pairs are there in each of the following molecules: (a) \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3},(\mathbf{b}) \mathrm{CO},(\mat
View solution Problem 22
How many electron domains are surrounding the central atoms which adopt the following geometries? (a) linear (b) trigonal planar (c) trigonal pyramidal (d) trig
View solution