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Chapter 9: Problem 34

Name the proper three-dimensional molecular shapes for each of the following molecules or ions, showing lone pairs as needed: $(\mathbf{a}) \mathrm{ClO}_{2}^{-}(\mathbf{b}) \mathrm{SO}_{4}^{2-}(\mathbf{c}) \mathrm{NF}_{3}(\mathbf{d}) \mathrm{CCl}_{2} \mathrm{Br}_{2}(\mathbf{e}) \mathrm{SF}_{4}^{2+}$

Short Answer

Expert verified
The molecular shapes for the given molecules and ions are as follows: (a) ClO₂⁻ has a linear shape, (b) SO₄²⁻ has a tetrahedral shape, (c) NF₃ has a trigonal pyramidal shape, (d) CCl₂Br₂ has a tetrahedral shape, and (e) SF₄²⁺ has a see-saw shape.

Step by step solution

01

(a) Clo₃⁻ ion: Determine the central atom and valence electrons

In the ClO₂⁻ ion, the central atom is chlorine (Cl) because oxygen (O) is more electronegative, so chlorine will have oxygen atoms surrounding it. Chlorine has 7 valence electrons, each oxygen atom has 6 valence electrons, and there is one extra electron due to the negative charge of the ion. Therefore, the total number of valence electrons is: \(7 + 2 \times 6 + 1 = 20\).
02

(a) Clo₃⁻ ion: Determine the bonding and lone electron pairs

The total number of electron pairs is 20 ÷ 2 = 10. Each oxygen bonded to chlorine contributes one bonding electron pair, thus using 2 electron pairs. The remaining 10 - 2 = 8 electron pairs are lone pairs and distributed evenly to the oxygen atoms. This means each oxygen atom has three lone pairs while the chlorine does not have any.
03

(a) Clo₃⁻ ion: Predict the molecular shape

Since Cl is surrounded by 2 atoms and 0 lone pairs, its electron pair geometry is linear. Therefore, the molecular shape of ClO₂⁻ is linear.
04

(b) SO₄²⁻ ion: Determine the central atom and valence electrons

In the SO₄²⁻ ion, the central atom is sulfur (S), surrounded by four oxygen (O) atoms. Sulfur has 6 valence electrons, each oxygen atom has 6 valence electrons, and there are two extra electrons due to the negative charge of the ion. Therefore, the total number of valence electrons is: \(6 + 4 \times 6 + 2 = 32\).
05

(b) SO₄²⁻ ion: Determine the bonding and lone electron pairs

The total number of electron pairs is 32 ÷ 2 = 16. Each oxygen atom contributes one bonding electron pair, so there are 4 bonding electron pairs. The other 16 - 4 = 12 electron pairs are lone pairs and are distributed evenly to the oxygen atoms. Each oxygen atom has three lone pairs.
06

(b) SO₄²⁻ ion: Predict the molecular shape

Since the sulfur atom is surrounded by 4 atoms and no lone pairs, its electron pair geometry is tetrahedral. Therefore, the molecular shape of SO₄²⁻ is tetrahedral.
07

(c) NF₃ molecule: Determine the central atom and valence electrons

In the NF₃ molecule, the central atom is nitrogen (N), surrounded by three fluorine (F) atoms. Nitrogen has 5 valence electrons and each fluorine atom has 7 valence electrons. Therefore, the total number of valence electrons is: \(5 + 3 \times 7 = 26\).
08

(c) NF₃ molecule: Determine the bonding and lone electron pairs

The total number of electron pairs is 26 ÷ 2 = 13. Each fluorine atom contributes one bonding electron pair, so there are 3 bonding electron pairs. The other 13 - 3 = 10 electron pairs are lone pairs. Nitrogen has one lone pair, and each fluorine atom has three lone pairs.
09

(c) NF₃ molecule: Predict the molecular shape

Since the nitrogen atom is surrounded by 3 atoms and 1 lone pair, its electron pair geometry is tetrahedral. However, due to the presence of a lone pair, the molecular shape of NF₃ is trigonal pyramidal.
10

(d) CCl₂Br₂ molecule: Determine the central atom and valence electrons

In the CCl₂Br₂ molecule, the central atom is carbon (C) surrounded by two chlorine (Cl) atoms and two bromine (Br) atoms. Carbon has 4 valence electrons, each chlorine atom has 7 valence electrons, and each bromine atom has 7 valence electrons as well. Therefore, the total number of valence electrons is: \(4 + 2 \times 7 + 2 \times 7 = 32\).
11

(d) CCl₂Br₂ molecule: Determine the bonding and lone electron pairs

The total number of electron pairs is 32 ÷ 2 = 16. Each chlorine and bromine atom shares a bonding electron pair with carbon, so there are 4 bonding electron pairs. The other 16 - 4 = 12 electron pairs are lone pairs and are distributed between the chlorine and bromine atoms. Each chlorine and bromine atom has three lone pairs.
12

(d) CCl₂Br₂ molecule: Predict the molecular shape

Since carbon is surrounded by 4 atoms and no lone pairs, its electron pair geometry is tetrahedral. Therefore, the molecular shape of CCl₂Br₂ is tetrahedral.
13

(e) SF₄²⁺ ion: Determine the central atom and valence electrons

In the SF₄²⁺ ion, the central atom is sulfur (S), which is surrounded by four fluorine (F) atoms. Sulfur has 6 valence electrons, each fluorine atom has 7 valence electrons, and there are two additional valence electrons due to the positive charge of the ion. Therefore, the total number of valence electrons is: \(6 + 4 \times 7 - 2 = 34\).
14

(e) SF₄²⁺ ion: Determine the bonding and lone electron pairs

The total number of electron pairs is 34 ÷ 2 = 17. Each fluorine atom contributes one bonding electron pair, so there are 4 bonding electron pairs. The other 17 - 4 = 13 electron pairs are lone pairs. Sulfur has one lone pair, and each fluorine atom has three lone pairs.
15

(e) SF₄²⁺ ion: Predict the molecular shape

Since the sulfur atom is surrounded by 4 atoms and 1 lone pair, its electron pair geometry is trigonal bipyramidal. However, due to the presence of one lone pair, the molecular shape of SF₄²⁺ is see-saw.

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Most popular questions from this chapter

What hybridization do you expect for the atom that is underlined in each of the following species? (a) $\underline{\mathrm{O}}_{2}^{-} ;(\mathbf{b}) \underline{\mathrm{N}} \mathrm{H}_{4}^{+} ;$ (c) \(\mathrm{SCN}^{-}\) (d) \(\underline{\mathrm{Br}} \mathrm{Cl}_{3}\)

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a \(\sigma\) bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals combine to form a \(\pi\) bond? Explain.

(a) Which geometry and central atom hybridization would you expect in the series $\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}{ }^{+} ?(\mathbf{b})$ What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated \(\mathrm{B}\) atom? (b) What is the electron configuration of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B-F bonds in \(\mathrm{BF}_{3}\) ? (d) What valence orbitals, if any, remain unhybridized on the \(\mathrm{B}\) atom in \(\mathrm{BF}_{3} ?\)

An \(\mathrm{AB}_{2}\) molecule is described as having a tetrahedral geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) linear, (ii) bent, (iii) trigonal planar, or (iv) tetrahedral?
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