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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
a) ClO2- has a bent molecular shape. b) SO4 2- has a tetrahedral molecular shape. c) NF3 has a trigonal pyramidal molecular shape. d) CCl2Br2 has a tetrahedral molecular shape. e) SF4 2+ has a seesaw (or distorted tetrahedron) molecular shape.

Step by step solution

01

a) ClO2-

Firstly, we must determine the central atom, which is Chlorine (Cl). Next, we'll calculate the total number of valence electrons: 7 (from Cl) + 2(6) (from two Oxygen atoms) + 1 (from the extra electron due to negative charge) = 20 valence electrons. The electron domain geometry can be found by determining the number of electron domains around the central atom. Here, we have three domains (two O atoms and one lone pair). The domains are arranged in a trigonal planar geometry. Since there is one lone pair and two bonded atoms, the molecular shape is bent.
02

b) SO4 2-

The central atom is Sulfur (S). We calculate the total valence electrons as follows: 6 (from S) + 4(6) (from four Oxygen atoms) - 2 (due to the 2- charge) = 32 valence electrons. In this case, we have four electron domains (four O atoms) surrounding the central sulfur atom, which means the Sulfur atom has a tetrahedral electron domain geometry. Since all the domains are occupied by oxygen atoms, the molecular shape is also tetrahedral.
03

c) NF3

The central atom is Nitrogen (N). The total valence electrons can be calculated as 5 (from N) + 3(7) (from Fluorine atoms) = 26 valence electrons. Nitrogen has four electron domains (three F atoms and one lone pair) with tetrahedral electron domain geometry. Since there is one lone pair and three bonded atoms, the molecular shape is trigonal pyramidal.
04

d) CCl2Br2

Here, the central atom is Carbon (C) with four valence electrons. The total valence electrons are 4 (from C) + 2(7) (from Chlorine atoms) + 2(7) (from Bromine atoms) = 32 valence electrons. There are four electron domains around the carbon atom (two Cl and two Br), and the electron domain geometry is tetrahedral. Since all domains are occupied by other atoms, the molecular shape is also tetrahedral.
05

e) SF4 2+

The central atom is Sulfur (S). We calculate the total valence electrons: 6 (from S) + 4(7) (from four Fluorine atoms) - 2 (due to the 2+ charge) = 30 valence electrons. With five electron domains (four F atoms and one lone pair) around the central sulfur atom, the electron domain geometry is trigonal bipyramidal. There is one lone pair and four bonded atoms; hence, the molecular shape is called a "seesaw" or "distorted tetrahedron."

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Most popular questions from this chapter

In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape? (a) \(\operatorname{SiH}_{4,}(\mathbf{b}) \mathrm{PF}_{3},(\mathbf{c}) \mathrm{HBr},(\mathbf{d}) \mathrm{HCN},(\mathbf{e}) \mathrm{SO}_{2}\)

(a) An AB \(_{6}\) molecule has no lone pairs of electrons on the A atom. What is its molecular geometry? (b) An AB \(_{4}\) molecule has two lone pairs of electrons on the A atom (in addition to the four B atoms). What is the electron-domain geometry around the A atom? (c) For the AB \(_{4}\) molecule in part (b), predict the molecular geometry.

Ammonia, \(\mathrm{NH}_{3},\) reacts with incredibly strong bases to produce the amide ion, NH \(_{2}\) . Ammonia can also react with acids to produce the ammonium ion, \(\mathrm{NH}_{4}^{+} .\) (a) Which species (amide ion, ammonia, or ammonium ion) has the largest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle? (b) Which species has the smallest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle?

Which of the following statements about hybrid orbitals is or are true? (i) After an atom undergoes sp hybridization, there is one unhybridized \(p\) orbital on the atom, (ii) Under \(s p^{2}\) hybridization, the large lobes point to the vertices of an equilateral triangle, and (iii) The angle between the large lobes of \(s p^{3}\) hybrids is \(109.5^{\circ} .\)

For each statement, indicate whether it is true or false. (a) The greater the orbital overlap in a bond, the weaker the bond. (b) The greater the orbital overlap in a bond, the shorter the bond. (c) To create a hybrid orbital, you could use the sorbital on one atom with a porbital on another atom. (d) Nonbonding electron pairs cannot occupy a hybrid orbital.
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