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Chapter 8: Problem 120

Predict the molecular structure (including bond angles) for each of the following. a. \(\mathrm{PCl}_{3}\) b. \(\mathrm{SCl}_{2}\) c. \(\mathrm{SiF}_{4}\)

Short Answer

Expert verified
The molecular structures and bond angles for the given compounds are as follows: a. PCl3 has a tetrahedral shape with a bond angle slightly less than 109.5掳 due to the presence of a lone pair. b. SCl2 has a bent shape with a bond angle less than 109.5掳 due to the presence of two lone pairs. c. SiF4 has a tetrahedral shape with a bond angle of 109.5掳.

Step by step solution

01

a. PCl3 molecular structure

To predict the molecular structure and bond angles of PCl3, follow these steps: 1. Draw the Lewis structure for PCl3: P is the central atom with 5 valence electrons and each Cl has 7 valence electrons. Lewis structure: P surrounded by 3 Cl atoms, with a single bond between P and each Cl atom. P has one lone pair of electrons. 2. Identify the shape of the molecule based on the number of electron domains P has 3 bond pairs (from P-Cl bonds) and 1 lone pair, which means there are 4 electron domains. According to the VSEPR theory, this molecular geometry is called tetrahedral. 3. Determine the bond angles based on the shape of the molecule The bond angles in a tetrahedral shape are idealistically 109.5掳. However, the lone pair slightly increases the repulsion and decreases the bond angles. Therefore, the bond angle for PCl3 is slightly less than 109.5掳.
02

b. SCl2 molecular structure

To predict the molecular structure and bond angles of SCl2, follow these steps: 1. Draw the Lewis structure for \(\mathrm{SCl}_{2}\): S is the central atom with 6 valence electrons and each Cl has 7 valence electrons. Lewis structure: S surrounded by 2 Cl atoms, with a single bond between S and each Cl atom. S has two lone pairs of electrons. 2. Identify the shape of the molecule based on the number of electron domains: S has 2 bond pairs (from S-Cl bonds) and 2 lone pairs, which means there are 4 electron domains. According to the VSEPR theory, this molecular geometry is called bent. 3. Determine the bond angles based on the shape of the molecule: The bond angles in a bent shape with four electron domains are idealistically 109.5掳. However, the lone pairs increase the repulsion and decrease the bond angles. Therefore, the bond angle for \(\mathrm{SCl}_{2}\) is less than 109.5掳.
03

c. SiF4 molecular structure

To predict the molecular structure and bond angles of \(\mathrm{SiF}_{4}\), follow these steps: 1. Draw the Lewis structure for \(\mathrm{SiF}_{4}\): Si is the central atom with 4 valence electrons and each F has 7 valence electrons. Lewis structure: Si surrounded by 4 F atoms, with a single bond between Si and each F atom. Si has no lone pairs of electrons. 2. Identify the shape of the molecule based on the number of electron domains: Si has 4 bond pairs (from Si-F bonds) and no lone pairs, which means there are 4 electron domains. According to the VSEPR theory, this molecular geometry is called tetrahedral. 3. Determine the bond angles based on the shape of the molecule: The bond angles in a tetrahedral shape are idealistically 109.5掳. Thus, the bond angle for \(\mathrm{SiF}_{4}\) is 109.5掳.

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Most popular questions from this chapter

Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) consists of a six- membered ring of carbon atoms with one hydrogen bonded to each carbon. Write Lewis structures for benzene, including resonance structures.

Use formal charge arguments to explain why CO has a much smaller dipole moment than would be expected on the basis of electronegativity

An alternative definition of electronegativity is $$\text { Electronegativity } = \text { constant } (\mathrm{I.E.}-\mathrm{E.A.})$$ where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 7 to calculate the \((\mathrm{I} . \mathrm{E} .-\mathrm{E} \cdot \mathrm{A} .)\) term for \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\) and \(\mathrm{I}\). Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are 1678, 1255, 1138, and 1007 kJ/mol, respectively. (Hint: Choose a constant so that the electronegativity of fluorine equals 4.0. Using this constant, calculate relative electronegativities for the other halogens and compare to values given in the text.)

Write the Lewis structure for \(\mathrm{O}_{2} \mathrm{F}_{2} \quad\left(\mathrm{O}_{2} \mathrm{F}_{2}\right.\) exists as \(\mathrm{F}-\mathrm{O}-\mathrm{O}-\mathrm{F} )\). Assign oxidation states and formal charges to the atoms in O2F2. This compound is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of \(\mathrm{O}_{2} \mathrm{F}_{2}\)?

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part c, why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not \(\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?\) What about MgO compared to \(\mathrm{MgO}_{2} ? \mathrm{Mg}_{2} \mathrm{O} ?\)
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