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Chapter 9: Problem 41

Write Lewis structures for these molecules: (a) ICl, (b) \(\mathrm{PH}_{3}\) (c) \(\mathrm{P}_{4}\) (each \(\mathrm{P}\) is bonded to three other P atoms), (d) \(\mathrm{H}_{2} \mathrm{~S},\) (e) \(\mathrm{N}_{2} \mathrm{H}_{4}\) (f) \(\mathrm{HClO}_{3},(\mathrm{~g}) \mathrm{COBr}_{2}\) (C is bonded to \(\mathrm{O}\) and \(\mathrm{Br}\) atoms ).

Short Answer

Expert verified
The Lewis structures for ICl, \(\mathrm{PH}_{3}\), \(\mathrm{P}_{4}\), \(\mathrm{H}_{2} \mathrm{S}\), \(\mathrm{N}_{2} \mathrm{H}_{4}\), \(\mathrm{HClO}_{3}\), and \(\mathrm{COBr}_{2}\) have been drawn successfully.

Step by step solution

01

Lewis Structure for ICl

First, we calculate the total number of valence electrons. Iodine (I) has 7 valence electrons and chlorine (Cl) has 7. Therefore, the total is \(7 + 7 = 14\). We represent every atom by its symbol and link each atom to the other by a single bond (which signifies 2 electrons). To satisfy the octet rule for each atom, we distribute the rest of the valence electrons as lone pairs (on each atom). Our Lewis structure should look like the following (E represents a lone pair of electrons): E-ICl-E.
02

Lewis Structure for \(\mathrm{PH}_{3}\)

Phosphorus (P) has 5 valence electrons and hydrogen (H) has 1. So, the total is \(5 + 1 \times 3 (for 3 hydrogen atoms) = 8\). Hydrogen atoms will always be terminal, so Phosphorus is the central atom. Each hydrogen atom forms a single bond with Phosphorus atom. The Lewis structure should be constructed as follows: P-H, H-P-H, H-P-H.
03

Lewis Structure for \(\mathrm{P}_{4}\)

Each Phosphorus atom has 5 valence electrons. For 4 phosphorus atoms, the total is \(5 \times 4 = 20\). Each phosphorus atom is bonded to the three other phosphorus atoms forming a P4 molecule. The Lewis structure should be a tetrahedral arrangement of the atoms with a bond between each pair of phosphorus atoms.
04

Lewis Structure for \(\mathrm{H}_{2} \mathrm{S}\)

Sulfur (S) has 6 valence electrons and hydrogen (H) has 1. So the total is \(6 + 1 \times 2 (for 2 hydrogen atoms) = 8\). Each hydrogen atom forms a single bond with the sulfur atom. The Lewis structure should be like H-S-H.
05

Lewis Structure for \(\mathrm{N}_{2} \mathrm{H}_{4}\)

Nitrogen (N) has 5 valence electrons and hydrogen (H) has 1. So, the total is \(5 \times 2 (for 2 nitrogen atoms) + 1 \times 4 (for 4 hydrogen atoms) = 14\). Each nitrogen atom forms a single bond with the two hydrogen atoms and a single bond with the other nitrogen atom. The Lewis structure should be: H-N-H, H-N-H.
06

Lewis Structure for \(\mathrm{HClO}_{3}\)

Chlorine (Cl) has 7 valence electrons, hydrogen (H) has 1, and oxygen (O) has 6. So, the total is \(7 + 1 + 3 \times 6 (for 3 oxygen atoms) = 26\). The chlorine atom is the central atom. It forms a single bond with the hydrogen atom and the three oxygen atoms. The Lewis structure should be O=Cl-(O-H)-O, O=Cl-(O-H)-O, O=Cl-(O-H)-O.
07

Lewis Structure for \(\mathrm{ COBr}_{2}\)

Carbon (C) has 4 valence electrons, oxygen (O) has 6, and bromine (Br) has 7. So, the total is \(4 + 6 + 2 \times 7 (for 2 bromine atoms) = 24\). The carbon atom is the central atom. It forms a single bond with the oxygen atom and the two bromine atoms. The Lewis structure should be: Br-C-Br, O=C-Br, O=C-Br.

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Most popular questions from this chapter

In the vapor phase, beryllium chloride consists of discrete molecular units \(\mathrm{BeCl}_{2}\). Is the octet rule satisfied for Be in this compound? If not, can you form an octet around Be by drawing another resonance structure? How plausible is this structure?

Match each of these energy changes with one of the processes given: ionization energy, electron affinity, bond enthalpy, standard enthalpy of formation. (a) \(\mathrm{F}(g)+e^{-} \longrightarrow \mathrm{F}^{-}(g)\) (b) \(\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{~F}(g)\) (c) \(\mathrm{Na}(g) \longrightarrow \mathrm{Na}^{+}(g)+e^{-}\) (d) \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{~F}_{2}(g) \longrightarrow \mathrm{NaF}(s)\)

A rule for drawing plausible Lewis structures is that the central atom is invariably less electronegative than the surrounding atoms. Explain why this is so.

Because fluorine has seven valence electrons \(\left(2 s^{2} 2 p^{5}\right),\) seven covalent bonds in principle could form around the atom. Such a compound might be \(\mathrm{FH}_{7}\) or \(\mathrm{FCl}_{7}\). These compounds have never been prepared. Why?

An ionic bond is formed between a cation \(\mathrm{A}^{+}\) and an anion \(\mathrm{B}^{-}\). How would the energy of the ionic bond [see Equation (9.2)\(]\) be affected by the following changes? (a) doubling the radius of \(\mathrm{A}^{+},\) (b) tripling the charge on \(A^{+},(\mathrm{c})\) doubling the charges on \(\mathrm{A}^{+}\) and \(\mathrm{B}^{-},\) (d) decreasing the radii of \(\mathrm{A}^{+}\) and \(\mathrm{B}^{-}\) to half their original values.
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