Write Lewis structures that obey the octet rule for each of the following molecules. a. \(\mathrm{CCl}_{4}\) c. \(\mathrm{SeCl}_{2}\) b. \(\mathrm{NCl}_{3}\) d. \(\mathrm{ICl}\) In each case, the atom listed first is the central atom.

To do this, we need to add the valence electrons of each atom in the molecule. Remember that in the periodic table, group 1 has 1 valence electron, group 2 has 2 valence electrons, and so on. a. \(\mathrm{CCl}_{4}\): Carbon (C) has 4 valence electrons and Chlorine (Cl) has 7 valence electrons each. So, the total valence electrons are \(4 + 4(7) = 32\). c. \(\mathrm{SeCl}_{2}\): Selenium (Se) has 6 valence electrons and Chlorine (Cl) has 7 valence electrons each. So, the total valence electrons are \(6 + 2(7) = 20\). b. \(\mathrm{NCl}_{3}\): Nitrogen (N) has 5 valence electrons and Chlorine (Cl) has 7 valence electrons each. So, the total valence electrons are \(5 + 3(7) = 26\). d. \(\mathrm{ICl}\): Iodine (I) has 7 valence electrons and Chlorine (Cl) has 7 valence electrons. So, the total valence electrons are \(7 + 7 = 14\).

We need to distribute the electrons in such a way that each atom is surrounded by eight electrons, either in lone pairs or bond pairs. a. \(\mathrm{CCl}_{4}\): Carbon (C) shares 1 electron with each of the 4 Chlorine (Cl) atoms, forming 4 single bonds. All 4 Chlorine atoms have 3 lone pairs, and Carbon has no lone pairs. c. \(\mathrm{SeCl}_{2}\): Selenium (Se) shares 1 electron with each of the 2 Chlorine (Cl) atoms, forming 2 single bonds. Both Chlorine atoms have 3 lone pairs. However, to complete the octet for Selenium, which has 6 valence electrons, it also needs to have one lone pair. b. \(\mathrm{NCl}_{3}\): Nitrogen (N) shares 1 electron with each of the 3 Chlorine (Cl) atoms, forming 3 single bonds. All 3 Chlorine atoms have 3 lone pairs, and Nitrogen has 1 lone pair. d. \(\mathrm{ICl}\): Iodine (I) shares 1 electron with the Chlorine (Cl) atom, forming a single bond. Both Iodine and Chlorine have 3 lone pairs each.

Now, we will draw the Lewis structures for each molecule, keeping in mind the number of bond pairs and lone pairs discussed in the previous step. a. \(\mathrm{CCl}_{4}\): ``` Cl | Cl -- C -- Cl | Cl ``` c. \(\mathrm{SeCl}_{2}\): ``` Cl | Se -- Cl ¦ ``` b. \(\mathrm{NCl}_{3}\): ``` Cl | Cl -- N -- Cl ¦ ``` d. \(\mathrm{ICl}\): ``` I -- Cl ¦ ¦ ```