Write Lewis structures that obey the octet rule for each of the following molecules. a. \(\mathrm{CCl}_{4} \quad\) c. \(\mathrm{SeCl}_{2}\) b. \(\mathrm{NCl}_{3} \quad\) d. \(\mathrm{ICl}\)

1. Identify the central atom: The central atom in this molecule is Carbon (C) because Carbon is less electronegative than Chlorine. 2. Count the total number of valence electrons: Carbon has 4 valence electrons, and each Chlorine has 7 valence electrons. Since there are 4 Chlorine atoms, there are a total of 4 + (4 × 7) = 32 valence electrons. 3. Place single bonds between central atom and outer atoms: Connect Carbon to each Chlorine with a single bond, which uses 2 valence electrons per bond. This uses a total of 4 × 2 = 8 valence electrons, leaving 24 electrons. 4. Fill the remaining valence electrons to reach octet for outer atoms: Each Chlorine needs 6 more valence electrons to have an octet. There are 4 Chlorine atoms, so we need 4 × 6 = 24 valence electrons. This exhausts all the remaining electrons. 5. Check if central atom has an octet: Carbon now has 4 bonds, each sharing 2 electrons, resulting in 8 electrons surrounding it, thus satisfying the octet rule. The Lewis structure for \(\mathrm{CCl}_{4}\) is: Cl | Cl-C-Cl | Cl

1. Identify the central atom: Nitrogen (N) is the central atom since it is less electronegative than Chlorine. 2. Count the total number of valence electrons: Nitrogen has 5 valence electrons, and each Chlorine has 7 valence electrons. With 3 Chlorine atoms, the total is 5 + (3 × 7) = 26 valence electrons. 3. Place single bonds between central atom and outer atoms: Connect Nitrogen to each Chlorine with a single bond, using 2 valence electrons per bond. We use a total of 3 × 2 = 6 valence electrons, leaving 20 electrons. 4. Fill the remaining valence electrons to reach octet for outer atoms: Each Chlorine needs 6 more valence electrons to have an octet. With three Chlorine atoms, we need 3 × 6 = 18 valence electrons, leaving 2 electrons. 5. Place any remaining electrons on the central atom: Place the 2 remaining valence electrons on Nitrogen. 6. Check if the central atom has an octet: Nitrogen has 3 bonds and a lone pair, totaling to 8 electrons satisfying the octet rule. The Lewis structure for \(\mathrm{NCl}_{3}\) is: Cl | Cl-N-Cl : : = lone pair of electrons

1. Identify the central atom: Selenium (Se) is the central atom as it is less electronegative than Chlorine. 2. Count the total number of valence electrons: Selenium has 6 valence electrons, and each Chlorine has 7 valence electrons. With 2 Chlorine atoms, the total is 6 + (2 × 7) = 20 valence electrons. 3. Place single bonds between central atom and outer atoms: Connect Selenium to each Chlorine with a single bond, using 2 valence electrons per bond. Using 2 × 2 = 4 valence electrons, leaving 16 electrons. 4. Fill the remaining valence electrons to reach octet for outer atoms: Each Chlorine needs 6 more valence electrons to have an octet. With 2 Chlorine atoms, we need 2 × 6 = 12 valence electrons, leaving 4 electrons. 5. Place any remaining electrons on the central atom: Place the 4 remaining valence electrons on Selenium as two lone pairs. 6. Check if the central atom has an octet: Selenium has 2 bonds and 2 lone pairs, totaling to 8 electrons satisfying the octet rule. The Lewis structure for \(\mathrm{SeCl}_{2}\) is: Cl | Se-Cl :: :: = 2 lone pairs of electrons

1. Identify the central atom: Both atoms are halogens, but Iodine (I) is less electronegative than Chlorine, so it will act as the central atom. 2. Count the total number of valence electrons: Iodine has 7 valence electrons, and Chlorine has 7 valence electrons, resulting in a total of 7 + 7 = 14 valence electrons. 3. Place single bonds between central atom and outer atoms: Connect Iodine to Chlorine with a single bond. This uses 2 valence electrons, leaving 12 electrons. 4. Fill the remaining valence electrons to reach octet for both atoms: Each atom needs 6 more valence electrons to have an octet. Both Iodine and Chlorine require 6 + 6 = 12 valence electrons, exhausting the remaining electrons. 5. Check if both atoms have an octet: Both Iodine and Chlorine have a bond and 3 lone pairs, totaling to 8 electrons each satisfying the octet rule. The Lewis structure for \(\mathrm{ICl}\) is: I-Cl : : : : : = lone pairs of electrons