Question

Draw Lewis structures showing all valence electrons for these molecules. (a) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (b) \(\mathrm{CS}_{2}\) (c) \(\mathrm{HCN}\)

Short answer

Question: Draw the Lewis structures of ethane (C2H6), carbon disulfide (CS2), and hydrogen cyanide (HCN). Answer: The Lewis structures for these molecules are as follows: Ethane (C2H6): H - C - C - H | | H H Carbon Disulfide (CS2): S = C = S Hydrogen Cyanide (HCN): H - C ≡ N

Step-by-step solution

Count the total number of valence electrons

Calculate the total number of valence electrons for each molecule by summing the valence electrons of the individual atoms. (a) \(\mathrm{C}_{2} \mathrm{H}_{6}\): Each carbon has 4 valence electrons, and each hydrogen has 1 valence electron. Total valence electrons = 2 * 4 (for Carbon) + 6 * 1 (for Hydrogen) = 8 + 6 = 14 electrons (b) \(\mathrm{CS}_{2}\): Carbon has 4 valence electrons, and each sulfur has 6 valence electrons. Total valence electrons = 4 (for Carbon) + 2 * 6 (for Sulfur) = 4 + 12 = 16 electrons (c) \(\mathrm{HCN}\): Hydrogen has 1 valence electron, carbon has 4 valence electrons, and nitrogen has 5 valence electrons. Total valence electrons = 1 (for Hydrogen) + 4 (for Carbon) + 5 (for Nitrogen) = 10 electrons

Arrange the atoms and connect with single bonds

Next, arrange the atoms for each molecule and connect them with single bonds. This uses up 2 electrons from the total valence electrons calculated in Step 1. (a) In \(\mathrm{C}_{2} \mathrm{H}_{6}\), the two carbon atoms are in the center and connected with a single bond. Each carbon atom is then connected to three hydrogen atoms with single bonds. Used electrons: 7 single bonds * 2 = 14 electrons (b) In \(\mathrm{CS}_{2}\), the carbon atom is in the center, and it is connected to two sulfur atoms with single bonds. Used electrons: 2 single bonds * 2 = 4 electrons (c) In \(\mathrm{HCN}\), the hydrogen atom is connected to the carbon atom, which is connected to the nitrogen atom. Used electrons: 2 single bonds * 2 = 4 electrons

Distribute remaining electrons as lone pairs or multiple bonds

Finally, distribute the remaining electrons as lone pairs or multiple bonds to satisfy the octet rule (except for hydrogen, which needs only 2 electrons). (a) In \(\mathrm{C}_{2} \mathrm{H}_{6}\), all the electrons (14) have been used to form single bonds, and no more electrons need to be distributed. (b) In \(\mathrm{CS}_{2}\), 12 electrons remain (16 - 4). We will use these electrons to form two double bonds between the carbon and the two sulfur atoms. This uses up all 12 electrons: \(\mathrm{C}=\mathrm{S}-\mathrm{C}=\mathrm{S}\) (c) In \(\mathrm{HCN}\), 6 electrons remain (10 - 4). We will use these electrons to form a triple bond between carbon and nitrogen, and a lone pair on the nitrogen atom. This completes the octet rule for all atoms: \(\mathrm{H}-\mathrm{C} \equiv \mathrm{N}\)