Assign an oxidation state to each atom in HCN. (Hint: Begin with a dot diagram.)

To begin, find the number of valence electrons for each atom in the molecule. Hydrogen (H) has 1 valence electron, carbon (C) has 4 valence electrons, and nitrogen (N) has 5 valence electrons. Add these together to find the total number of valence electrons: \(1+4+5 = 10\). Now, let's draw the HCN Lewis structure. Place the Carbon in the center of the molecule as it has the lowest electronegativity. Connect hydrogen to carbon and nitrogen to carbon with single bonds. Now there are \(2\) electrons left from the \(10\) valence electrons; we put both electrons on the nitrogen atom as lone pairs. The Lewis structure for HCN looks like: ``` H - C ≡ N: ``` The lines represent bonds, with a triple bond between carbon and nitrogen.

To assign an oxidation state to each atom in the molecule, follow these rules: 1. The oxidation state of an element in its elemental form is 0. 2. The oxidation state of hydrogen is usually +1 when bonded to non-metals. 3. The oxidation state of oxygen is usually -2 in compounds. 4. The oxidation state of the remaining elements is assigned such that the sum of all oxidation states equals the overall charge of the molecule or ion. First, assign the oxidation state of hydrogen as +1. Next, we need to assign the oxidation state for carbon and nitrogen. Since the HCN molecule is neutral, the sum of oxidation states must equal 0. Let the oxidation state of carbon be x and that of nitrogen be y. So, we have the equation: \(1 + x + y = 0\) Since carbon is forming a triple bond with nitrogen, we can infer that nitrogen will have a higher oxidation state than carbon. Therefore, assign the oxidation state of nitrogen as -3. This gives us: \(1 + x - 3 = 0\) Now, solve for x: \(x = 3 - 1 = 2\) So, the oxidation state of carbon is +2. In summary, the oxidation states for each atom in the HCN molecule are: - Hydrogen (H): +1 - Carbon (C): +2 - Nitrogen (N): -3