Fluorouracil is a compound administered to cancer patients as a part of chemotherapy. Assign an oxidation state to every atom:

Here are some general rules regarding oxidation numbers: 1. Elements in their natural state have an oxidation state of 0. 2. Fluorine always has an oxidation state of -1 in compounds. 3. Oxygen usually has an oxidation state of -2 in compounds (except in peroxides where it is -1). 4. Hydrogen usually has an oxidation state of +1 in compounds (except in metallic hydrides where it is -1). Now we will assign oxidation states to the atoms in fluorouracil. Step 2: Draw the Lewis structure of the fluorouracil

Before we can calculate the oxidation states, we need to draw the Lewis structure of the molecule. 1. Calculate the total number of valence electrons in the molecule: Fluorine (F) has 7, Oxygen (O) has 6, Nitrogen (N) has 5, and Hydrogen (H) has 1. The total number of valence electrons in the molecule is 19. 2. Arrange the atoms in the structure: C-N-H ring, with O bonding to C, and F bonding to N. 3. Place a single bond between all connected atoms and fill in the remaining valence electrons as lone pairs. Step 3: Assign oxidation states based on the Lewis structure and the general rules

Now that we have the Lewis structure, we can use it to calculate the oxidation states: 1. Hydrogen (H) bonded to nitrogen has an oxidation state of +1 as per rule 4. 2. Nitrogen (N) bonded to hydrogen, carbon, and fluorine has an oxidation state of -3 since it donates 3 electrons to other atoms. 3. Fluorine (F) bonded to the nitrogen atom has an oxidation state of -1 as per rule 2. 4. Oxygen (O) bonded to carbon has an oxidation state of -2 as per rule 3. 5. Carbon (C) bonded to oxygen and nitrogen has an oxidation state of +4 since it receives 4 electrons from other atoms. The oxidation states of each atom in the fluorouracil molecule are as follows: 1. Hydrogen (H): +1 2. Nitrogen (N): -3 3. Fluorine (F): -1 4. Oxygen (O): -2 5. Carbon (C): +4