Using Lewis Formulas, assign the oxidation states to the atoms in:
(a) CH3S–SCH3(l)
Assume sulfur is more electronegative than carbon.
To assign the oxidation states to the atoms in CH3S–SCH3(l), we need to consider the electronegativity and the bonding patterns of each element.
In the compound CH3S–SCH3(l), we have two carbon atoms (C), one sulfur atom (S), and two hydrogen atoms (H).
According to the assumption that sulfur is more electronegative than carbon, we can assign the oxidation state for each element as follows:
1. Hydrogen (H):
Hydrogen usually has an oxidation state of +1 when bonded to nonmetals. Therefore, each hydrogen atom has an oxidation state of +1.
2. Carbon (C):
Since carbon typically has four valence electrons, it will share electrons with other atoms. In CH3S–SCH3(l), each carbon atom is bonded to three hydrogen atoms and one sulfur atom.
Considering that hydrogen has an oxidation state of +1, the carbon atoms will have an oxidation state of -1. This is because the oxidation states of each hydrogen atom sum up to +3, and the oxidation state of the carbon atom must balance out the total charge of the molecule, which is 0.
3. Sulfur (S):
Given that sulfur is more electronegative than carbon, it will carry a negative oxidation state. To determine its exact value, we need to consider that the overall charge of the molecule is 0.
Since we have two carbon atoms with an oxidation state of -1 and each hydrogen atom has an oxidation state of +1, we can calculate the oxidation state of sulfur.
2*(-1) + 2*(+1) + oxidation state of sulfur = 0
-2 + 2 + oxidation state of sulfur = 0
oxidation state of sulfur = 0
The oxidation state of sulfur in CH3S–SCH3(l) is 0.
Therefore, the oxidation states for the atoms in CH3S–SCH3(l) are as follows:
Carbon (C): -1
Hydrogen (H): +1
Sulfur (S): 0
To assign oxidation states using Lewis Formalism, we need to follow a set of guidelines:
1. Determine the overall charge of the molecule or ion: In this case, the molecule is CH3S–SCH3, where the negative charge indicates an extra electron.
2. Assign the oxidation state for each atom in the molecule or ion by considering the following rules:
a. The oxidation state of an atom in its elemental form is always 0. Therefore, both carbon (C) and sulfur (S) atoms in this molecule are initially assigned an oxidation state of 0.
b. Assign the oxidation state of atoms bonded to more electronegative elements: In this case, sulfur (S) is more electronegative than carbon (C). So, the S atom is assigned a negative oxidation state (-2).
c. Assign the oxidation state of atoms bonded to less electronegative elements: In this molecule, both carbon (C) atoms are bonded to sulfur (S). Since carbon is less electronegative than sulfur, each C atom is assigned a positive oxidation state (+2) in order for the molecule to be neutral overall.
d. Calculate the oxidation state based on valence electrons. In general, the sum of the oxidation states for all the atoms in a neutral molecule should add up to zero, while for ions, the sum should equal the charge.
Now, let's assign the oxidation states of the atoms in CH3S–SCH3:
For the Sulfur (S) atom:
- Sulfur (S) is bonded to three hydrogen (H) atoms, which are less electronegative. Therefore, each hydrogen contributes +1 to the oxidation state of sulfur (S): (+1) x 3 = +3
- The overall charge of the molecule is -1. Therefore, the oxidation state of sulfur (S) can be calculated as follows:
-1 = (-2) + (+3)
Since (-2) + (+3) = +1, the oxidation state of sulfur (S) is +1.
For the Carbon (C) atoms:
- Each carbon (C) atom is bonded to one sulfur (S) atom, which is more electronegative. Thus, the contribution from sulfur (S) is -2 for each C atom.
- The overall charge is -1, so the oxidation state of carbon (C) can be calculated as follows:
-1 = (+2) + (-2) + (+2)
Since (+2) + (-2) + (+2) = +2, the oxidation state of carbon (C) is +2.
Therefore, the oxidation states of the atoms in CH3S–SCH3 are:
Sulfur (S) atom: +1
Carbon (C) atoms: +2