Can someone check my answers?
What is the oxidation state of each element in K2Cr2O7?
K +1
Cr +6
O -2
What is the oxidation state of each element in SO32–?
S +4
O -6
OK on 1.
On 2 EACH element for O is -2. The total is -6 for 3 O atoms.
To determine the oxidation state of an element in a compound, we need to assign a charge or oxidation state to each element based on a set of rules. Let's go through the oxidation states of each element in the given compounds.
For K2Cr2O7:
- The oxidation state of K (potassium) is +1, as it is an alkali metal.
- The oxidation state of Cr (chromium) can be determined by considering that the total charge of the compound is zero. Since there are two potassium ions each with a +1 charge and seven oxygen ions each with a -2 charge (2 x 1 + 7 x -2 = 0), we can calculate the oxidation state of Cr: 2x +1 + 2x + (-2) x 7 = 0. By solving this equation, we find that the oxidation state of Cr is +6.
- The oxidation state of O (oxygen) is always -2 in most compounds, so in this case, it is -2. There are seven oxygen atoms in the compound, each with an oxidation state of -2.
Therefore, the oxidation state of each element in K2Cr2O7 is:
K +1
Cr +6
O -2
For SO32–:
- The oxidation state of S (sulfur) can be determined by considering that the total charge of the compound is -2. Based on the charge, the oxidation state can be calculated: x + (-2) x 3 = -2. By solving this equation, we find that the oxidation state of S is +4.
- The oxidation state of O (oxygen) is always -2 in most compounds, so in this case, it is -2. There are three oxygen atoms in the compound, each with an oxidation state of -2.
Therefore, the oxidation state of each element in SO32– is:
S +4
O -2
To check the oxidation state of each element in a compound, you need to determine the possible oxidation states for each element and then assign the most appropriate oxidation state based on the compound's overall charge and the known oxidation states of other elements.
1. K2Cr2O7:
- The oxidation state of potassium (K) in most compounds is +1, as it readily donates one electron.
- The oxidation state of oxygen (O) in most compounds is -2.
- Since the compound has a neutral charge (0), the sum of the oxidation states must add up to zero.
- With this information, you can assign the oxidation state of chromium (Cr) using the formula: K2(+1) + Crx + 7(-2) = 0.
- Solving for x, you get x = +6. Therefore, the oxidation state of Cr in K2Cr2O7 is +6.
2. SO32–:
- The oxidation state of oxygen (O) in most compounds is -2.
- The compound has a negative charge (-2), so the sum of the oxidation states must add up to -2.
- With this information, you can assign the oxidation state of sulfur (S) using the formula: Sx + 3(-2) = -2.
- Solving for x, you get x = +4. Therefore, the oxidation state of S in SO32– is +4.
To check if your answers are correct, you can compare them with the oxidation states derived from the rules and formula provided.