What are the oxidation states of the following elements? (a) Cr in K2Cr2O7 (b) Sulfur in Li2S2O3
The Correct Answer and Explanation is:
Let’s calculate the oxidation states of the elements in the given compounds:
(a) Chromium (Cr) in K2Cr2O7 (Potassium dichromate)
- K (Potassium): Potassium typically has an oxidation state of +1 because it is an alkali metal.
- O (Oxygen): Oxygen typically has an oxidation state of -2 in most compounds (except in peroxides and superoxides).
Now, let’s use these values to find the oxidation state of Cr:
- The overall charge of the molecule K2Cr2O7 is neutral, so the sum of all oxidation states must be 0.
- We have 2 potassium ions (each with a +1 charge), 2 chromium atoms (which we’ll call xxx for their oxidation state), and 7 oxygen atoms (each with a -2 charge).
The equation for the sum of oxidation states is:2(+1)+2(x)+7(−2)=02(+1) + 2(x) + 7(-2) = 02(+1)+2(x)+7(−2)=0
Simplifying this:2+2x−14=02 + 2x – 14 = 02+2x−14=02x−12=02x – 12 = 02x−12=02x=122x = 122x=12x=+6x = +6x=+6
Thus, the oxidation state of Cr in K2Cr2O7 is +6.
(b) Sulfur (S) in Li2S2O3 (Lithium thiosulfate)
- Li (Lithium): Lithium has an oxidation state of +1, as it is an alkali metal.
- O (Oxygen): Again, oxygen typically has an oxidation state of -2.
Now, let’s determine the oxidation state of sulfur (S):
- The molecule Li2S2O3 is neutral, so the sum of oxidation states must be 0.
- There are 2 lithium ions (each with a +1 charge), 2 sulfur atoms (which we’ll call xxx for their oxidation state), and 3 oxygen atoms (each with a -2 charge).
The equation for the sum of oxidation states is:2(+1)+2(x)+3(−2)=02(+1) + 2(x) + 3(-2) = 02(+1)+2(x)+3(−2)=0
Simplifying this:2+2x−6=02 + 2x – 6 = 02+2x−6=02x−4=02x – 4 = 02x−4=02x=42x = 42x=4x=+2x = +2x=+2
Thus, the oxidation state of sulfur in Li2S2O3 is +2.
Summary:
- (a) The oxidation state of Cr in K2Cr2O7 is +6.
- (b) The oxidation state of sulfur in Li2S2O3 is +2.
