What is the oxidation number of Cr in the following compound?
K2Cr2O7
My answer, (since there are no multiple choice selections) is " 6 "
To determine the oxidation number of chromium (Cr) in the compound K2Cr2O7, we need to assign oxidation numbers to all the atoms in the compound.
Step 1: Start with the known oxidation numbers for other elements.
In K2Cr2O7, potassium (K) has a known oxidation number of +1 since it belongs to Group 1 of the periodic table.
Step 2: Determine the oxidation number for oxygen (O).
Oxygen typically has an oxidation number of -2 in compounds, and since there are seven oxygens in K2Cr2O7, the total oxidation number contribution from the oxygen atoms is -2 x 7 = -14.
Step 3: Determine the overall charge of the compound.
K2Cr2O7 is a neutral compound, meaning the sum of the oxidation numbers must add up to zero. Since there are two potassium atoms, each with an oxidation number of +1, the total contribution from potassium is +2. Adding this to the total oxidation number of oxygen (-14) gives us -14 + 2 = -12.
Step 4: Determine the oxidation number for chromium (Cr).
To find the oxidation number of chromium, we need to solve for x in the equation: +2 (from K) + x (from Cr) + (-12) (from O) = 0.
Simplifying the equation, we get:
2 + x - 12 = 0
x - 10 = 0
x = +10
Therefore, the oxidation number of chromium (Cr) in K2Cr2O7 is +6, not 6 as you mentioned in your answer.