Answer:
A. + 6.
B. + 3.
Step-by-step explanation:
A. K₂CrO₄:
- To calculate the oxidation no. of an element in a compound:
We have that the sum of the oxidation no. of different elements in the compound multiplied by its no. is equal to the overall charge of the compound.
∵ 2(oxidation no. of K) + (oxidation no. of Cr) + 4(oxidation no. of O) = 0.
oxidation no. of K = + 1, oxidation no. of O = - 2.
∴ 2(+ 1) + (oxidation no. of Cr) +4(- 2) = 0.
∴ 2 + (oxidation no. of Cr) - 8 = 0.
∴ (oxidation no. of Cr) - 6 = 0.
∴ oxidation no. of Cr = + 6.
B. Cr₂O₃:
∵ 2(oxidation no. of Cr) + 3(oxidation no. of O) = 0.
oxidation no. of O = - 2.
∴ 2(oxidation no. of Cr) + 3(- 2) = 0.
∴ 2(oxidation no. of Cr) - 6 = 0.
∴ 2(oxidation no. of Cr) = + 6.
∴ (oxidation no. of Cr) = + 6/2 = + 3.