Question

In chromite ore, the oxidation number of iron, and chromium are respectively.
A. +3, +2
B. +3, +6
C. +2, +6
D. +2, +3

Answer 1

The formula of chromite ore is $FeC\mathop r\nolimits_2 \mathop O\nolimits_4$ or we can say its actual composition is $FeO.C\mathop r\nolimits_2 \mathop O\nolimits_3$. Oxidation no. is a number of an element in chemical combination which represents the number of electrons lost or gained.

Complete step by step answer:
As we already mentioned that chromite ore exists as $FeO.C\mathop r\nolimits_2 \mathop O\nolimits_3$ . For calculating oxidation no. of Fe and Cr, we will separate $FeO$ and $C\mathop r\nolimits_2 \mathop O\nolimits_3$, as these two are complete formulas. As we know, that oxidation no. or oxidation state of oxygen is -2 and in a complete formula, having no charge on it, the sum of oxidation no. of all the atoms is zero. First, we will calculate the oxidation no. of Fe. In $FeO$, we know that oxidation no. of O is 0 and let's say oxidation no. of Fe is x.
So, $x + \left( { - 2} \right) = 0$
$x = 2$
So, the oxidation no. of Fe in chromite ore is 2. Now, we will calculate oxidation no. of
Cr from $C\mathop r\nolimits_2 \mathop O\nolimits_3$. Let's say oxidation no. of Cr is y.
So, $\left( y \right)2 + \left( { - 2} \right)3 = 0$
$\left( y \right)2 = 6$
And,$y = \dfrac{6}{2} = 3$ .
So, oxidation no. of Cr is +3.
So, the correct option will be D. +2 and +3.

Note:
We can cross check our answer by putting the value of oxidation no. of Fe and Cr in
$FeC\mathop r\nolimits_2 \mathop O\nolimits_4$.
So, oxidation no. of$Fe + 2Cr + 4O = 0$ .
$2 + 2\left( 3 \right) + 4\left( { - 2} \right) = 2 + 6 - 8 = 0$ .