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Question: A transition metal M can exist in two oxidation states +2 and +3. It forms an oxide whose experiment...

A transition metal M can exist in two oxidation states +2 and +3. It forms an oxide whose experimental formula is given by MxO{{M}_{x}}O where x < 1. Then the ratio of metal ions in +3 state to those in +2 state in oxide is given by:
(A)- 1−x1+x\dfrac{1-x}{1+x}
(B)- 2(x−1)(2−3x)\dfrac{2(x-1)}{(2-3x)}
(C)- 1+x21+\dfrac{x}{2}
(D)- 2(x−1)3x−2\dfrac{2(x-1)}{3x-2}

Explanation

Solution

The ratio of the two oxidation states present in the metal ion can be calculated by the balancing of the charge over the compound. As from the stoichiometry of the formula, the amount of the metal ions present can be determined.

Complete answer:
It is given that the oxide of transition metal, M is in the form of MxO{{M}_{x}}O, where the metal exists in two oxidation states. From the stoichiometry of the formula, it is seen that in one mole of the metal oxide, it has ‘x’ moles of the metal ion and one mole of the oxide.
In the metal ion, there are (+2) and (+3) oxidation states. Let the number of moles of (+2) state be ‘y’, then the number of moles of (+3) state will be x-y.
Then, by balancing the charge over the compound, My+2 Mx−y+3 OM_{y}^{+2}\,M_{x-y}^{+3}\,O . As the overall charge of the compound is zero, we will obtain a relation between x and y as follows:
2y+3(x−y)+(−2)×1=02y+3(x-y)+(-2)\times 1=0
2y+3x−3y−2=02y+3x-3y-2=0
y=3x−2y=3x-2 --------- (a)
So, now the ratio of the two ions in the metal oxide will be:
M+3M+2=x−yy\dfrac{{{M}^{+3}}}{{{M}^{+2}}}=\dfrac{x-y}{y}
Substituting value of y, from equation (a) in above equation, we will get,
M+3M+2=x−(3x−2)3x−2=−2x+23x−2=−2(x−1)3x−2=2(x−1)2−3x\dfrac{{{M}^{+3}}}{{{M}^{+2}}}=\dfrac{x-(3x-2)}{3x-2}=\dfrac{-2x+2}{3x-2}=\dfrac{-2(x-1)}{3x-2}=\dfrac{2(x-1)}{2-3x}
But as given, that x < 1, so the ration can also be written as, M+3M+2=2(1−x)3x−2\dfrac{{{M}^{+3}}}{{{M}^{+2}}}=\dfrac{2(1-x)}{3x-2}.

Therefore, the ratio of metal ions in +3 state to those in +2 state in the metal oxide is given by option (B)- 2(x−1)(2−3x)\dfrac{2(x-1)}{(2-3x)}.

Note:
The transition metals belonging to the d-block show variable oxidation states due to the presence of the incomplete filling in its d-orbital. Thus, having a difference of unity in the oxidation states is possible.