QJEEM13Ch9dfe: JEE Main 2013 Chemistry - Class 11 JEEM13Ch9dfe Solution: Experimentally it was found that a metal oxide has formula M…

Given metal oxide = M_0.98O
We know oxidation number of O = (-2)
Now assume oxidaton no of M = $$x$$

$$\therefore$$ 0.98$$x$$ + 1$$\times$$ (-2) = 0
$$\Rightarrow x=$$ $$\frac{200}{98}$$
This represent the charge in one atom of M. As you can see the charge of M is in the range $$2<\frac{200}{98}<3$$.
So we can say in M mixture of M⁺² and M⁺³ present.

Assume total no of atoms present in M is 100.
Let M⁺³ present in M = y atoms
So M⁺² present in M = (100 - y) atoms

In 1 atom of M⁺³ charge present = +3
So in y atoms of M⁺³ charge present = +3y

Similarly in 1 atom of M⁺² charge present = +2
So in (100 - y) atoms of M⁺² charge present = +2(100 - y)

$$\therefore$$ Total charge = 200 - 2y + 3y = 200 + y

In 100 atoms of M total charge = 200 + y
So in 1 atoms of M total charge = $$\frac{200+y}{100}$$

Earlier we found that charge in one atom of M is = $$\frac{200}{98}$$

So we can write,
$$\frac{200+y}{100}$$ = $$\frac{200}{98}$$
$$\Rightarrow y=4.08$$

So Option (B) is correct.