Question

When an acidified solution of $N{a_2}Mo{O_n}$ (atomic mass of Mo=96) is electrolyzed, ${O_2}$ gas is liberated corresponding to a volume of 0.112 L at STP and mass of Mo deposited is 0.32 g. Then, the formula of the salt and oxidation state of Mo is:
A. $N{a_2}MoO,0$
B. $N{a_2}Mo{O_4}, + 6$
C. $N{a_2}Mo{O_2}, + 2$
D. $N{a_2}Mo{O_3}, + 4$

Answer 1

Hint: Oxidation number also called oxidation state, the total number of electrons that an atom either gains or losses in order to form a chemical bond with another atom. Oxidation states simplify the whole process of working out what is being oxidized and what is being reduced in redox reactions. However, for the purpose of finding out the oxidation state.

Complete step by step answer:
Here we have found out the formula of the salt and oxidation state of Mo. We will find the value by the given values as given below:
  N factor of ${O_2} = 4$
So, the calculation,
 $ = \dfrac{{0.112}}{{\dfrac{{22.4}}{4}}}$
\[ = 0.02\]
 Eq. of ${O_2} = $ Eq. of Mo
 $0.02 = \dfrac{{0.32}}{{\dfrac{{96}}{x}}}$
$ \Rightarrow x = 6$
Now the formula of salt is: $N{a_2}Mo{O_n}$
\[O = 2\left( 1 \right) + 6 + n\left( { - 2} \right)\]
\[ \Rightarrow n = 4\]
So, the formula of salt is $N{a_2}Mo{O_4}$ and the oxidation state is 6.

Here, Option (D) is the correct answer.

Note: ESR method that catalyst of epoxidation of olefins by ethylbenzene hydroperoxides is a compound of molybdenum in the oxidation state \[ + 5\] which, when it reacts with ethylbenzene hydroperoxides, is converted into oxidation state \[ + 6\] of Mo. The electrochemical behaviors of molybdenum and its oxides, both in bulk and thin film dimensions, are critical because of their widespread applications in steel and solar cells.