Question

One mole of X₂H₄ releases 10 moles of electrons to form a compound Y. What should be the oxidation number of X in the compound Y?

• A. +3
• B. -3
• C. -6
• D. +1

Answer 1

Answer: (A)

+3

Answer 2

The initial compound is X₂H₄. We assume the oxidation state of hydrogen is +1. Let the oxidation state of X in X₂H₄ be $Ox_i(X)$.

The sum of oxidation states in a neutral molecule is zero:

$2 \times Ox_i(X) + 4 \times (+1) = 0$

$2 \times Ox_i(X) + 4 = 0$

$2 \times Ox_i(X) = -4$

$Ox_i(X) = -2$

One mole of X₂H₄ releases 10 moles of electrons to form compound Y. This means that the total oxidation state of the atoms being oxidized increases by 10. Assuming that only element X is oxidized and hydrogen remains in the +1 oxidation state in compound Y, the total increase in oxidation state is due to the two X atoms in X₂H₄.

Let the oxidation state of X in compound Y be $Ox_f(X)$. The change in oxidation state for one X atom is $Ox_f(X) - Ox_i(X) = Ox_f(X) - (-2) = Ox_f(X) + 2$.

Since there are two X atoms in one molecule (or one mole) of X₂H₄, the total change in oxidation state for the two X atoms is $2 \times (Ox_f(X) + 2)$.

This total change in oxidation state is equal to the number of electrons released, which is 10 moles for one mole of X₂H₄.

Total change in oxidation state = Number of electrons released

$2 \times (Ox_f(X) + 2) = 10$

Now, we solve for $Ox_f(X)$:

$Ox_f(X) + 2 = \frac{10}{2}$

$Ox_f(X) + 2 = 5$

$Ox_f(X) = 5 - 2$

$Ox_f(X) = +3$

Thus, the oxidation number of X in compound Y is +3.

Alternatively, we can consider the total oxidation state of X atoms.

Initial total oxidation state of X in X₂H₄ = $2 \times Ox_i(X) = 2 \times (-2) = -4$.

Let the final oxidation state of X in Y be $Ox_f(X)$. The final total oxidation state of the two X atoms in Y is $2 \times Ox_f(X)$.

The release of 10 electrons corresponds to a total increase in oxidation state of +10.

Final total oxidation state of X = Initial total oxidation state of X + Total increase in oxidation state

$2 \times Ox_f(X) = -4 + 10$

$2 \times Ox_f(X) = +6$

$Ox_f(X) = \frac{+6}{2}$

$Ox_f(X) = +3$