Question

Identify the compound formed from elements X, Y, Z having oxidation states $+2$, $+5$, $-2$ respectively.

• A. $\text{X}(\text{Y}_4\text{Z})$
• B. $\text{X}_3(\text{YZ}_4)_2$
• C. $\text{X}_3(\text{YZ}_2)_2$
• D. $\text{XYZ}_2$

Hint:

Think of this as a classic criss-cross valency problem! Treat the polyatomic cluster $(\text{YZ}_4)$ as a single radical unit. Its net charge is $+5 + 4(-2) = -3$. Criss-crossing the valencies of $\text{X}^{2+}$ and $(\text{YZ}_4)^{3-}$ immediately delivers the formula $\text{X}_3(\text{YZ}_4)_2$!

Answer 1

The Correct Option is B

Step 1: State the neutrality rule.
A real compound is electrically neutral, so the oxidation states of all atoms, weighted by how many there are, must add to zero.
Step 2: List the charges.
X is $+2$, Y is $+5$, Z is $-2$.
Step 3: Test option (1) $\text{X}(\text{Y}_4\text{Z})$.
$(+2) + 4(+5) + (-2) = +20$, not zero, so it is rejected.
Step 4: Test option (2) $\text{X}_3(\text{YZ}_4)_2$.
First the group $\text{YZ}_4$ has charge $(+5) + 4(-2) = -3$. Then $3(+2) + 2(-3) = 6 - 6 = 0$. This balances.
Step 5: Test option (3) $\text{X}_3(\text{YZ}_2)_2$.
Group $\text{YZ}_2$ is $(+5) + 2(-2) = +1$, and $3(+2) + 2(+1) = +8$, not zero.
Step 6: Test option (4) $\text{XYZ}_2$.
$(+2) + (+5) + 2(-2) = +3$, not zero.
Step 7: Choose the balanced formula.
Only option (2) sums to zero, so it is the answer.
\[ \boxed{\text{X}_3(\text{YZ}_4)_2} \]