Question

Which among the following formulae is correctly represented according to stock notation?

• A. Fe (II) $\text{Cl}_3$
• B. Mn ( II ) $\text{O}_2$
• C. Au (III)Cl
• D. Sn ( IV ) $\text{Cl}_4$

Hint:

To verify Stock notation instantly, ignore the written Roman numeral for a second, look at the subscript of the monoanionic ligand (like $\text{Cl}^-$) to find the metal's true charge, and see if it matches the numeral. For $\text{SnCl}_4$, the 4 chlorines mean a $+4$ charge, which matches (IV) perfectly!

Answer 1

The Correct Option is D

Step 1: State the Stock notation rule.
The Roman numeral in parentheses must equal the true oxidation state of the metal in that compound.
Step 2: Test Fe(II)Cl3.
Each Cl is $-1$, three Cl give $-3$, so iron must be $+3$. The notation says (II), which is wrong; it should be (III).
Step 3: Test Mn(II)O2.
Each O is $-2$, two O give $-4$, so manganese must be $+4$. The notation says (II), wrong; it should be (IV).
Step 4: Test Au(III)Cl.
One Cl is $-1$, so gold must be $+1$. The notation says (III), wrong; it should be (I).
Step 5: Test Sn(IV)Cl4.
Four Cl give $-4$, so tin must be $+4$. The notation says (IV), which exactly matches. Correct.
Step 6: Conclude.
Only $\text{Sn(IV)Cl}_4$ has the Roman numeral matching the calculated oxidation state.
\[ \boxed{\text{Sn(IV)Cl}_4,\ \text{option (4)}} \]