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: Understanding the Question:
The question asks us to identify the chemical formula that correctly indicates the oxidation state of the central metal atom using Stock notation.

Step 2: Key Formula or Approach:
In Stock notation, the oxidation state of the metal is specified by a Roman numeral enclosed in parentheses immediately following the metal name or symbol in the molecular formula. The value of this Roman numeral must perfectly match the actual calculation of the metal's oxidation state in that specific compound.

Step 3: Detailed Explanation:
Let's test each compound by computing the true oxidation state of the metal: 1.

Fe (II) $\text{Cl}_3$: Chlorine has a standard charge of $-1$. For $\text{FeCl}_3$, the equation is $x + 3(-1) = 0 \implies x = +3$. The correct notation should be $\text{Fe(III)Cl}_3$. Hence, option (A) is incorrect.
2.

Mn ( II ) $\text{O}_2$: Oxygen has a standard charge of $-2$. For $\text{MnO}_2$, $x + 2(-2) = 0 \implies x = +4$. The correct notation should be $\text{Mn(IV)O}_2$. Hence, option (B) is incorrect.
3.

Au (III)Cl: For AuCl, $x + 1(-1) = 0 \implies x = +1$. The correct notation should be Au(I)Cl. Hence, option (C) is incorrect.
4.

Sn ( IV ) $\text{Cl}_4$: For $\text{SnCl}_4$, $x + 4(-1) = 0 \implies x = +4$. The Roman numeral given is (IV), which accurately represents the $+4$ oxidation state of Tin. Therefore, this formula is correctly represented.

Step 4: Final Answer:
The correctly represented formula is Sn ( IV ) $\text{Cl}_4$, which corresponds to option (D).