Question

Using stock notation, match the oxidation number against metal in its compound.}

• A. (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
• B. (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)
• C. (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)
• D. (i)-(d), (ii)-(a), (iii)-(b), (iv)-(c)
• E. (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)

Hint:

In oxides, oxygen is usually \(-2\). Use total charge \(=0\) for neutral compounds to find the metal oxidation state.

Answer 1

The Correct Option is A

Find oxidation states: For \(MnO\): \[ x + (-2)=0 \Rightarrow x=+2 \] For \(MnO_2\): \[ x + 2(-2)=0 \Rightarrow x=+4 \] For \(Fe_2O_3\): \[ 2x + 3(-2)=0 \Rightarrow 2x=6 \Rightarrow x=+3 \] For \(Tl_2O\): \[ 2x + (-2)=0 \Rightarrow 2x=2 \Rightarrow x=+1 \] So matching is: \[ MnO \to II,\quad MnO_2 \to IV,\quad Fe_2O_3 \to III,\quad Tl_2O \to I \]
Hence, the correct answer is: \[ \boxed{(A)} \]