Match List I with List II:

Choose the correct answer from the option given below:

i-c, ii-d, iii-b, iv-a

The correct answer is option 1. i-c, ii-d, iii-b, iv-a.

To match the compounds with their respective oxidation states of the central metal atoms, let us calculate the oxidation states for each:

i. \(Cr_2O_7^{2-}\) (Dichromate ion)

The dichromate ion \(Cr_2O_7^{2-}\) consists of two chromium atoms and seven oxygen atoms. The overall charge of the ion is -2. The oxidation state of oxygen is typically -2.

Let the oxidation state of chromium be \(x\). The sum of the oxidation states in the dichromate ion is:

\(2x + 7(-2) = -2 \)

\(2x - 14 = -2 \)
\( 2x = 12 \)
\(x = +6 \)

So, the oxidation state of chromium in \(Cr_2O_7^{2-}\) is +6.

ii. \(MnO_4^-\) (Permanganate ion)

The permanganate ion \(MnO_4^-\) consists of one manganese atom and four oxygen atoms. The overall charge of the ion is -1. The oxidation state of oxygen is -2.

Let the oxidation state of manganese be \(x\). The sum of the oxidation states in the permanganate ion is:

\(x + 4(-2) = -1 \)

\(x - 8 = -1 \)

\(x = +7 \)

So, the oxidation state of manganese in \(MnO_4^-\) is +7.

iii. \(VO_3^-\) (Metavanadate ion)

The metavanadate ion \(VO_3^-\) consists of one vanadium atom and three oxygen atoms. The overall charge of the ion is -1. The oxidation state of oxygen is -2.

Let the oxidation state of vanadium be \(x\). The sum of the oxidation states in the metavanadate ion is:

\(x + 3(-2) = -1 \)

\(x - 6 = -1 \)

\(x = +5 \)

So, the oxidation state of vanadium in \(VO_3^-\) is +5.

iv. \(FeF_6^{3-}\) (Hexafluoroferrate(III) ion)

The hexafluoroferrate(III) ion \(FeF_6^{3-}\) consists of one iron atom and six fluorine atoms. The overall charge of the ion is -3. The oxidation state of fluorine is -1.

Let the oxidation state of iron be \(x\). The sum of the oxidation states in the hexafluoroferrate(III) ion is:

\(x + 6(-1) = -3 \)

\(x - 6 = -3 \)

\(x = +3 \)

So, the oxidation state of iron in \(FeF_6^{3-}\) is +3.

The correct answer is: Option 1: i-c, ii-d, iii-b, iv-a.