Question

Match List I with List II for the oxidation state of central atoms:

List I	List II
(A) Cr2O72−	(I) +3
(B) MnO4−	(II) +5
(C) VO3−	(III) +7
(D) FeF63−	(IV) +6

Choose the correct answer from the options given below:

• A. (A)- (I), (B)- (II), (C)- (III), (D)- (IV)
• B. (A)- (IV), (B)- (III), (C)- (II), (D)- (I)
• C. (A)- (I), (B)- (II), (C)- (IV), (D)- (III)
• D. (A)- (IV), (B)- (I), (C)- (III), (D)- (II)

Answer 1

The Correct Option is B

To determine the oxidation state of the central atoms in the given chemical species, we need to use the rules for calculating oxidation numbers. Let's analyze each compound one by one:

1. Cr₂O₇²⁻ (Dichromate ion):
   Let the oxidation state of Cr be x.
   We know the oxidation state of O is -2.
   The equation is:
   \((2 \cdot x) + (7 \cdot -2) = -2\)
   \(2x - 14 = -2\)
   \(2x = 12\)
   \(x = +6\)
   Thus, the oxidation state of Cr in Cr₂O₇²⁻ is +6.
2. MnO₄⁻ (Permanganate ion):
   Let the oxidation state of Mn be x.
   The oxidation state of O is -2.
   The equation is:
   \(x + (4 \cdot -2) = -1\)
   \(x - 8 = -1\)
   \(x = +7\)
   Thus, the oxidation state of Mn in MnO₄⁻ is +7.
3. VO₃⁻ (Metavanadate ion):
   Let the oxidation state of V be x.
   The oxidation state of O is -2.
   The equation is:
   \(x + (3 \cdot -2) = -1\)
   \(x - 6 = -1\)
   \(x = +5\)
   Thus, the oxidation state of V in VO₃⁻ is +5.
4. FeF₆³⁻ (Hexafluoroferrate(III) ion):
   Let the oxidation state of Fe be x.
   The oxidation state of F is -1.
   The equation is:
   \(x + (6 \cdot -1) = -3\)
   \(x - 6 = -3\)
   \(x = +3\)
   Thus, the oxidation state of Fe in FeF₆³⁻ is +3.

Now that we have calculated the oxidation states, we can match List I with List II:

• (A) Cr₂O₇²⁻ - (IV) +6
• (B) MnO₄⁻ - (III) +7
• (C) VO₃⁻ - (II) +5
• (D) FeF₆³⁻ - (I) +3

Therefore, the correct answer is: (A)- (IV), (B)- (III), (C)- (II), (D)- (I).

Answer 2

Based on the information given in the image, the correct answer is Option B. Here's a breakdown of the reasoning:

(A) Cr₂O₇²⁻ (Dichromate ion):

• The oxidation state of Cr in Cr₂O₇²⁻ is +6 (since the overall charge of the ion is -2 and the oxidation states of oxygens are -2).
• This matches Option II in List II.

(B) MnO₄⁻ (Permanganate ion):

• The oxidation state of Mn in MnO₄⁻ is +7 (since the overall charge of the ion is -1 and the oxidation states of oxygens are -2).
• This matches Option III in List II.

(C) VO₃⁻ (Vanadate ion):

• The oxidation state of V in VO₃⁻ is +5 (since the overall charge of the ion is -1 and the oxidation states of oxygens are -2).
• This matches Option II in List II.

(D) FeF₆³⁻ (Ferrate ion):

• The oxidation state of Fe in FeF₆³⁻ is +3 (since the overall charge of the ion is -3 and each fluorine has an oxidation state of -1).
• This matches Option I in List II.

Thus, the correct matching is Option B:
(A) - (IV), (B) - (III), (C) - (II), (D) - (I).