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 ascertain the oxidation state of the central atoms in the provided chemical species, we apply the established rules for calculating oxidation numbers. Each compound will be examined sequentially:

1. Cr₂O₇²⁻ (Dichromate ion):
   Designate the oxidation state of Cr as x.
   The oxidation state of O is recognized as -2.
   The derived equation is:
   \((2 \cdot x) + (7 \cdot -2) = -2\)
   \(2x - 14 = -2\)
   \(2x = 12\)
   \(x = +6\)
   Consequently, the oxidation state of Cr in Cr₂O₇²⁻ is +6.
2. MnO₄⁻ (Permanganate ion):
   Designate the oxidation state of Mn as x.
   The oxidation state of O is -2.
   The derived equation is:
   \(x + (4 \cdot -2) = -1\)
   \(x - 8 = -1\)
   \(x = +7\)
   Therefore, the oxidation state of Mn in MnO₄⁻ is +7.
3. VO₃⁻ (Metavanadate ion):
   Designate the oxidation state of V as x.
   The oxidation state of O is -2.
   The derived 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):
   Designate the oxidation state of Fe as x.
   The oxidation state of F is -1.
   The derived equation is:
   \(x + (6 \cdot -1) = -3\)
   \(x - 6 = -3\)
   \(x = +3\)
   Hence, the oxidation state of Fe in FeF₆³⁻ is +3.

Following the calculation of the oxidation states, we proceed to match List I with List II:

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

The conclusive pairing is: (A)- (IV), (B)- (III), (C)- (II), (D)- (I).