**Match Column I with Column II (Substance vs n-factor):**
**Column I:**
(A)
(B)
(C)
(D)
**Column II:**
(p)
(q)
(r)
(s)

Question

**Match Column I with Column II (Substance vs n-factor):**
**Column I:**
(A) $Fe_2(C_2O_4)_3$
(B) $Fe(HC_2O_4)_2$
(C) $KHC_2O_4.H_2C_2O_4$
(D) $Fe_{0.9}O \rightarrow Fe_2O_3$
**Column II:**
(p) $n=6$
(q) $n=5$
(r) $n=4$
(s) $n=0.7$

Prashant Kotian · Accepted Answer

Correct answer: A. **Logic:**
(A) is +3 (no change). 3 oxalates lose .
(B) (1) and 2 oxalates lose 4. Total .
(C) Reducing agent has 2 oxalate units. Total .
(D) average O.S. = 20/9. Final = 3. Change per atom = 7/9. For 0.9 atoms = .

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