Redox Reactions — NEET UG practice

1. In acidic medium, $\mathrm{KMnO}_4$ oxidises oxalic acid completely, so we first write the balanced redox reaction between them.

$$2\mathrm{MnO}_4^- + 5\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 + 6\mathrm{H}^+ \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}$$

2. Since this is a titration, the law of equivalence applies — the number of milliequivalents of $\mathrm{KMnO}_4$ equals the number of milliequivalents of oxalic acid at the end point.

$$V_1 N_1(\mathrm{KMnO}_4) = V_2 N_2(\mathrm{oxalic\ acid})$$

3. The n-factor of $\mathrm{KMnO}_4$ in acidic medium is 5 (it gains 5 electrons per formula unit), and the n-factor of oxalic acid is 2 (it loses 2 electrons). Given $V(\mathrm{KMnO}_4) = V(\text{oxalic acid}) = 10\ \text{mL}$ and $M(\text{oxalic acid}) = 0.25\ \text{M}$, we substitute:

$$10 \times (M \times 5) = 10 \times (0.25 \times 2)$$

4. Solving for $M$, the molarity of $\mathrm{KMnO}_4$:

$$M = \frac{10 \times 0.5}{10 \times 5} = \frac{5}{50} = 0.10\ \text{M}$$

Hence, the strength of $\mathrm{KMnO}_4$ solution is 0.10 M, so the correct answer is Option B.

1. To check each option, we must calculate the oxidation state of the underlined atom using the rule that oxygen is usually $-2$ and hydrogen is usually $+1$ (except in metal hydrides where it is $-1$), and the sum of oxidation states equals the overall charge (zero for neutral molecules).

2. Checking option A: for $\mathrm{P}_4\mathrm{O}_6$,

$$4x + 6(-2) = 0 \Rightarrow x = +3$$

for $\mathrm{Cl}_2\mathrm{O}_7$,

$$2x + 7(-2) = 0 \Rightarrow x = +7$$

Since $+3$ is not greater than $+7$, the claimed order P > Cl is wrong, so option A is incorrect.

3. Checking option B: for $\mathrm{P}_4\mathrm{O}_{10}$,

$$4x + 10(-2) = 0 \Rightarrow x = +5$$

for $\mathrm{SO}_3$,

$$x + 3(-2) = 0 \Rightarrow x = +6$$

Since $+5$ is not greater than $+6$, this order is also wrong.

4. Checking option C: for $\mathrm{N}_2\mathrm{O}_5$,

$$2x + 5(-2) = 0 \Rightarrow x = +5$$

for $\mathrm{Al}_2\mathrm{O}_3$,

$$2x + 3(-2) = 0 \Rightarrow x = +3$$

for $\mathrm{H}_2\mathrm{S}$ (here S is more electronegative than H, so H is $+1$),

$$2(+1) + x = 0 \Rightarrow x = -2$$

Therefore the order is

$$+5 > +3 > -2$$

which is indeed strictly decreasing — this option is correct.

5. Checking option D: for $\mathrm{PbO}_2$, Pb $= +4$; for $\mathrm{N}_2\mathrm{O}_3$, N $= +3$; for $\mathrm{SO}_3$, S $= +6$. Since $+6 > +4$, the claimed order is wrong.

Hence, only option C gives a genuinely decreasing sequence of oxidation states, so the correct answer is Option C.

1. Alkali metals always show an oxidation state of $+1$ in their compounds, since they readily lose their single valence electron. Therefore in $\mathrm{KO}_2$,

$$\text{oxidation state of K} = +1$$

2. In $\mathrm{H}_2\mathrm{O}_2$, oxygen is present as a peroxide linkage ($\mathrm{O–O}$), which gives each oxygen atom an oxidation state of $-1$ instead of the usual $-2$. Taking hydrogen as $+1$,

$$2(+1) + 2x = 0 \Rightarrow x = -1$$

3. In $\mathrm{H}_2\mathrm{SO}_4$, hydrogen is $+1$ and oxygen is $-2$ (normal oxide, no peroxide link here), so we solve for sulfur:

$$2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$$

Hence, the oxidation states of the underlined elements are $+1$, $-1$, and $+6$ respectively, matching Option C.