Identify the process in which change in the oxidation state is five :

Redox Reactions PYQ — JEE Main 2021 · 25 Jul · Shift 2 · Q14 with Solution

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Correct answer: B

We need to find the process in which the change in oxidation state is 5.
Check each option by calculating oxidation states of the relevant atom.

Option A: $Cr_2O_7^{2-} \to 2Cr^{3+}$

Let oxidation state of Cr in $Cr_2O_7^{2-}$ be $x$ .

$2x + 7(-2) = -2$ $2x - 14 = -2$ $2x = 12$ $x = +6$

In product, Cr is $+3$ .

So change per Cr atom:

$+6 \to +3$ $\Delta = 3$

So this is not 5.

Option B: $MnO_4^- \to Mn^{2+}$

Let oxidation state of Mn in $MnO_4^-$ be $x$ .

$x + 4(-2) = -1$ $x - 8 = -1$ $x = +7$

In product, Mn is $+2$ .

So change:

$+7 \to +2$ $\Delta = 5$

So this matches.

Option C: $CrO_4^{2-} \to Cr^{3+}$

Let oxidation state of Cr in $CrO_4^{2-}$ be $x$ .

$x + 4(-2) = -2$ $x - 8 = -2$ $x = +6$

In product, Cr is $+3$ .

So change:

$+6 \to +3$ $\Delta = 3$

So this is not 5.

Option D: $C_2O_4^{2-} \to 2CO_2$

Let oxidation state of each C in $C_2O_4^{2-}$ be $x$ .

$2x + 4(-2) = -2$ $2x - 8 = -2$ $2x = 6$ $x = +3$

In $CO_2$ , carbon is:

$x + 2(-2) = 0$ $x = +4$

So change:

$+3 \to +4$ $\Delta = 1$

So this is not 5.

Therefore, the only process with oxidation state change equal to 5 is:

$MnO_4^- \to Mn^{2+}$

So the correct option is B.

Redox Reactions question

Redox Reactions question

Option A: $Cr_2O_7^{2-} \to 2Cr^{3+}$

Option B: $MnO_4^- \to Mn^{2+}$

Option C: $CrO_4^{2-} \to Cr^{3+}$

Option D: $C_2O_4^{2-} \to 2CO_2$

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Redox Reactions question

Redox Reactions question

Option A: $Cr_2O_7^{2-} \to 2Cr^{3+}$

Option B: $MnO_4^- \to Mn^{2+}$

Option C: $CrO_4^{2-} \to Cr^{3+}$

Option D: $C_2O_4^{2-} \to 2CO_2$

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Option A: Cr2O72−→2Cr3+Cr_2O_7^{2-} \to 2Cr^{3+}Cr2​O72−​→2Cr3+

Option B: MnO4−→Mn2+MnO_4^- \to Mn^{2+}MnO4−​→Mn2+

Option C: CrO42−→Cr3+CrO_4^{2-} \to Cr^{3+}CrO42−​→Cr3+

Option D: C2O42−→2CO2C_2O_4^{2-} \to 2CO_2C2​O42−​→2CO2​

More from Redox Reactions

Option A: Cr2O72−→2Cr3+Cr_2O_7^{2-} \to 2Cr^{3+}Cr2​O72−​→2Cr3+

Option B: MnO4−→Mn2+MnO_4^- \to Mn^{2+}MnO4−​→Mn2+

Option C: CrO42−→Cr3+CrO_4^{2-} \to Cr^{3+}CrO42−​→Cr3+

Option D: C2O42−→2CO2C_2O_4^{2-} \to 2CO_2C2​O42−​→2CO2​

More from Redox Reactions

Option A: $Cr_2O_7^{2-} \to 2Cr^{3+}$

Option B: $MnO_4^- \to Mn^{2+}$

Option C: $CrO_4^{2-} \to Cr^{3+}$

Option D: $C_2O_4^{2-} \to 2CO_2$

Option A: $Cr_2O_7^{2-} \to 2Cr^{3+}$

Option B: $MnO_4^- \to Mn^{2+}$

Option C: $CrO_4^{2-} \to Cr^{3+}$

Option D: $C_2O_4^{2-} \to 2CO_2$