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D and F Block Elements question

2022 · 26 Jul · Shift 1 · Q5

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D and F Block Elements
2022 · 26 Jul · Shift 1 · Q5

D and F Block Elements question

2022 · 26 Jul · Shift 1 · Q5

JEE MainChemistryD and F Block ElementsMCQ+4 / −1

The dark purple colour of

\mathrm{KMnO}_{4}

disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is :

A
5
B
1
C
7
D
2

View written solutionFree

Correct answer: A

Identify the oxidation state of manganese in $\mathrm{KMnO_4}$

In $\mathrm{KMnO_4}$ :
- $\mathrm{K} = +1$
- $\mathrm{O} = -2$ each, so $4\mathrm{O} = -8$
Let oxidation state of Mn be $x$ .

$x + 1 - 8 = 0$ $x - 7 = 0$ $x = +7$

So, manganese is in the $+7$ oxidation state in $\mathrm{KMnO_4}$ .
Find the oxidation state of manganese after reaction in acidic medium

In acidic medium, permanganate ion $\mathrm{MnO_4^-}$ is reduced to $\mathrm{Mn^{2+}}$ .

Thus, final oxidation state of manganese is:

$+2$
Calculate the overall change in oxidation number

Initial oxidation state of Mn $= +7$

Final oxidation state of Mn $= +2$

Change in oxidation number:

$7 - 2 = 5$
Conclusion

The oxidation number of manganese decreases by 5 units.

Therefore, the correct option is:

$\boxed{\text{A: } 5}$

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