Calculate the oxidation number of the underlined atom in each of the following: (a) $\underline{\text{Mn}}$ in $\text{KMnO}_4$, (b) $\underline{\text{Cr}}$ in $\text{K}_2\text{Cr}_2\text{O}_7$, (c) $\underline{\text{S}}$ in $\text{H}_2\text{SO}_4$, (d) $\underline{\text{N}}$ in $\text{HNO}_3$.

Calculate the oxidation number of the underlined atom in each of the following: (a) underlineMn in KMnO₄, (b)

Exercise · Q4

Q. Calculate the oxidation number of the underlined atom in each of the following:

(a)

\underline{\text{Mn}}

\text{KMnO}_4

(b)

\underline{\text{Cr}}

\text{K}_2\text{Cr}_2\text{O}_7

(c)

\underline{\text{S}}

\text{H}_2\text{SO}_4

(d)

\underline{\text{N}}

\text{HNO}_3

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✓ Free question

(a) $\text{KMnO}_4$ : $\text{K}=+1$ , four O at $-2$ each gives $-8$ . Neutral formula: $(+1) + x + (-8) = 0 \Rightarrow x = +7$ . So $\text{Mn} = +7$ .\n\n(b) $\text{K}_2\text{Cr}_2\text{O}_7$ : two K give $+2$ , seven O give $-14$ . Neutral formula: $2 + 2x - 14 = 0 \Rightarrow 2x = 12 \Rightarrow x = +6$ . So $\text{Cr} = +6$ .\n\n(c) $\text{H}_2\text{SO}_4$ : two H give $+2$ , four O give $-8$ . Neutral formula: $2 + x - 8 = 0 \Rightarrow x = +6$ . So $\text{S} = +6$ .\n\n(d) $\text{HNO}_3$ : one H gives $+1$ , three O give $-6$ . Neutral formula: $1 + x - 6 = 0 \Rightarrow x = +5$ . So $\text{N} = +5$ . [!ANSWER] (a) $\text{Mn} = +7$ ; (b) $\text{Cr} = +6$ ; (c) $\text{S} = +6$ ; (d) $\text{N} = +5$ .

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set HE2181 markMCQ
$Q. In which compound oxygen give +2 oxidation state?$
(a) H2O
(b) Na2O
(c) OF2
(d) MgO
›Reveal solutionSolution

Oxidation number is assigned to the MORE electronegative atom as negative; fluorine is the only element more electronegative than oxygen, so OF2 is the one compound where oxygen is positive (+2).

Oxygen is normally the second most electronegative element (after fluorine), so in almost all its compounds oxygen is assigned the negative oxidation state (usually -2, as in H2O, Na2O and MgO, where oxygen is more electronegative than H, Na and Mg respectively).

Fluorine, however, is the most electronegative element of all. In OF2, fluorine (electronegativity ~4.0) is more electronegative than oxygen (~3.5), so by convention the more electronegative atom (F) is assigned -1 each, and oxygen must balance the charges: O + 2(-1) = 0 (neutral molecule), so oxidation number of O = +2.

Check the other options: H2O (O = -2, since O > H in electronegativity), Na2O (O = -2), MgO (O = -2). Only OF2 gives O a positive oxidation state.

✓Final answer
$(C) OF2 — oxygen is +2 here, the one common exception where oxygen shows a positive oxidation state.$
CBSE 2020Set HE8221 mark
$Q. Fill in the blank: Oxidation state of phosphorous in orthophosphoric acid is ______.$
›Reveal solutionSolution

In H₃PO₄, using H = +1 and O = −2, the oxidation number of P works out to +5 by the sum-of-charges rule.

Orthophosphoric acid has the formula H $_3$ PO $_4$ . To find the oxidation state of phosphorus, apply the rule that the algebraic sum of oxidation numbers of all atoms in a neutral molecule equals zero, using the standard oxidation numbers H = +1 and O = −2.

Let the oxidation number of P be $x$ :

$3(+1) + x + 4(-2) = 0$

$3 + x - 8 = 0$

$x = +5$

✓Final answer
$The oxidation state of phosphorus in orthophosphoric acid (H₃PO₄) is +5 .$

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Q.Calculate the oxidation number of the underlined atom in each of the following:

Concept understanding — Oxidation Number

Q.Calculate the oxidation number of the underlined atom in each of the following:

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