Which combination is odd with respect to oxidation numbers of ${\rm{S}},{\rm{Cr}},{\rm{N}}$ and ${\rm{H}}$ respectively:

${\rm{N}}{{\rm{H}}_3},{\rm{NH}}_4^ + ,{{\rm{N}}_3}{\rm{H}},{\rm{NO}}_2^ - $

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_5},{{\rm{H}}_2}{{\rm{S}}_2}{{\rm{O}}_8},{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4},{\rm{S}}{{\rm{F}}_6}$

${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7},{{\rm{K}}_2}{\rm{Cr}}{{\rm{O}}_4},{\rm{Cr}}{{\rm{O}}_5},{\rm{Cr}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}$

${\rm{Ca}}{{\rm{H}}_2},{\rm{NaH}},{\rm{LiH}},{\rm{Mg}}{{\rm{H}}_2}$

The ox.no. of S in N${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_4}{{\rm{O}}_6}$ is :

${\rm{ + 5}}\,\,{\rm{and}}\,{\rm{0}}\,\,{\rm{(two}}\,{\rm{S}}\,{\rm{have}}\,{\rm{ + 5}}\,\,{\rm{and}}\,{\rm{the}}\,{\rm{other}}\,{\rm{two}}\,{\rm{S}}\,{\rm{have}}\,{\rm{0)}}$

$ + 2\,and\, + 3$ (two S have $ + 2$ and other two have $ + 3$)

$ + 2\,and\, + 3$ (three S have $ + 2$ and one S has $ + 3$)

Assertion :- Oxidation number of $Cr$ in $CrO_5$ is +6. Reason :- In $CrO_5,$ four oxygen atoms are involed in peroxide linkage.

If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion

If Assertion is True but the Reason is False

If both Assertion & Reason are False

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NEET • CHEMISTRY • REDOX REACTIONS

Redox reactions practice · Mixed difficulty

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31 Hard

31 of 184

Application +4 | -1

The oxidation number of arsenic in arsenate is :

View answer and explanation

Correct option: A

+ 5

Option A (

+ 5

) is correct because the oxidation number of arsenic in arsenate (

\mathrm{AsO}_4^{3 - }

) is +5.

32 Medium

32 of 184

Analyzing +4 | -1

Which of the following chemical reactions depicts the oxidising behaviour of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

View answer and explanation

Correct option: A

2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}

Option A (

2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}

) is correct because H2SO4 oxidizes HI to I2, and itself gets reduced to SO2, showing its oxidizing behavior.

33 Medium

33 of 184

Application +4 | -1

The oxidation state of Ni in

\mathrm{Ni}{\left( {\mathrm{CO}} \right)_4}

is :

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because carbonyl (CO) is a neutral ligand, so the oxidation state of Ni in the complex is zero.

34 Medium

34 of 184

Application +4 | -1

Which compound has oxidation number of carbon equal to zero?

View answer and explanation

Correct option: D

{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}

Option D (C6H12O6) is correct because the oxidation number of carbon in glucose is zero, calculated by balancing the oxidation states of hydrogen and oxygen.

35 Hard

35 of 184

Application +4 | -1

O.N. of

\mathrm{V}

\mathrm{Rb}_4 \mathrm{Na}_{[}\left[\mathrm{HV}_{10} \mathrm{O}_{28}\right]

is-

View answer and explanation

Correct option: B

+5

Option B (+5) is correct because Rb has +1, Na has +1, H has +1, and O has -2, balancing the charge of the complex ion.

36 Medium

36 of 184

Analyzing +4 | -1

Assuming complete ionization, same moles of which of the following compounds will require the least
Amount of acidified

\mathrm{KMn}{\mathrm{O}_\mathrm{4}}

for complete oxidation?

View answer and explanation

Correct option: C

\mathrm{FeS}{\mathrm{O}_\mathrm{4}}

Option C is correct because FeSO4 requires the least amount of KMnO4 for complete oxidation as it involves the oxidation of only Fe2+ to Fe3+.

37 Medium

37 of 184

Application +4 | -1

The oxidation number of

Fe

{\mathrm{K}_4}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}

is :

View answer and explanation

Correct option: A

+ 2

Option A (+2) is correct because K has +1 and CN has -1, balancing the overall charge of the complex.

38 Medium

38 of 184

Application +4 | -1

Nitrogen has fractional oxidation number in :

View answer and explanation

Correct option: C

\mathrm{H}{\mathrm{N}_3}

Option C (

\mathrm{H}{\mathrm{N}_3}

) is correct because the average oxidation state of nitrogen in hydrazoic acid is -1/3, which is a fraction.

39 Medium

39 of 184

Analyzing +4 | -1

Which combination is odd with respect to oxidation numbers of

\mathrm{S},\mathrm{Cr},\mathrm{N}

and

\mathrm{H}

respectively:

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{H}_3},\mathrm{NH}_4^ + ,{\mathrm{N}_3}\mathrm{H},\mathrm{NO}_2^ -

Option C (

\mathrm{N}{\mathrm{H}_3},\mathrm{NH}_4^ + ,{\mathrm{N}_3}\mathrm{H},\mathrm{NO}_2^ -

) is the odd combination because the oxidation state of H is +1 in the first three compounds, but it is not present in the last compound.

40 Hard

40 of 184

Application +4 | -1

An element

A

in a compound ABD has oxidation number

{A^{n - }}

. It is oxidized by

\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }

in acidic medium. In the experiment

1.68 \times {10^{ - 3}}

mole of

{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

were used for

3.26 \times {10^{ - 3}}

mole of

ABD

. The new oxidation number of

A

after oxidation is :

View answer and explanation

Correct option: B

3 - n

Option B (

3 - n

) is correct because the oxidation state of A increases by 3 due to the reaction with dichromate, thus the new oxidation number is

n - + 3

41 Hard

41 of 184

Analyzing +4 | -1

Equal volumes of

1 \mathrm{M}

each of

\mathrm{KMnO}_4

and

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

are used to oxidise

\mathrm{Fe}

(II) solution in acidic medium. The amount of

\mathrm{Fe}

oxidized will be

View answer and explanation

Correct option: C

more with

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

Option C is correct because K2Cr2O7 oxidizes more Fe(II) than KMnO4, given equal volumes and molarities, due to its higher number of electrons transferred per mole.

42 Hard

42 of 184

Application +4 | -1

The ox.no. of S in N

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_4}{\mathrm{O}_6}

is :

View answer and explanation

Correct option: D

\mathrm{ + 5}\,\,\mathrm{and}\,\mathrm{0}\,\,\mathrm{(two}\,\mathrm{S}\,\mathrm{have}\,\mathrm{ + 5}\,\,\mathrm{and}\,\mathrm{the}\,\mathrm{other}\,\mathrm{two}\,\mathrm{S}\,\mathrm{have}\,\mathrm{0)}

Option D is correct because the structure of Na2S4O6 contains two sulfur atoms with a +5 oxidation state and two sulfur atoms with a 0 oxidation state.

43 Easy

43 of 184

Understanding +4 | -1

In the aluminothermic process, aluminium acts as :

View answer and explanation

Correct option: C

A reducing agent

Option C (A reducing agent) is correct because aluminum loses electrons to reduce the metal oxide in the aluminothermic process.

44 Medium

44 of 184

Application +4 | -1

The eq. wt. of

\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4}

in ,

\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4} \to \mathrm{F}{\mathrm{e}^{3 + }} + 2\mathrm{C}{\mathrm{O}_2}\mathrm{is}:

View answer and explanation

Correct option: B

mol. wt./3

Option B (mol. wt./3) is correct because the equivalent weight is the molecular weight divided by the change in oxidation number, which is 3 in this case.

45 Easy

45 of 184

Understanding +4 | -1

The highest oxidation state of Mn is shown by

View answer and explanation

Correct option: B

KMn{o_4}

Option B (

KMn{o_4}

) is correct because in

KMn{o_4}

, Mn has an oxidation state of +7, which is its highest possible oxidation state.

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46 Hard

46 of 184

Analyzing +4 | -1

Assertion :- Oxidation number of

Cr

CrO_5

is +6.

Reason :- In

CrO_5,

four oxygen atoms are involed in peroxide linkage.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion

Assertion and Reason are true, and the Reason is a correct explanation of the Assertion. The presence of peroxide linkages in CrO₅ leads to a lower average oxidation state calculation, but the actual oxidation state of Cr is +6.

47 Easy

47 of 184

Understanding +4 | -1

Which acts as reducing agent as well as oxidising agent?

View answer and explanation

Correct option: A

{\mathrm{O}_3}

Option A (

{\mathrm{O}_3}

) is correct because ozone can act as both an oxidizing agent (accepting electrons) and a reducing agent (donating electrons).

48 Easy

48 of 184

Analyzing +4 | -1

Which of the following has the highest oxidation number

View answer and explanation

Correct option: D

KI{O_4}

Option D (KIO4) is correct because iodine has the highest oxidation number of +7 in this compound.

49 Medium

49 of 184

Analyzing +4 | -1

Statement-I: In

{HClO}_4

, Chlorine has the oxidation number of +4.
Statement-II:

{HClO}_4

(perchloric) acid has two peroxide linkages.

View answer and explanation

Correct option: B

Both the statements are incorrect

Both statements are incorrect. In HClO₄, chlorine has an oxidation number of +7, and perchloric acid does not have peroxide linkages.

50 Hard

50 of 184

Analyzing +4 | -1

Assertion :- Bromide ion is serving as a reducing agent in the reaction.

\mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{MnO}_2(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})

Reason :- Oxidation number of

\mathrm{Br}

increases from

-1

+5.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion

Option A is correct because the bromide ion (Br-) is indeed serving as a reducing agent as its oxidation number increases from -1 to +5, thus causing the reduction of MnO4-.

51 Medium

51 of 184

Analyzing +4 | -1

Statement-I: O.N. of carbon in

\mathrm{H}-\mathrm{C} \equiv \mathrm{N}

is +4.

Statement-II: Carbon always shows an O.N. of +4.

View answer and explanation

Correct option: B

Both the statements are incorrect

Both statements are incorrect. The oxidation number of carbon in HCN is -1, and carbon can exhibit various oxidation states, not always +4.

52 Medium

52 of 184

Remembering +4 | -1

The nature of oxides

{\mathrm{V}_\mathrm{2}}{\mathrm{O}_\mathrm{3}}

and

\mathrm{CrO}

is indexed as

X'

and

'Y'

type respectively. The correct set of

X

and

Y

View answer and explanation

Correct option: D

X = basic, y = basic

Option D (X = basic, y = basic) is correct because oxides of metals in lower oxidation states are generally basic.

53 Easy

53 of 184

Application +4 | -1

\mathrm{HCO}_3^ -

contains carbon in the oxidation state:

View answer and explanation

Correct option: C

Option C (4) is correct because the oxidation state of carbon in

\mathrm{HCO}_3^ -

is calculated to be +4 based on the oxidation states of other elements.

54 Hard

54 of 184

Understanding +4 | -1

{Co}({s})+{Cu}^{2+}({aq}) \longrightarrow {Co}^{2+}({aq})+{Cu}({s})

The above reaction is

View answer and explanation

Correct option: C

Redox reaction

Option C (Redox reaction) is correct because the reaction involves both oxidation (Co to Co2+) and reduction (Cu2+ to Cu).

55 Hard

55 of 184

Analyzing +4 | -1

In the reaction,

2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{Cu}_2 \mathrm{~S} \longrightarrow 6 \mathrm{Cu}+\mathrm{SO}_2

, sulphur of

\mathrm{Cu}_2 \mathrm{~S}

helps copper both in

\mathrm{Cu}_2 \mathrm{~S}

itself and

\mathrm{Cu}_2 \mathrm{O}

to decrease its oxidation number, therefore sulphur of

\mathrm{Cu}_2 \mathrm{~S}

View answer and explanation

Correct option: B

reductant

Option B (reductant) is correct because the sulphur in Cu2S causes a decrease in the oxidation number of copper, thus acting as a reducing agent.

56 Easy

56 of 184

Analyzing +4 | -1

Which one of the following reactions is not an example of redox reaction

View answer and explanation

Correct option: D

HCl + {H_2}O \to {H_3}{O^ + } + C{l^ - }

Option D (

HCl + {H_2}O \to {H_3}{O^ + } + C{l^ - }

) is correct because there is no change in oxidation states of any element, hence it's not a redox reaction.

57 Easy

57 of 184

Understanding +4 | -1

An element with electronic arrangement at 2, 8, 18, 1 will exhibit the following oxidation states

View answer and explanation

Correct option: B

Option D (+ 1 only) is correct because the element has one valence electron, which it will lose to achieve a stable configuration, resulting in a +1 oxidation state.

58 Easy

58 of 184

Application +4 | -1

In which of the following compounds Nitrogen exhibits the highest oxidation state

View answer and explanation

Correct option: A

{N_3}H

Option A (

{N_3}H

) is correct because the average oxidation state of nitrogen in hydrazoic acid is -1/3, which is the highest among the given options.

59 Medium

59 of 184

Remembering +4 | -1

The oxidation state of a metal in a compound is represented according to the notation which is known as

View answer and explanation

Correct option: C

Stock notation

Option C (Stock notation) is correct because it is the notation used to represent the oxidation state of a metal in a compound.

60 Easy

60 of 184

Analyzing +4 | -1

Which of the following change represents a disproportionation reaction

View answer and explanation

Correct option: C

Both 1 and 2

Option C (Both 1 and 2) is correct because in both reactions, the same element undergoes both oxidation and reduction.

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