Oxidation no. of $P$ in ${{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_5},{{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_6},{{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_7}$ are respectively :

$ + 3,{\rm{ }} + 4,{\rm{ }} + 5$

$ + 3,{\rm{ }} + 5,{\rm{ }} + 4$

$ + 4,{\rm{ }} + 3,{\rm{ }} + 5$

$ + 5,{\rm{ }} + 3,{\rm{ }} + 4$

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JEE • CHEMISTRY • REDOX REACTIONS

Redox reactions practice · Medium

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Showing 151-171 of 171 questions

151 Medium

151 of 171

Understanding +4 | -1

In which case

+ 1

oxidation state is stable than

+ 3

View answer and explanation

Correct option: C

\mathrm{Tl}

Option C (

\mathrm{Tl}

) is correct because Thallium (Tl) exhibits a stable +1 oxidation state due to the inert pair effect, which makes the +3 oxidation state less stable.

152 Medium

152 of 171

Understanding +4 | -1

It is found that V forms a double salt isomorphous with Mohr’s salt. The oxidation number of V in this compound is :

View answer and explanation

Correct option: B

+ \mathrm{ }2

Option B (

+ \mathrm{ }2

) is correct because Mohr's salt has the formula

\mathrm{FeS}{\mathrm{O}_4}\mathrm{.}{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{S}{\mathrm{O}_4}.6{\mathrm{H}_2}\mathrm{O}

, and if V forms a similar double salt, it replaces Fe with the same oxidation state of +2.

153 Medium

153 of 171

Analyzing +4 | -1

In the following reaction

4 {P}+3 {KOH}+3 {H}_2 {O} \longrightarrow 3 {KH}_2 {PO}_2+{PH}_3

View answer and explanation

Correct option: A

Phosphorus is both oxidised and reduced.

The correct answer is A because the oxidation state of phosphorus changes from 0 to +1 in KH2PO2 (oxidation) and from 0 to -3 in PH3 (reduction).

154 Medium

154 of 171

Application +4 | -1

The value of

x

on the

{\left[ {{\text{Ni}}{{\left( {{\text{CN}}} \right)}_4}} \right]^x}

View answer and explanation

Correct option: B

– 2

Option B (– 2) is correct because the overall charge of the complex ion is calculated by considering the oxidation state of Nickel and the charges of the cyanide ligands.

155 Medium

155 of 171

Understanding +4 | -1

Given

E^{\Theta}

(i)

{Mg}^{2+} / {Mg}({s}), {E}^{\Theta}=-2.36

(ii)

{Ag}^{+} / {Ag}

(s),

{E}^{\Theta}=0.80

(iii)

{Al}^{3+} / {Al}({s}), {E}^{\Theta}=-1.66

(iv)

{Cu}^{2+} / {Cu}

(s),

{E}^{\Theta}=0.52

Out of the above given elements which is the strongest oxidising agent and which is the weakest oxidising agent ?

View answer and explanation

Correct option: B

(ii) is the strongest whereas (i) is the weakest oxidising agent

Option B is correct because the strongest oxidizing agent has the highest reduction potential, and the weakest has the lowest. Ag+/Ag has the highest, and Mg2+/Mg has the lowest.

156 Medium

156 of 171

Application +4 | -1

Oxidation no. of

P

{\mathrm{H}_4}{\mathrm{P}_2}{\mathrm{O}_5},{\mathrm{H}_4}{\mathrm{P}_2}{\mathrm{O}_6},{\mathrm{H}_4}{\mathrm{P}_2}{\mathrm{O}_7}

are respectively :

View answer and explanation

Correct option: C

+ 3,\mathrm{ } + 4,\mathrm{ } + 5

Option C (+3, +4, +5) is correct because it accurately calculates the oxidation number of P in each compound based on the standard rules for assigning oxidation numbers.

157 Medium

157 of 171

Analyzing +4 | -1

Assertion :-

\mathrm{MnO}_4^{-}

is always reduced to

\mathrm{Mn}^{+2}.

Reason :- Decrease in oxidation number or gaining of electron means oxidation

View answer and explanation

Correct option: D

If both Assertion & Reason are False

The correct answer is D (If both Assertion & Reason are False) because MnO4- can be reduced to different oxidation states, and decrease in oxidation number is reduction, not oxidation.

158 Medium

158 of 171

Analyzing +4 | -1

Given:

\begin{aligned} & {E}_{\frac{1}{2} {Cl}_2 / {Cl}^{-}}^{{o}}=1.36 {~V}, {E}_{{Cr}^{3+} / {Cr}}^{{o}}=-0.74 {~V}, \\ & {E}_{{Cr}_2 {O}_7^{2-} / {Cr}^{3+}}^{{o}}=1.33 {~V}, {E}_{{MnO}_4^{-} / {Mn}^{2+}}^{{o}}=1.51 {~V} \end{aligned}

The correct order of reducing power of the species

\left({Cr}, {Cr}^{3+}, {Mn}^{2+}\right.

and

\left.{Cl}^{-}\right)

will be

View answer and explanation

Correct option: A

{Mn}^{2+}<{Cl}^{-}<{Cr}^{3+}<{Cr}

The correct answer is A (

{Mn}^{2+}<{Cl}^{-}<{Cr}^{3+}<{Cr}

) because the reducing power increases as the reduction potential decreases.

159 Medium

159 of 171

Analyzing +4 | -1

How many of the following shows disproportionation reactions?

\mathrm{H}_2 \mathrm{O}_2, \mathrm{Ag}, \mathrm{Cu}^{+}, \mathrm{K}^{+}, \mathrm{F}_2, \mathrm{Cl}_2, \mathrm{ClO}_3^{-}

View answer and explanation

Correct option: B

The correct answer is 4 because H_2O_2, Cu⁺, Cl_2, and ClO_3^- can undergo disproportionation reactions.

160 Medium

160 of 171

Understanding +4 | -1

Standard reduction potentials at

25^{\circ} \mathrm{C}

\mathrm{Li}^{+} / \mathrm{Li}, \mathrm{Ba}^{2+} / \mathrm{Ba}, \mathrm{Na}^{+} / \mathrm{Na}

and

\mathrm{Mg}^{2+} / \mathrm{Mg}

are

-3.05,-2.90,-2.71

and

-2.37

volt respectively. Which one of the following is the strongest oxidizing agent?

View answer and explanation

Correct option: B

\mathrm{Mg}^{2+}

Option B (

\mathrm{Mg}^{2+}

) is correct because the strongest oxidizing agent is the one with the highest reduction potential.

161 Medium

161 of 171

Application +4 | -1

When

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

is converted into

{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4},

the change in oxidation number of chromium is

View answer and explanation

Correct option: A

Option A (0) is correct because when K2Cr2O7 is converted to K2CrO4, the oxidation state of chromium remains +6 in both compounds.

162 Medium

162 of 171

Understanding +4 | -1

The most common oxidation state of an element is

- 2

. The number of electrons present in its outermost shell is :

View answer and explanation

Correct option: C

Option C (6) is correct because an element with a common oxidation state of -2 typically has 6 valence electrons in its outermost shell, as it needs to gain 2 electrons to achieve a stable octet.

163 Medium

163 of 171

Analyzing +4 | -1

Which of the following reactions are disproportionation reaction?

\mathrm{(a})\,2\mathrm{C}{\mathrm{u}^ + }\, \to \,\mathrm{C}{\mathrm{u}^{2 + }}\, + \,\mathrm{C}{\mathrm{u}^0}\,

\mathrm{(b})\,\mathrm{3MnO}_\mathrm{4}^{\mathrm{2} - }\mathrm{ + 4}{\mathrm{H}^\mathrm{ + }}\, \to \,\mathrm{2MnO}_\mathrm{4}^ - \,\mathrm{ + }\,\mathrm{Mn}{\mathrm{O}_\mathrm{2}}\,\mathrm{ + }\,\mathrm{2}{\mathrm{H}_\mathrm{2}}\mathrm{O}

\mathrm{(c)}\,\mathrm{2KMn}{\mathrm{O}_\mathrm{4}}\xrightarrow{\Delta }\,{\mathrm{K}_\mathrm{2}}\mathrm{Mn}{\mathrm{O}_\mathrm{4}}\mathrm{ + Mn}{\mathrm{O}_\mathrm{2}}\,\,\mathrm{ + }\,{\mathrm{O}_\mathrm{2}}

\mathrm{(d)}\,\mathrm{2MnO}_\mathrm{4}^ - \, + \,\mathrm{3M}{\mathrm{n}^{2 + }}\, + \,\mathrm{2}{\mathrm{H}_\mathrm{2}}\mathrm{O}\, \to \,5\mathrm{Mn}{\mathrm{O}_2}\mathrm{ + 4}{\mathrm{H}^ \oplus }

Select the correct option from the following

View answer and explanation

Correct option: A

(a) and (b) only

Option A ((a) and (b) only) is correct because reactions (a) and (b) are disproportionation reactions where the same element is both oxidized and reduced.

164 Medium

164 of 171

Analyzing +4 | -1

Find the incorrect statement

View answer and explanation

Correct option: D

All of these

The correct answer is D (All of these) because higher reduction potential corresponds to a stronger oxidizing agent, lower oxidation potential means a stronger reducing agent, and the oxidation state of oxygen in O3 is 0.

165 Medium

165 of 171

Understanding +4 | -1

Which of the following changes is/are observed. When a clean metallic rod of zinc is placed in a solution of copper nitrate.

View answer and explanation

Correct option: D

Both 1 and 2

Option D (Both 1 and 2) is correct because zinc reduces copper ions to copper metal, which deposits on the zinc rod, and the zinc rod dissolves as it is oxidized.

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166 Medium

166 of 171

Analyzing +4 | -1

In which reaction the underlined substance has been reduced?

View answer and explanation

Correct option: C

Steam + \mathrm{ }iron\mathrm{ } \to \mathrm{ }hydrogen\mathrm{ } + \mathrm{ }iron\mathrm{ }oxide

Option C (

Steam + \mathrm{ }iron\mathrm{ } \to \mathrm{ }hydrogen\mathrm{ } + \mathrm{ }iron\mathrm{ }oxide

) is correct because the oxidation state of oxygen in steam changes, indicating reduction.

167 Medium

167 of 171

Application +4 | -1

Among

\mathrm{N}{\mathrm{H}_3},\mathrm{HN}{\mathrm{O}_{3,}}\mathrm{Na}{\mathrm{N}_3}

and

\mathrm{M}{\mathrm{g}_3}{\mathrm{N}_2};

the number of molecules having nitrogen in negative oxidation state is

View answer and explanation

Correct option: C

Option C (3) is correct because NH3, NaN3 and Mg3N2 have nitrogen in negative oxidation state.

168 Medium

168 of 171

Remembering +4 | -1

Which of the following oxidation state is the most common among the lanthanoides :

View answer and explanation

Correct option: D

Option D (3) is correct because the +3 oxidation state is the most stable and common oxidation state among the lanthanoids due to the stability of the resulting electronic configurations.

169 Medium

169 of 171

Understanding +4 | -1

In which of the following method, it is assumed that there is a complete transfer of electron from a less electronegative atom to more electronegative atom?

View answer and explanation

Correct option: B

Oxidation number method

Option B (Oxidation number method) is correct because the oxidation number method assumes complete electron transfer from less to more electronegative atoms.

170 Medium

170 of 171

Analyzing +4 | -1

How many of the following species / reagent can reduce

\mathrm{Fe}^{+3}(a q)

in to

\mathrm{Fe}^{+2}(a q)

at normal conditions

\left(\mathrm{NH}_4\right)_2 \mathrm{~S}, \mathrm{HI}, \mathrm{Sn}^{+2}(a q), \mathrm{CN}^{-}(a q), \mathrm{NaNO}_2, \mathrm{SO}_2, \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3, \mathrm{SCN}^{-}(a q)

, acidified

\mathrm{NaIO}_3

View answer and explanation

Correct option: A

Option A (5) is correct because the following species can reduce

\mathrm{Fe}^{+3}(aq)

\mathrm{Fe}^{+2}(aq)

(\mathrm{NH}_4)_2 \mathrm{~S}, \mathrm{HI}, \mathrm{Sn}^{+2}(aq), \mathrm{SO}_2, \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3

171 Medium

171 of 171

Understanding +4 | -1

Which halide is not oxidised by

{MnO}_2

View answer and explanation

Correct option: A

F

The correct answer is A (F) because fluorine is the strongest oxidizing agent and cannot be oxidized by MnO2.

Revision notes and key takeaways

Review REDOX REACTIONS in your textbook or class notes, then use this set to check your understanding.

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