MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 9

In which of the following compounds iron has lowest oxidation state :-

0%
$FeSO_4(NH_4)_2SO_4.6H_2O$
0%
$K_4[Fe(CN)_6]$
0%
$[Fe(CO)_5]$
0%
$Fe_{0.94}O$

Heated saw dust catches fire when a drop of concentrated nitric acid is added to it. This is due to

0%
dehydration
0%
oxidation
0%
reduction
0%
dehydrogenation

Which of the following show non-zero multiple oxidation state?

0%
$S$
0%
$O$
0%
$Zn$
0%
$H$

Explanation

The possible oxidation states of elements are :

$S\rightarrow -2,-1,0,+1,+2,+3,+4,+5,+6$

$O\rightarrow -2,-1,+1,+2$

$Zn\rightarrow -2,+1,+2$

$H\rightarrow -1,+1$

Out of the above elements, only S shows non-zero multiple oxidation states.

Four

$Cl_{2}$ molecules undergo a loss and gain of

$6$ mole of electrons to form two oxidation states of

$Cl$ in auto redox change. What are the

$+ve$ and

$-ve$ oxidation states of

$Cl$ in the change:

0%
$Cl^{5+},Cl^{o}$
0%
$Cl^{7+},Cl^{-}$
0%
$Cl^{3+},Cl^{o}$
0%
$Cl^{3+},Cl^{-}$

Explanation

The overall reaction is:

$4{ Cl }_{ 2 }\rightarrow 2{ Cl }^{ +3 }+6{ Cl }^{ - }$

Here total number of atom is balanced and total charge is balance.

Change of oxidation state is +3 and -1.

In which of the following compounds, the oxidation state of

$I-$ atom is highest?

0%
${KI}_{3}$
0%
$KI{O}_{4}$
0%
$KI{O}_{3}$
0%
$All\ of\ the \ above$

Explanation

$KI_3 \Rightarrow I_3^- + K^+$

$3X= -1$

$\Rightarrow X= -1/3$

$KIO_4 \Rightarrow IO_4^- + K^+$

$X + 4 \times {(-2)}= -1$

$\Rightarrow X= +7$

$KIO_3 \Rightarrow {IO_3}^- + K^+$

$X + 3 \times {(-2)}= -1$

$\Rightarrow X= +5$

Option B is correct.

Sulphur has highest oxidation state in:

0%
${ H }_{ 2 }S$
0%
${ { H }_{ 2 }SO }_{ 4 }$
0%
$Na_{ 2 }{ S }_{ 2 }{ O }_{ 3 }$
0%
$Na_{2}S_{4}O_{6}$

Oxidation state of oxygen in

$Cr{ O }_{ 5 }$ is :

0%
-1
0%
-2
0%
Both $A$ and $B$
0%
$-\frac { 1 }{ 2 }$

Explanation

$CrO_5$ is chromium with four peroxy oxygen and one coordinated oxygen.

It is a butterfly type structure in which central chromium surrounds with 4 peroxy cyclic oxygen and 1 coordinate oxygen. In this, the oxidation number of chromium is +6.

$CrO_5$ has 2 pairs in peroxide. It implies that 4 oxygen is in

$-1$ oxidation state and one in

$-2$ oxidation state.

Hence, option

$(C)$ is correct.

1437955_1076655_ans_849291d1aa484faeb06fde9b53860076.JPG

What is the oxidation number of chromium in the dimeric hydroxo bridged species :-

0%
$+6$
0%
$+4$
0%
$+3$
0%
$+2$

Explanation

let

$'Cr'$ oxidation state

$\rightarrow x$

$2x-2=+4$

$2x=6$

$x=3$

Each chromium shows

$+3$ oxidation state.

The sum of the oxidation numbers of all the carbons in

$C_6H_5CHO$ is :

0%
0
0%
+2
0%
+4
0%
-4

Explanation

oxidation state of

$C$ in

$C_6H_5CHO$ is

$6x+5+x+1+(-2)=0$

$7x+6-2=0$

$7x+4=0$

$7x=-4$

as per the given question sum of oxidation states of all C atoms is sum therefore answer becomes

$-4$

A solution contains

${Fe}^{+2},{Fe}^{+3}$ and

${I}^{-}$ ions. This solution was treated with iodine at

${35}^{o}C$ . Then the favourable redox reaction is:

(Given that

${ E }_{ { Fe }^{ +3 }/{ Fe }^{ +2 } }^{ o }=+0.77V;\quad { E }_{ { I }_{ 2 }/{ I }^{ - } }^{ o }=0.536V$ )

0%
${I}_{2}$ will be reduced to ${I}^{-}$
0%
there will be no redox reaction
0%
${I}^{-}$ will be oxidised to ${I}_{2}$
0%
${Fe}^{+}$ will be oxidised to ${Fe}^{+3}$

Explanation

${ 2I }^{ - }\rightarrow { I }_{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$

${ E }_{ Oxidation }^{ 0 }=-0.536V$

${ Fe }^{ 3+ }+e^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$

$E^{ 0 }_{ Reduction }=-0.77V$
----------------------------------------------------------------------------------------------------------------------

${ 2Fe }^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I }_{ 2 }$

$E^{ 0 }={ E }_{ oxidation }^{ 0 }+E_{ Reduction }^{ 0 },+ve$

Hence the reaction will take place.

${ 2I }^{ - }\rightarrow { I }_{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$

${ E }^{ 0 }_{ Oxidation }=-0.536V$

${ Fe }^{ 3+ }+{ e }^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$

$E_{ Reduction }^{ 0 }=-0.77V$
------------------------------------------------------------------------------------------------------------------------

$2Fe^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I }_{ 2 };$

${ E }^{ 0 }={ E }_{ oxidation }^{ 0 }+{ E }_{ reduction }^{ 0 };+ve$

Average oxidation number of iodine in

$KI_3$ ?

0%
$+1/3$
0%
$-1/3$
0%
$+3$
0%
$-1$

Explanation

Criven :

$-K I_{3}$

Here I

$_{3}$ is present as

$I_{3}$

So, the negative charge is on theree Iodine

atoms . Average onidation number of Iodine

$=-\dfrac{1}{3}$

B) is covrect answer.

The oxidation number of nitrogen in

$NCI_3$ is ?

0%
$+3$
0%
$-3$
0%
$Zero$
0%
- $1/3$

Explanation

Oxidation number is the apparent charge on an atom of an element present in a molecule.In $NCl_3$ , Electronegativity of Nitrogen is 3 while that of Chlorine is 3.2. Thus Chlorine being more electronegative has oxidation state=-1.

Sum of oxidation numbers of all elements present in a neutral compound is zero.

So, O.N. of Nitrogen+ 3(O.N. of $Cl$ )=0

O.N. of $N^{-3}=0$

O.N. of $N=+3$

${ KMnO }_{ 4 }$ can be prepared from

${ K }_{ 2 }MnO_{ 4 }$ as per the reaction,

${ 3MnO }_{ 2 }^{ 2- }+{ 2H }_{ 2 }O\rightleftharpoons { 2MnO }_{ 4 }^{ - }+{ MnO }_{ 2 }{ +4OH }^{ - }$

The reaction can go to completion by removing

${ OH }^{ - }$ ions by adding:

0%
$KOH$
0%
${ CO }_{ 2 }$
0%
$SO$
0%
$HCl$

Explanation

Since

${OH}^{-}$ are generated from water

$\left( {H}_{2}O \right)$ which is a weak acid, therefore a weak acid like

$C{O}_{2}$ should be used to remove it because-

A strong acid reverse the reaction $\left( HCl \right)$ ,
$KOH$ increases the concentration of ${OH}^{-}$ , thus again shifts the reaction in the backward direction.
$C{O}_{2}$ combines with ${OH}^{-}$ to give carbonate which is easily removed.
$S{O}_{2}$ reacts with water to give strong acid, so it cannot be used.

$3 {K}_{2}Mn{O}_{4} + 2 C{O}_{2} \longrightarrow 2 KMn{O}_{4} + Mn{O}_{2} + 2 {K}_{2}C{O}_{3}$

Hence the reaction can go to completion by removing

${OH}^{−}$ ions by adding

$C{O}_{2}$ .

Of the following elements, which one has the same oxidation state in all of its compounds?

0%
Hydrogen
0%
Fluorine
0%
Carbon
0%
Oxygen

Explanation

Possible oxidation states of the elements are :-

Hydrogen

$=$

$-1,+1$

Fluorine

$=$

$-1$

Carbon

$=$

$-4,-3,-2,-1,0,+1,+2,+3,+4$

Oxygen

$=$

$-2,-1,0,+1,+2$

Here, only fluorine shows a single oxidation state in all of its compounds.

Oxidation number of sodium in sodium amalgam is:

0%
+2
0%
+1
0%
-3
0%
0

The oxidation state of Cr is

$[Cr(NH_3)_4Cl_2]$ is?

0%
$+3$
0%
$+2$
0%
$+1$
0%
$0$

Explanation

$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\right]$

We know that

$C 1=-1$

$\mathrm{NH}_{3}=0$

So onidation state of

$\underline{Cr}$

$\begin{array}{l}x+4 \times 0+2(-1)=0\\x=+2\end{array}$

Corret Answer

$-\underline{B}$

Which of the following shows highest oxidation number in combined state?

0%
$O$
0%
$Ru$
0%
Both $(A)$ and $(B)$
0%
None of the above

Explanation

The possible oxidation states shown by :-

$(O)$ Oxygen

$\rightarrow$

$-2,-1,0,+1,+2$

$(Ru)$ Ruthenium

$\rightarrow$

$-4,-2,0,+1,+2,+3,+4,+5,+6,+7,+8$

Therefore, ruthenium

$(Ru)$ shows highest oxidation number i.e.

$+8$ in combined state.

What is the oxidation state of

$Co$ in

$[Co(H_2O)_5Cl]^{2+}$ ?

0%
$+2$
0%
$+3$
0%
$+1$
0%
$+4$

Explanation

$\underset { x }{ [Co } \underset { 0 }{ { \left( { H }_{ 2 }O \right) }_{ 5 } } \underset { -1 }{ { Cl] }^{ 2+ } }$

Let,

$x$ be the oxidation state of Co.

Now,

$x+5\times 0-1=2;\quad \Rightarrow x=+3$

${ I }_{ 2 }+KI\rightarrow { KI }_{ 3 }\quad$
In the above reaction:

0%
only oxidation takes place
0%
only reduction takes place
0%
both of the above
0%
none of these

Explanation

${ \overset { 0 }{ I } }_{ 2 }+K\overset { -1 }{ I } \longrightarrow K{ \overset { -1 }{ I } }_{ 3 }$

Here, only

${ \overset { 0 }{ I } }_{ 2 }$ is reduced to

$K{ \overset { -1 }{ I } }_{ 3 }$ and no other effect takes place.

Therefore, only reduction takes place.

Oxidation state of oxygen in hydrogen peroxide is?

0%
-1
0%
+1
0%
0
0%
-2

Explanation

Hydrogen peroxide

$={ H }_{ 2 }{ O }_{ 2 }$

$\overset { +1 }{ H } -\overset { -1 }{ O } -\overset { -1 }{ O } -\overset { +1 }{ H }$

Here, the oxidation states of hydrogen is

$+1$ and that of oxygen atoms is

$-1$ .

Oxidation number of sulphur in

$Na_2S_2O_3$ would be :

0%
+4
0%
+2
0%
-2
0%
0

Explanation

${ Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }\Rightarrow$

Oxidation number of

${ S }_{ 1 }=-2$

Oxidation number of

${ S }_{ 2 }=+1+1+2+2=+6$

Average oxidation number of

$S=\dfrac { -2+6 }{ 2 } =+2$

1051788_1097390_ans_4c9a8496da4d4c4ba852bb472a85c22a.jpg

Which of the following is not correctly matched?

0%
$CrO_5$ ; oxidation number of $Cr =$ $+10$
0%
$Fe_3O_4$ ; oxidation state of $Fe = -\dfrac{8}{3}$
0%
$Na - Hg$ ; oxidation number of $Na$ is = $+1$
0%
None of the above

Explanation

${ CrO }_{ 5 }$ :- Let oxidation number of

$Cr$ be

$x$

$\therefore$

$x+5(-2)=0$

$\therefore$

$\boxed { x=+10 }$

${ Fe }_{ 3 }{ O }_{ 4 }$ :- Let oxidation of

$Fe$ by

$y$

$\therefore$

$3y+4(-2)=0$

$\therefore$

$\boxed { y=+8/3 }$

$Na-Hg$ :- The amalgam is not a compound but mixture. Therefore, the metal has

$O$ oxidation number.

Therefore,

$(B)$ and

$(C)$ are not correctly matched.

Oxidation number of gold metal is

0%
+1
0%
-1
0%
0
0%
All of these

$4Zn+10H{NO}_{3}\rightarrow 4Zn{({NO}_{3})}_{2}+{NH}_{4}{NO}_{3}+3{H}_{2}O$
In this reaction one mole of

$H{NO}_{3}$ is reduced by

0%
$26g$ $Zn$
0%
$64g$ $Zn$
0%
$128g$ $Zn$
0%
$256g$ $Zn$

For the half reaction

$B(s) \rightarrow B^{2+} + 2e^{-}$

$E_{1}^{\circ} = - 0.44 V$
and

$B_{2}+\rightarrow B^{3+} + e^{-}$

$E_{2}^{\circ} = 1.3 V$
What is

$E^{\circ}$ for the reaction,

$3e^{-} + B^{+3} \rightarrow B(s)$

0%
$0.86$ V
0%
$0.14$ V
0%
$-0.14$ V
0%
$-0.28$ V

Explanation

$B(s)\rightarrow { B }^{ 2 }+{ 2e }^{ - }\quad \quad \quad { E }_{ 1 }^{ 0 }=-0.44V$

$- i$

${ B }^{ 2+ }\rightarrow { B }^{ 3+ }+{ e }^{ - }\quad \quad \quad { E }_{ 1 }^{ 0 }=1.3V$

$- ii$

${ B }^{ 3+ }+{ 3e }^{ - }\rightarrow B(s)\quad \quad \quad \quad { E }_{ 3 }^{ 0 }=?$

$- iii$

iii

$=- (II+I)$

Or,

$-3\times { E }_{ 3 }^{ 0 }\times F=-\left\{ -1\times 1.3\times F+0.44\times 2\times F \right\}$

$\therefore { E }_{ 3 }^{ 0 }=-0.14V$

The oxidation number of S is

$(CH_{3})_{2}SO$ is ?

0%
1
0%
2
0%
0
0%
3

Explanation

1)Structure: $S$ is double-bonded to $O$ ( $— 2$ electrons) and is bonded to $2 C$ ( $+2$ electrons).

2) $+2 + -2 = 0. S$ has a $0$ Oxidation number in $(CH_3)_2SO$ .

The oxidation number of S in

$(CH_{3})_{2}SO$ is ?

0%
1
0%
2
0%
0
0%
3

Explanation

$S$ is double-bonded to $O (— 2$ electrons $)$ and is bonded to $2 C (+2$ electrons).

$+2 + -2 = 0$ .

$S$ has a $0$ Oxidation number in $(CH_3)_2SO$ .

Highest positive oxidation state shown by:

0%
Cl
0%
Os
0%
Mn
0%
Cr

Explanation

Osmium exhibits oxidation states from $0 - +8$ in its compounds, with the exception of $+1$ .

well-characterized and stable compounds contain the element in $+2, +3, +4, +6,$ and $+8$ states.

There are also carbonyl and organometallic compounds in the low oxidation states $−2, 0, +1$ .

The elements having lowest and highest oxidation number respectively are:

0%
$Sc,Fe$
0%
$Sc,Mn$
0%
$Ag,Mn$
0%
$Fe,Mn$

A gas X at 1 atm is bubbled through the solution containing a mixture of 1MY and 1MZ at 25 degrees. If the order of reduction potential is Z > Y > X, then

0%
Y will oxidise X not Z
0%
Y will oxidise Z not X
0%
Y will oxidise both X and Y
0%
Y will reduce both X and Y

Explanation

The reduction tendencies is

$Z > Y > X$ .Hence,Y can oxidise X and not Z

The complex compound in which metal is not present in zero oxidation state?

0%
${ Mn }_{ 2 }{ \left( CO \right) }_{ 10 }$
0%
$\left[ Ni{ \left( CO \right) }_{ 4 } \right]$
0%
$\left[ Cr{ \left( { C }_{ 6 }{ H }_{ 6 } \right) }_{ 2 } \right]$
0%
$K\left[ { PtCl }_{ 3 }\left( { C }_{ 2 }{ H }_{ 4 } \right) \right]$

Explanation

$A.$ $Mn_2{(CO)}_{10}\Longrightarrow 2x+(10\times 0)=0\Longrightarrow x=0,$ ( $x=$ oxidation state of metal.)

$B.$ $[Ni{(CO)}_{4}]\Longrightarrow x+(4\times 0)=0\Longrightarrow x=0,$ ( $x=$ oxidation state of metal.)

$C.$ $[Cr{(C_6H_6)}_{2}]\Longrightarrow x+(2\times 0)=0\Longrightarrow x=0,$ ( $x=$ oxidation state of metal.)

$D.$ $K[PtCl_3{(C_2H_4)}]\Longrightarrow +1+x+3\times (-1)+0=0\Longrightarrow x=+2,$ ( $x=$ oxidation state of metal.)

$\therefore$ $Pt$ is in $+2$ Oxidation state.

$\therefore$ Option $D.$ is the correct answer.

How many oxygen atoms in blue per chomate

${ CrO }_{ 5 }$ are having

$-1$ oxidation state?

0%
1
0%
2
0%
4
0%
5

In which of the following compounds, the metal is in the lower oxidation state

0%
$[Co( NH_{3})_{5}Br_{2}]SO_{4}$
0%
$Fe_{3} [Fe( CN)_{6}]_{2}$
0%
$[Mn_{2}(CO)_{10}]$
0%
$K[ PtCI_{3}(C_{2} H_{4})]$

Oxidation numbers of chlorine atoms in CaOCl

$_2$ are

0%
0,0
0%
-1,-1
0%
-1+1
0%
None of these

Explanation

$CaOCl_2$ , Cl has -1 and +1 oxidation states.

Hence, the correct option is C.

1412131_1130582_ans_547696ba75b048e5ab4df027a48d0086.PNG

Oxidation numbers of P in

$P{ O }_{ 4 }^{ 3- }$ , of S in

$S{ O }_{ 4 }^{ 2- }$ and that of Cr in

$C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }$ are respectively:

0%
+3,+6 and +5
0%
+5,+3and +6
0%
-3,+6 and +6
0%
+5,+6 and +6

Explanation

(I)

$^xPO_4^{3-} \Rightarrow x+4\times (-2)=-3$

$\Rightarrow x= -3+8=+5$

$\Rightarrow x=+5$

Oxidation number of

$P=+5$

(II)

$^xSO_4^{2-} \Rightarrow x+4\times (-2)= -2$

$\Rightarrow x= -2+8$

$\Rightarrow x=+6$

Oxidation number of

$S=+6$

(III)

$^x{Cr_2}O_7^{2-} \Rightarrow 2x+7\times (-2)=-2$

$\Rightarrow 2x=-2+14$

$\Rightarrow 2x=12$

$\Rightarrow x=\cfrac {12}{2}=+6$

Fe shows an oxidation state of

$+1$ in:

0%
$Fe(CO)_5$
0%
$[Fe(H_2O)_5 NO]SO_4$
0%
$Fe_4[Fe(CN)_6]$
0%
$FeCl_4^-$

Explanation

In the standard brown ring test for the nitrate ion, the brown ring complex is:

$[Fe(H_2O)_5(NO)]^{2+}$

In this compound, the nitrosyl ligand is positively charged, and iron is in a +1 oxidation state.

Hence option B is correct answer.

Which is not a redox reaction?

0%
$BaO_{2}+H_{2}SO_{4}\rightarrow BaSO_{4}+H_{2}O_{2}$
0%
$2BaO_+O{_2}\rightarrow 2BaO_{2}$
0%
$4KClO_{3}\rightarrow 4KClO_{2}+2O_{2}$
0%
$SO_{2}+2H_{2}S\rightarrow 2H_{2}O+3S$

Explanation

Here $BaO_2+H_2SO_4\longrightarrow BaSO_4+H_2O_2$ is a double displacement reaction where the cations and anions of the $2$ reactants switch places forming $2$ new compounds.

All other options involve change in oxidation status. Hence they are redox reactions.

In which of the compounds does hydrogen have an oxidation state of - 1 ?

0%
$CH_4$
0%
$NH_3$
0%
$HCI$
0%
$CaH_2$

The correct order of decreasing oxidation number of P in compounds is:

0%
${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}{\kern 1pt} {\text{ < }}{\kern 1pt} {{\text{H}}_{\text{4}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\kern 1pt} {\text{ < }}{\kern 1pt} {{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}{\kern 1pt} {\text{ < }}{\kern 1pt} {{\text{P}}_{\text{4}}}$
0%
${{\text{H}}_{\text{4}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\kern 1pt} {\text{>}}{\kern 1pt} {{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}{\kern 1pt} > {\kern 1pt} {{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_2}{\kern 1pt} {\text{ > }}{\kern 1pt} {{\text{P}}_{\text{4}}}$
0%
${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}{\kern 1pt} < {\kern 1pt} {{\text{H}}_{\text{4}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\kern 1pt} {\text{ < }}{\kern 1pt} {{\text{P}}_{\text{4}}}{\kern 1pt} {\kern 1pt} {\text{ < }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_2}$
0%
${{\text{H}}_{\text{4}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\kern 1pt} {\text{ < }}{\kern 1pt} {{\text{P}}_{\text{4}}}{\kern 1pt} {\kern 1pt} {\text{ < }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_2}{\kern 1pt} {\text{ < }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}{\kern 1pt}$

Ratio of number of oxygen atoms which show -1 and -2 oxidation states in the compound

${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$ is

0%
1 : 3
0%
1 : 2
0%
3 : 1
0%
None of these

Oxidation number of fluorine in

${F}_{2}O_2$ is:

0%
$+1$
0%
$+2$
0%
$-1$
0%
$-2$

Which of the following equation depict the oxidising nature of

$H_{2}O_{2}$ ?

0%
$2MnO^{-}_{4}+6H^{+}+5H_{2}O_{2}\rightarrow 2Mn^{2}+8H_{2}O+5O_{2}$
0%
$2Fe^{3+}+2H^{+}+H_{2}O_{2}\rightarrow 2Fe^{2}+2H_{2}O+O_{2}$
0%
$2I^{-}+2H^{+}+H_{2}O_{2}\rightarrow I_{2}+2H_{2}O$
0%
$KIO_{4}+H_{2}O_{2}\rightarrow KIO_{3}+H_{2}O+O_{2}$

Explanation

Oxidising nature of

$H_2O_2$ means it oxidises other substance and itself gets reduced. In such reactions

$O_2$ is not evolved.

In (a),

$MnO_4^-$ gets reduced from

$Mn^{7+}$ to

$Mn^{2+}$ .

In (b),

$Fe^{3+}$ gets reduced to

$Fe^{2+}$

In (c),

$I^-$ gets oxidised to

$I_2$ .

In (d),

$KIO_4$ gets reduced from

$I^{7+}$ to

$I^{5+}$ .

Therefore in option (C) is correct Answer.

The oxidation number of Cr is +6 in:

0%
$K_{2}CrO_{4}$
0%
$K_{2}Cr_{2}O_{7}$
0%
$KCrO_{3}CI$
0%
All of these

Explanation

Chromium is hexavalent in this compound - each one has an oxidation state of

$+6$ to balance out.

$-2 × 3$ oxygen ions and

$+1$ potassium ions and

$Cl$ has

$-1$ (the net total has to be zero for a neutral compound).

The oxidation state of Fe in

$\left[ Fe\left( { H }_{ 2 }{ O } \right) _{ 5 }NO \right] { SO }_{ 4 }$ is :

0%
$+1$
0%
$+2$
0%
$+3$
0%
$+4$

Explanation

$NO$ in brown ring complex has

+ 1

oxidation state.

Let oxidation state of Fe be

x

Therefore,

$x + 5(0)+(1)+(-2)=0$ =

$x+ 1-2= 0$

$x= +1$

Hence, the oxidation state of iron in the brown ring complex is

+ 1

Thus option A is correct.

Given that

${ Cl }_{ 2\left( g \right) }+{ 2e }^{ - }\longrightarrow { 2Cl }^{ - }\left( aq \right)$ is the reduction half-reaction for the overall reaction

$2Ag\left( s \right) +{ Cl }_{ 2 }\left( g \right) \longrightarrow 2AgCl\left( s \right)$ .

What is the oxidation half reaction?

0%
$Ag\left( s \right) \longrightarrow Ag^{ + }\left( aq \right) +e^{ - }$
0%
$Ag\left( s \right) +{ Cl }^{ - }\left( aq \right) \longrightarrow Ag\left( s \right) +e^{ - }$
0%
$Ag\left( s \right) +{ Cl }_{ 2 }\left( g \right) +e^{ - }\longrightarrow AgCl\left( s \right) +Cl^{ - }\left( aq \right)$
0%
${ 2Cl }^{ - }\left( aq \right) \longrightarrow { Cl }_{ 2 }\left( g \right) +2e^{ - }$

The oxidation state(s) of

$Cl$ in

$CaO_{2}Cl_{2}$ (bleaching powder) is/are:

0%
+1 only
0%
-1 only
0%
+1 and -1
0%
none of these

Explanation

In bleaching powder,

$Ca(OCl)Cl$ , the two

$Cl$ atoms are in different oxidation states i.e., one

$Cl–$ having oxidation number of

$–1$ and the other as

$OCl–$ having oxidation number of

$+1.$

Hence, option C is correct.

1042060_1142130_ans_4f6286daee9b44b0a3fd25299a2a70fa.JPG

The oxidation state of

$Xe$ in

$XeO_{3}$ and bond angle in it are:

0%
$+6,\ 109^{o}$
0%
$+8,\ 103^{o}$
0%
$+6,\ 103^{o}$
0%
$+8,\ 120^{o}$

Explanation

$XeO_3$ , oxidation state of oxygen is

$-2$ .

So,

$x+3(-2)=0$

$\Rightarrow x-6=0$

$\Rightarrow x=6$

where,

$x$ is oxidation state of

$Xe$ .

$O$

$O=\overset {||}{Xe}=O$

Based on structure, it has a tetrahedral shape with bond angle nearly

$109^o$ .

When

$SO_2$ is passed in acidified potassium dichromate solution, the oxidation number of

$S$ is changed from:

0%
$+4$ to zero
0%
$+4$ to $+2$
0%
$+4$ to $+6$
0%
$+6$ to $+4$

Explanation

The reaction of sulphur dioxide with acidified $K_2Cr_2O_7$ in acidic medium is

$3SO_2 + Cr_2O_7^{2-} + 2 H^{+} \rightarrow 3 SO_4^{2-} + 2 Cr^{3+} + H_2O$

So Oxidation state of Sulphur in $SO_2$ is $+4$ which is changed to $+6$ in $SO_4^{2-}$ .

Hence option $C$ is correct.

Oxidation number of N in HNO

$_3$ is :

0%
$-3.5$
0%
$+3.5$
0%
$-6$
0%
none of these

Explanation

Oxidation number of hydrogen =

$+1$

Oxidation number of oxygen=

$-2$

Let oxidation number of nitrogen=

$x$

Since the net charge in the molecule is zero,

$1+x+3\times(-2)=0$

$\Rightarrow x=+5$

Hence D is correct

Which one of the following is a redox reaction ?

0%
$H_{2}+Br_{2} \rightarrow 2HBr$
0%
$2NaCl+H_{2}SO_{4} \rightarrow Na_{2}SO_{4}+2HCl$
0%
$HCl+AgNO_{3} \rightarrow AgCl+HNO_{3}$
0%
$NaOH+HCl \rightarrow NaCl+H_{2}O$

Explanation

Solution:

O.N decreased by

$1$ (Reduction)

(A)

$H_2+Br_2\rightarrow 2HBr$

O.N increased by

$1$ (oxidation)

(B)

$2NaCl+H_2SO_4\rightarrow Na_2SO_4+2HCl$

No oxidation & No reduction occurs

(C)

$HCl+AgNO_3\rightarrow AgCl+HNO_3$

No oxidation & No reduction occurs

(D)

$NaOH+HCl\rightarrow NaCl+H_2O$

No reduction & No oxidation occurs.

Hence, the redox reaction occurs in A.

1120350_1207765_ans_3720c391cb0741539701e642ea9bea20.jpg

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 9 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 9 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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