MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 4

To an acidic solution of an anion, a few drops of

$KMn{ O }_{ 4 }$ solution are added. Which of the following, if present, will not decolourise the

$KMn{ O }_{ 4 }$ solution?

0%
$C{ O }_{ 3 }^{ 2- }$
0%
$N{ O }_{ 2 }^{ \ominus }$
0%
${ S }^{ 2- }$
0%
${ Cl }^{ \ominus }$

Explanation

As we know, oxidation state of

$C$ in

$C{O}_{3}^{-2}$ is +4, which is maximum. So, it does not have any tendency to undergo oxidation and it will not able to reduce

$KMnO_4$ . Therefore, it will not decolourise the

$KMnO_4$ solution.

A metal ion

${M}^{3+}$ loses three electrons. Its oxidation number will be:

0%
$+3$
0%
$+6$
0%
$0$
0%
$-3$

Explanation

The reaction is as follows:

${ M }^{ 3+ }\longrightarrow { M }^{ 6+ }+3{ e }^{ - }$

So, the new oxidation of M is

$+6$ .

Hence, the correct option is

$B$

The oxidation number of

$Cl$ in

$CaO{Cl}_{2}$ is :

0%
$-1$ and $+1$
0%
$+2$
0%
$-2$
0%
none

Explanation

In the given compound

$CaOCl_2$ , one

$Cl$ is directly attached to

$Ca$ and other to

$O$ .

$O{ Cl }^{ \ominus }: -2+x = -1 \Rightarrow x = +1$

${ Cl }^{ \ominus }: -1$

Oxidation states of

$Cl$ are

$-1$ and

$+1$ .

Hence, option A is correct.

In which of the following compounds, the oxidation state of transition metal is zero?

0%
$Cr{ O }_{ 5 }$
0%
${ Fe }_{ 3 }{ O }_{ 4 }$
0%
$FeS{ O }_{ 4 }$
0%
$Fe{ \left( CO \right) }_{ 5 }$

Explanation

$Fe{ \left( CO \right) }_{ 5 }$ , the oxidation state of transition metal is zero.
(A) In

$Cr{ O }_{ 5 }$ , the oxidation state of metal is +6.
(B) In

${ Fe }_{ 3 }{ O }_{ 4 }$ , the oxidation state of metal is

$\displaystyle \frac {8}{3}$ .
(C) In

$FeS{ O }_{ 4 }$ , the oxidation state of metal is +2.
(D) In

$Fe{ \left( CO \right) }_{ 5 }$ , the oxidation state of metal is zero.

In which of the following, the highest oxidation state is not possible?

0%
${ \left[ Xe{ O }_{ 6 } \right] }^{ 4- }$
0%
$Xe{ F }_{ 8 }$
0%
$Os{ O }_{ 4 }$
0%
$Ru{ O }_{ 4 }$

Explanation

$Xe$ shows

$+8$ oxidation state in

$Xe{F}_{8}$ but it does not exist because of steric hindrance of

$8 F$ atoms.

The oxidation state of iodine in

${ H }_{ 4 }I{ O }_{ 6 }^{ \ominus }$ is :

0%
$+7$
0%
$-1$
0%
$+5$
0%
$+1$

Explanation

Let

$x$ be the oxidation state of

$I$ in

$H_4IO_6^\ominus$ .
Since, the overall charge on the complex is

$-1$ , the sum of oxidation states of all elements in it should be equal to

$-1$ .
Therefore,

$+4+x + 6(-2) = -1$
or,

$x = +7$
Hence, the oxidation state of

$I$ is

$+7$ .

The oxidation states of sulphur in the anions

$S{ O }_{ 3 }^{ 2- },{ S }_{ 2 }{ O }_{ 4 }^{ 2- }$ and

${ S }_{ 2 }{ O }_{ 6 }^{ 2- }$ follow the order :

0%
${ S }_{ 2 }{ O }_{ 4 }^{ 2- } < S{ O }_{ 3 }^{ 2- } < { S }_{ 2 }{ O }_{ 6 }^{ 2- }$
0%
$S{ O }_{ 3 }^{ 2- } < { S }_{ 2 }{ O }_{ 4 }^{ 2- } < { S }_{ 2 }{ O }_{ 6 }^{ 2- }$
0%
${ S }_{ 2 }{ O }_{ 4 }^{ 2- } < { S }_{ 2 }{ O }_{ 6 }^{ 2- } < S{ O }_{ 3 }^{ 2- }$
0%
${ S }_{ 2 }{ O }_{ 6 }^{ 2- } < { S }_{ 2 }{ O }_{ 4 }^{ 2- } < S{ O }_{ 3 }^{ 2- }$

Explanation

Let the oxidation number of

$S$ be

$x$ .
The oxidation number of oxygen is

$-2$ .

${ S }_{ 2 }{ O }_{ 6 }^{ 2- }:2x-12=-2\Rightarrow x=5$

$S{ O }_{ 3 }^{ 2- }:x-6=-2\Rightarrow x=4$

${ S }_{ 2 }{ O }_{ 4 }^{ 2- }:2x-8=-2\Rightarrow x=3$
Hence, the correct order is

${ S }_{ 2 }{ O }_{ 4 }^{ 2- } < S{ O }_{ 3 }^{ 2- } < { S }_{ 2 }{ O }_{ 6 }^{ 2- }$ .

The oxidation number of underlined element in

$\displaystyle K\left [ \underline{Co}\left ( C_{2}O_{4} \right )_{2}.\left ( NH_{3} \right )_{2} \right ]$ compound is :

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

Explanation

Let

$x$ be the oxidation number of Co in

$\displaystyle K\left [ \underline{Co}\left ( C_{2}O_{4} \right )_{2}.\left ( NH_{3} \right )_{2} \right ]$ .
The overall charge on the complex is

$0$ . Therefore, sum of oxidation state sof all elements should be equal to zero.
Charge on

$K$ is

$+1$ .
Oxidation state of

$C_2O_4$ is

$-2$ .
Oxidation state of

$NH_3$ is

$0$ .
Therefore,

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

$x= +3$

Hence, the correct option is A.

$\displaystyle HCN$ oxidation number of carbon is?

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

Explanation

$HCN =H^+ + CN^{-}$

$x-3=-1$

$x=-1+3$

$x=+2$

${? +O_2\rightarrow 2K_2O}$

0%
$K$
0%
${K_2}$
0%
$2K$
0%
$4K$

Explanation

$?+O_2\longrightarrow 2K_2O$

Since in the product side there are

$4$ atoms of potassium. so, to have a balanced equation, reactant side should also have

$4K$ .

Which of the following reactions is/ are not intermolecular redox reaction?

0%
$PbO_2+H_2O\rightarrow PbO+H_2O_2$
0%
$2KClO_3\rightarrow 2KCl+3O_2$
0%
$(NH_4)_2Cr_2O_7\rightarrow N_2+Cr_2O_3+4H_2O$
0%
$NH_4NO_2\rightarrow N_2+2H_2O$

Explanation

(i) Inter molecular Redox:In such redox reactions, one molecule of reactant is oxidized whereas molecule of other reactant is reduced.

$NO_3^{-} + H_2S + H_2O + H^{+} \rightarrow NH_4^{+} + HSO_4^{-}$

$NO_3^{-}\rightarrow NH_4^{+}$ (this is oxidized)

$H_2S\rightarrow HSO_4^{-}$ ( this is reduced)

(2) Intra molecular Redox:

One atom of a molecule is oxidized and other atom of same molecule is reduced then it is intramolecular redox reaction.

$2Mn_2O_7\rightarrow 4MnO_2 + 3O_2$

So options B,C & D are intramolecular reactions only option A is inter molecular reaction.

Hence options B,C & D are correct.

Which of the following reactions is not oxidation-reduction reaction?

0%
$H^{\bigoplus}+OH^{\circleddash}\rightarrow H_2O$
0%
$\displaystyle\frac{1}{2}H_2+\frac{1}{2}Cl_2\rightarrow HCl$
0%
$CaCO_3\rightarrow CaO+CO_2$
0%
$A$ and $C$

Explanation

The reactions are as follows:

$H^{\bigoplus}+OH^{\circleddash}\rightarrow H_2O$
It is a neutralisation reaction.

$\displaystyle\dfrac{1}{2}H_2+\dfrac{1}{2}Cl_2\rightarrow HCl$
It is a redox reaction.

$CaCO_3\rightarrow CaO+CO_2$
It is a decomposition reaction.

The oxidation state of chromium in

$\left[ Cr{ \left( PP{ h }_{ 3 } \right) }_{ 3 }{ \left( CO \right) }_{ 3 } \right]$ is :

0%
$+3$
0%
$+8$
0%
$0$
0%
$+5$

Explanation

Let

$x$ be the oxidation state of

$Cr$ in

$[Cr(PPh_3)_3(CO)_3$ .
Since, the overall charge on the complex is

$0$ , the sum of oxidation states of all elements in it should be equal to

$0$ .
Therefore,

$x + 3(0) + 3(0) = 0$
or,

$x = 0$ .
Hence, the oxidation state of

$Cr$ in

$[Cr(PPh_3)_3(CO)_3$ is

$0$ .

Which of the following leads to redox reaction?

0%
$AgN{ O }_{ 3 }+HCl$
0%
$KOH+HCl$
0%
$KI+{ Cl }_{ 2 }$
0%
$N{ H }_{ 3 }+HCl$

Explanation

$KI + Cl_2 \rightarrow KCl_2 + \dfrac{1}{2} I_2$

Now, in $KI$ the O.S. of $I$ is $-1$ .

This changes to 0 (zero) in $I_2$ molecule formation.

$Cl$ has 0(zero) O.S. in $Cl_2$ and it changes to $-\dfrac{1}{2}$ in $KCl_2$ .

So, as iodine is increasing it's O.S. ( $-1$ to $0)$ , $Cl$ is decreasing its O.S. ( $0$ to $-\dfrac{1}{2}$ ) so this must be redox reaction.

Which of the following reactions does not involve either oxidation or reduction?

0%
$V{ O }^{ 2+ }\longrightarrow { V }_{ 2 }{ O }_{ 3 }$
0%
$Na\longrightarrow { Na }^{ \oplus }$
0%
$Cr{ O }_{ 4 }^{ 2- }\longrightarrow { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$
0%
${ Zn }^{ 2+ }\longrightarrow Zn$

Explanation

Following reaction does not involve either oxidation or reduction.

$Cr{ O }_{ 4 }^{ 2- }\longrightarrow { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$
The oxidation state of Cr in both side is same i.e.

$+6$ .

Find out oxidation state of

$Mn$ in acid medium.

0%
$+5$
0%
$+4$
0%
$+2$
0%
$+3$

Explanation

The ratio

$\displaystyle 100 : 33.3 : 20 = 1 : \dfrac {1}{3} : \dfrac {1}{5}$

In acidic medium, the equivalent mass

$\displaystyle = \dfrac { { Formula\ mass\ of\ KMnO_4}}{5}$
The oxidation state of Mn changes from +7 to +2.

In neutral medium, the equivalent mass

$\displaystyle = \dfrac { { Formula\ mass\ of\ KMnO_4}}{3}$
The oxidation state of Mn changes from +7 to +4.

In alkaline medium, the equivalent mass

$\displaystyle = \dfrac { { Formula\ mass\ of\ KMnO_4}}{1}$
The oxidation state of Mn changes from +7 to +6.

The oxidation number of

$C$ in

$HNC$ is :

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

The oxidation number of

$Fe$ in

${ Na }_{ 2 }\left[ Fe{ \left( CN \right) }_{ 5 }NO \right]$ is :

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

Explanation

The oxidation number of

$Fe$ in

${ Na }_{ 2 }\left[ Fe{ \left( CN \right) }_{ 5 }NO \right]$ is calculated as follows:

${ Na }_{ 2 }\left[ Fe{ \left( CN \right) }_{ 5 }NO \right] \longrightarrow2Na^{+} +$ [Fe

$(CN)_5NO]^{-2}$

Now, to determine the O.S. of

$Fe$ in

$[Fe(CN)_5NO]^{-2}$ , let O.S. of

$Fe = x$ .

So,

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

$x -5 +1 =2$

$x = +2$

As cyanide has

$-1$ charge and

$NO$ has

$+1$ charge.

The oxidation states of the most electronegative elements in the products of the reaction between

$Ba{O}_{2}$ and

${H}_{2}S{O}_{4}$ are, respectively :

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

Explanation

The reaction between

$BaO_2$ and

$H_2SO_4$ is as follows:

$Ba{ O }_{ 2 }+{ H }_{ 2 }S{ O }_{ 4 }\left( dil \right) \longrightarrow BaS{ O }_{ 4 }+{ H }_{ 2 }{ O }_{ 2 }$
Oxidation state of

$O$ in

${ H }_{ 2 }{ O }_{ 2 }=-1$
Oxidation state of

$O$ in

$BaS{ O }_{ 4 }=-2$

The oxidation state of nitrogen is correctly given for:

0%
$\begin{matrix} Compound & Oxidation\quad state \\ \left[ CO{ \left( N{ H }_{ 3 } \right) }_{ 5 }Cl \right] { Cl }_{ 2 } & 0 \end{matrix}$
0%
$\begin{matrix} Compound & Oxidation\quad state \\ N{ H }_{ 2 }OH & -2 \end{matrix}$
0%
$\begin{matrix} Compound & Oxidation\quad state \\ { \left( { N }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }S{ O }_{ 4 } & +2 \end{matrix}$
0%
$\begin{matrix} Compound & Oxidation\quad state \\ { Mg }_{ 3 }{ N }_{ 2 } & -3 \end{matrix}$

Explanation

${ \left[ { Co }^{ +3 }\overset { 0 }{ { \left( N{ H }_{ 3 } \right) }_{ 5 } } { Cl }^{ 1- } \right] }^{ +2 }{ Cl }_{ 2 }^{ -1\times 2 }$

$N{ H }_{ 3 }:x+3=0\Rightarrow x=-3$ ; oxidation state of

$N=-3$

$N{ H }_{ 2 }OH:x+3-2=0\Rightarrow x=-1$
Oxidation state of

$N=-1$

${ \left( { N }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }^{ +1\times 2 }S{ O }_{ 4 }^{ 2- }:{ N }_{ 2 }{ H }_{ 5 }^{ \oplus }:2x+5=+1\Rightarrow x=-2$
Oxidation number of

$N=-2$

$\overset { +2\times 3 }{ { Mg }_{ 3 } } \overset { -3\times 2 }{ { N }_{ 2 } }$ : Oxidation state of

$N=-3$

Among the following, identify the species with an atom in

$+6$ oxidation state.

0%
$Mn{ O }_{ 4 }^{ \ominus }$
0%
${ \left[ Cr{ \left( CN \right) }_{ 6 } \right] }^{ 3- }$
0%
${ \left[ Ni{ F }_{ 6 } \right] }^{ 2- }$
0%
$Cr{ O }_{ 2 }{ Cl }_{ 2 }$

Explanation

$\bf{Hint:}$ The number of electrons accepted or donated by an atom to complete its octet is known as oxidation state.

$\bf{Correct \ answer:}$ Option

$D$

$\bf{Explanation \ for \ correct \ option:}$

In the

$CrO_{2}Cl_{2}$ , let oxidation state of

$Cr$ is

$x$ . As we know oxygen in oxide form has oxidation state

$-2$ and chlorine has

$-1$ . So, the oxidation state of

$Cr$ will be

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

$x+[(-4) + (-2)]=0$

$x-6=0$

$x=+6$

Thus, in the above compound, chromium

$[Cr]$ is present in

$+6$ oxidation state.

$\bf{Explanation \ for \ incorrect \ options:}$

In the $MnO_{4}{}^{-}$ , let oxidation state of $Mn$ is $x$ . As we know oxygen in oxide form has an oxidation state $-2$ . So, the oxidation state of $Mn$ will be

$x+[4\times(-2)]=-1$
$x-8=-1$
$x=-1+8$
$x=+7$

Thus, in the above compound, chromium $[Mn]$ is present in the $+7$ oxidation state.

In the $Cr[(CN)_{6}]^{3-}$ , let oxidation state of $Cr$ is $x$ . As we know $CN^{-}$ has an overall charge of $-1$ . So, the oxidation state of $Cr$ will be

$x-6=-3$
$x=-3+6$
$x=+3$

Thus, in the above compound, chromium $[Cr]$ is present in $+3$ oxidation state.

In the $[NiF_{6}]^{2-}$ , let oxidation state of $Ni$ is $x$ . As we know $F$ has an oxidation state of $-1$ . So, the oxidation state of $Ni$ will be

$x-6=-2$
$x=-2+6$
$x=+4$

Thus, in the above compound, chromium $[Ni]$ is present in the $+4$ oxidation state.

Which one of the following statements is not correct?

0%
Oxidation number of $S$ in $(NH_4)_2S_2O_8$ is $+6$ .
0%
Oxidation number of $Os$ in $OsO_4$ is $+8$ .
0%
Oxidation number of $S$ in $H_2SO_5$ is $+8$ .
0%
Oxidation number of $O$ in $KO_2$ is $-\frac{1}{2}$ .

Explanation

Oxidation number of

$S$ in

$(NH_4)_2S_2O_8$ is

$+6$ (figure).
Oxidation number of

$S$ in

$H_2SO_5$ is

$+6$ (figure).
Peroxide linkages are present in the above two in which each

$O$ has oxidation number of

$-1$ .
Oxidation number of

$Os$ in

$OsO_4$ is

$+8$ while that of

$O$ in

$KO_2$ is

$-\frac{1}{2}$ .
Hence, option C in incorrect.
The structures are as follows:

Oxidation number of

$S$ in

${S_2O_3}^{2-}$ is:

0%
-2
0%
+2
0%
+6
0%
0

Explanation

Let Oxidation number of S in

$S_2O_3^{2-}$ be x.
Thus,
2x+(-2*3)= -2
2x-6= -2
2x= -2+6
2x=4
x=4\2
x=2
so the oxidation state of sulfur is +2

Which of the following compounds has phosphorus atom in

$+1$ oxidation state?

0%
$H_3PO_3$
0%
$H_3PO_4$
0%
$H_3PO_2$
0%
$H_4P_2O_7$

Explanation

For the oxidation state of

$P$ :

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

$\therefore 3 + x - 4 = 0$

$\therefore x - 1 = 0$

$\therefore x = 1.$

$FeCr_2O_4$ the oxidation numbers of Fe and Cr are:

0%
+ 2 and + 3
0%
0 and +2
0%
+2 and +6
0%
+3 and +6

Explanation

$FeCr_2O_4$

$Fe: x+6+(4\times -2)=0$

$\Rightarrow x=8-6$

$\Rightarrow x=+2$ .

$Cr: 2+2x+(4\times -2)=0$

$\Rightarrow 2x=8-2$

$\Rightarrow x=6/2$

$\Rightarrow x=+3$ .

A balanced chemical equation is in accordance with:

0%
Avogadro's Law
0%
Law of constant proportion
0%
Law of conservation of mass
0%
Law of gaseous volumes

Determine the. oxidation state of chromium in the finfll product formed by the reaction between KI and acidified potassium dichromate solution.

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

Explanation

K would not be expected to be in the final NET equation since K+ is a spectator ion; i.e., it does not react. When you put KI into water it dissolves and splits up into K+ and I- ions. Only the I- participates in the redox reaction. It wouldn't matter whether you used KI or NaI.
Oxidation:

$2I^- \rightarrow I^2 + 2e^-$
Reduction:

$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

The final product formed is :

$6I^- + Cr_2O_7^{2-} + 14H^+ ----> 3I_2 + 2Cr^{3+} + 7H_2O$
Calculating the oxidation state of chromium we get + 3

Identify X in the equation given below.

$FeCl_3 + SO_2 + H_2O \rightarrow FeCl_2 + HCl + X$

0%
$H_2 S$
0%
$SO_3$
0%
$H_2SO_4$
0%
$H_2SO_3$

Explanation

The balanced equation can be written as:

$2FeCl_{3} + SO_{2} + 2H_{2}O \rightarrow 2FeCl_{2} + 2HCl + H_{2}SO_{4}$
Thus X =

$H_{2}SO_{4}$

The reaction

$ZnO + C \rightarrow Zn+ CO$ is an example of a

0%
Combination reaction
0%
Reduction-oxidation reaction
0%
Displacement reaction
0%
Decomposition Reaction

Sulphur dioxide bleaches colouring matter by ___________.

0%
dehydration
0%
oxidation
0%
decomposition
0%
reduction

Explanation

Sulphur dioxide bleaches coloring matter by reduction. Bleaching action of sulphur dioxide is not permanent.

$SO_2(g) + 2H_2O \rightarrow H_2SO_4(aq) + 2[H^+]$

When an oxidising agent reacts, it consumes electrons. In the following changes, which one of the oxidising agents has consumed maximum number of electrons per mole?

0%
$Cr{O}_{4}^{2-}\longrightarrow {Cr}^{3+}$
0%
$Mn{O}_{2}\longrightarrow {Mn}^{3+}$
0%
$Mn{O}_{4}^{2-}\longrightarrow {Mn}^{2+}$
0%
$Mn{O}_{4}^{-}\longrightarrow {Mn}^{2+}$

Explanation

Oxidation number of

$Mn$ in

$Mn{O}_{4}$ is

$+7$
Therefore, conversion of

${Mn}^{2+}$ from

${Mn}^{+7}$ will have to accept

$5$ electrons, which is highest.

Identify redox reaction among the following:

0%
acid-base neutralization
0%
precipitation reaction
0%
metal displacement reaction
0%
all the above

Explanation

Metal displacement reactions are redox reactions in which both oxidation and reduction takes place.Example;

$Mg + ZnSO_{4} \rightarrow MgSO_{4} + Zn$

Here Mg is undergoing oxidation with an increase in oxidation number from 0 to +2 and Zn is undergoing reduction with a decrease in oxidation number.

Which one is not correct for the reaction?

$(Cn)_2 + H_2O \rightarrow HCN+ HOCN?$

0%
It is an auto redox change.
0%
Oxidation number of C in $(CN)_2, HCN$ and $HOCN$ are +3, +2 and +4 respectively.
0%
Oxidation number of n in $(CN)_2, HCN$ and $HOCN$ are +3, +2 and +4 respectively.
0%
The resultant solution is acidic.

Explanation

$(CN)_2 + H_2O \rightarrow HCN+ HOCN$
+3 +2 +4
So, its a autoredox reaction.
And oxidation number of N in

$(CN)_2, HCN$ and

$HOCN$ are +3, +2 and +4 respectively.

Oxidation number of carbon and total number atoms in carbon suboxide are, respectively :

0%
$+5, \:4$
0%
$+\frac{4}{3},\:3$
0%
$+\frac{4}{3},\: 5$
0%
$+4,\: 5$

Explanation

The formula of carbon suboxide is

$C_3O_2[O=C=C=C=O]$ .
So, it has total

$5$ atoms.
Let

$x$ be the oxidation state of

$C$ .

$3x-4=0; x=+\frac{4}{3}$
Hence, the oxidation state of

$C$ is

$+\frac{4}{3}$ .

The oxidation state of

$Cr$ in

$K_{2}Cr_{2}O_{7}$ is:

0%
$+4$
0%
$+3$
0%
$+6$
0%
$+5$

Explanation

$K_{2}Cr_{2}O_{7}$

Let the oxidation state of

$Cr$ is

$x$ .

$2(+1) + 2x + 7(-2) = 0$

$+2 + 2x - 14 = 0$

$2x - 12 = 0$

$2x=12$

$x = +6$

The oxidation states of iodine in

$HIO_{4}, H_{3}IO_{5}$ and

$H_{5}IO_{6}$ are respectively

0%
$+1, +3, +7$
0%
$+7, +7, +3$
0%
$+7, +7, +7$
0%
$+7, +5, +3$

Explanation

An oxidation state is a number that is assigned to an element in a chemical combination. This number represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element.

The total sum of the oxidation state of all atoms in any given species is equal to the net charge on that species. In neutral species, the total sum of the oxidation state of all atoms is

$0$ . The oxidation state of

$H$ is

$+1$ in a compound.The oxidation state of

$O$ is

$-2$ in a compound. Calculating the oxidation states of

$I$ , we get

$H_{5}IO_{6}; 5 + x - 12 = 0$ or

$x = +7$

$H_{3}IO_{5}; 3 + x - 10 = 0$ or

$x = +7$

$HIO_{4}; 1 + x - 8 = 0$ or

$x = +7$
Thus, the correct answer is

$+7, +7, +7$

In which of the following compounds, nitrogen exhibits highest oxidation state ?

0%
$N_3H$
0%
$NH_2OH$
0%
$N_2H_4$
0%
$NH_3$

The oxidation number of C atom in

$CH_{2}Cl_{2}$ and

$CCl_{4}$ are respectively :

0%
$-2$ and $-4$
0%
$0$ and $-4$
0%
$0$ and $4$
0%
$2$ and $4$

Explanation

Let the oxidation number of C be

$x$ .

$CH_{2}Cl_{2}$ , the oxidation number of

$C$ atom is:

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

$x + 2 - 2 = 0$

$x = 0$

$CCl_{4}$ , the oxidation number of

$C$ atom is:

$x + 4(-1) = 0$

$x - 4 = 0$

$x = 4$

Hence, option

$C$ is correct.

Identify the strongest acids out of the following list.

0%
$HBr{O}_{4}$
0%
$HOCl$
0%
$H{NO}_{2}$
0%
${H}_{3}{PO}_{3}$

Explanation

As the oxidation state of central atom increases, acidity increases. So, the strongest acid is the one which has maximum value of oxidation number of the central atom.

$HBr{ O }_{ 4 }$ ; O.N of

$Br=+7$

$HOCl$ ; O.N of

$Cl=+1$

$H{NO}_{2}$ ; O.N of

$N=+3$

${H}_{3}{PO}_{3}$ ; O.N of

$P=+3$
As, oxidation state of bromine is +7 , which is highest in all .Hence, the strongest acid is

$HBr{ O }_{ 4 }$ .

The oxidation number of

$Cr$ in

$K_{2}Cr_{2}O_{7}$ is :

0%
$+3$
0%
$-3$
0%
$+6$
0%
$-6$

Explanation

The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion. The oxidation number of simple ions is equal to the charge on the ion. The metals in Group IA

$(K)$ atom has an oxidation number of

$+1$ . Oxygen usually has an oxidation number of

$-2$ .

Oxidation no of

$Cr$ in

$K_{2}Cr_{2}O_{7}$

$= 2 \times 1 + 2 \times x + 7\times (-2) = 0$

$x = \dfrac {14 - 2}{2} = +6$

In the reaction

$2Cr + 3Ni^{2+}\rightarrow 3Ni + 2Cr^{3+}$ , the species oxidized is:

0%
$Cr$
0%
$Ni^{2+}$
0%
$Cr^{3+}$
0%
$Ni$
0%
none of the above

Explanation

Loss of electrons is oxidation. In reaction

$Cr$ is loosing electrons to form

$Cr^{3+}$ while

$\displaystyle Ni^{2+}$ gains 2 electrons to form

$Ni$ . Therefore

$Cr$ is oxidized while

$Ni^{2+}$ is reduced in the reaction.

Which one of the following have the highest oxidation number for chlorine?

0%
$HClO_{2}$
0%
$KClO_{3}$
0%
$HCl$
0%
$KClO_{4}$
0%
$CaCl_{2}$

Explanation

The oxidation number of

$Cl$ in the given compounds is as shown below.

$\displaystyle HClO_2$ +3

$\displaystyle KClO_3$ +5

$\displaystyle HCl$ -1

$\displaystyle KClO_4$ +7

$\displaystyle CaCl_2$ -1
Thus, the highest oxidation number of chlorine is in

$\displaystyle KClO_4$

The reducing agent in the reaction

$Al + Fe^{3+}\rightarrow Al^{3+}+ Fe$ is:

0%
$Fe$
0%
$Al$
0%
$Al^{3+}$
0%
$Fe^{3+}$
0%
none of the above

Explanation

In the reaction

$\displaystyle Al + Fe^{3+}\rightarrow Al^{3+}+ Fe$ , the oxidation number of

$Al$ increases from 0 to +3.

$Al$ accepts 3 electrons to form

$\displaystyle Al^{3+}$ ion.
Hence,

$Al$ is oxidized (loss of electrons is oxidation) and acts as reducing agent.

$Al$ reduces ferric ions to

$Fe$ metal. The oxidation number of iron decreases from +3 to 0.
Ferric ion gets 3 electrons to form iron metal. The gain of electrons is a reduction.

A compound contains

$X, Y\ and\ Z$ atoms. The oxidation states of

$X, Y\ and\ Z$ are

$+2, + 2\ and\ -2$ respectively. The possible formula of the compound is:

0%
$XYZ_2$
0%
$Y_2(XZ_3)_2$
0%
$X_3(Y_4Z)_2$
0%
$X_3(YZ_4)_3$

Potassium dichromate is a good oxidising agent in acidic medium, the oxidation state of chromium changes by :

0%
2
0%
3
0%
4
0%
5

Explanation

Potassium dichromate is a powerful oxidizing-agent in acidic medium.

$Cr_2O_7^{2-}+14H^+ +6e^-\longrightarrow 2Cr^{3+} +7H_2O$

Oxidation number convert

$+6$ to

$+3$ .

So, the answer is B.

$KH$ , the oxidation number of hydrogen is:

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

Explanation

$KH$ , the oxidation number of hydrogen is -1 and the oxidation number of

$K$ is +1.
When

$H$ is directly attached to strongly electropositive metals such as

$K$ , the

$H$ atom is present in the form of hydride ion

$\displaystyle (H^-)$ with an oxidation number of -1.

Among the following, the correct order of acidity is

0%
$HClO_{3} < HClO_{4} < HClO_{2} < HClO$
0%
$HClO < HClO_{2} < HClO_{3} < HClO_{4}$
0%
$HClO_{2} < HClO < HClO_{3} < HClO_{4}$
0%
$HClO_{4} < HClO_{2} < HClO < HClO_{3}$

Explanation

The increasing order of acidity is

$\displaystyle HClO < HClO_{2} < HClO_{3} < HClO_{4}$
+1 +3 +5 +7

The acidic character of the oxyacids of halogens increases with an increase in the oxidation number of halogens.

The sum of the coefficients in the balanced equation of

$Co + Ni^{2+} \rightarrow Co^{3+} + Ni$ , is:

0%
$5$
0%
$2$
0%
$3$
0%
$10$
0%
$13$

Explanation

$\displaystyle 2Co + 3Ni^{2+} \rightarrow 2Co^{3+} + 3Ni$
This is balanced chemical equation.
The sum of the coefficients in the balanced equation

$=\displaystyle 2+3+2+3=10$

Statement 1: Addition of the oxidation numbers of all the elements in the formula of a compound must be 0.
Statement 2: Generally oxidation number of oxygen is -2 for most of the compounds.

0%

Both the statements 1 and 2 are correct and statement 2 is not correct explanation of statement 1.
0%

Both the statements 1 and 2 are correct.
0%

Both the statements 1 and 2 are correct and statement 2 is correct explanation of statement 1.
0%

Statement 1 is not correct but statement 2 is correct.
0%

Statement 1 is correct but statement 2 is not correct.

Explanation

Statement 1: Addition of the oxidation numbers of all the elements in the formula of a compound must be 0.

For example, in water molecule, each H atom has oxidation number of +1 and oxygen atom has oxidation number of -2.
Addition of the oxidation numbers of all the elements in

$H_2O = 2(+1) -2 = 0$

Statement 2: Generally oxidation number of oxygen is -2 for most of the compounds. There are certain compounds in which the oxidation number of oxygen is not -2.

For example, in hydrogen peroxide

$H_2O_2$ , the oxidation number of oxygen is -1.

Both the statements 1 and 2 are correct.

What is the oxidation state of calcium in calcium carbonate (

$\displaystyle \ { Ca } { CO }_{ 3 }$ )?

0%
+6
0%
+5
0%
+4
0%
+1
0%
+2

Explanation

The oxidation state of calcium in calcium carbonate

$\displaystyle CaCO_3$ is +2.
Let,

$X$ be the oxidation state of

$Ca$ .

$\displaystyle X + 4 + 3(-2) = 0 \\ X +4 -6 = 0 \\ X -2 = 0 \\ X = +2$

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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