MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 3

Which has maximum oxidation number of the underlined atom in the following?

0%
$\underline { Mn } { O }_{ 4 }^{ 2- }$
0%
$\underline { Cr } { O }_{ 5 }$
0%
$\underline { Cr } { O }_{ 2 }{ Cl }_{ 2 }$
0%
Equal

Explanation

Oxidation state of the underlined atom in each case is given,
(a)

$\underline { Mn } { O }_{ 4 }^{ 2- }$ ;

$x=6$

(b)

$\underline { Cr } { O }_{ 5 }$ ;

$x=6$

(c)

$\underline { Cr } { O }_{ 2 }{ Cl }_{ 2 }$ ;

$x=6$
Hence, all are equal.

In a reaction,

$4$ moles of electrons are transferred to

$1$ mole of

${HNO}_{3}$ . The possible product obtained due to reduction is :

0%
$0.4$ mol of ${ N }_{ 2 }$
0%
$0.5$ mol of ${ N }_{ 2 }O$
0%
$1$ mol of $N{ O }_{ 2 }$
0%
$1$ mol of ${ NH }_{ 3 }$

Explanation

${ H{ NO }_{ 3 } } +4{ e }^{ - }\longrightarrow Product$
+5 +1
In

${ N }_{ 2 }O$ , oxidation number of

$N=+1$ .
2 moles of

$HNO_3$ produces 1 mole of

$N_2O$ .
So,

$0.5$ mol of

${ N }_{ 2 }O$ are formed by reduction of

$1$ mol of

$H{ NO }_{ 3 }$ with

$4$ mol of electrons.

When

$PbS$ is treated with warm dilute

${HNO}_{3}$ , the products are most likely to be :

0%
${ Pb }^{ 2+ },\:S,\:N{ O }_{ 2 }$
0%
${ Pb }^{ 2+ },\:S,\:N{ O }$
0%
$PbO,\:S,\:NO$
0%
$Pb{ O }_{ 2 },\:S{ O }_{ 4 }^{ 2- },\:{ N }_{ 2 }$

Explanation

The complete reaction is as follows :

$PbS + HNO_3 \rightarrow Pb(NO_3)_2 + NO + S + H_2O$
Hence, the products are most likely to be

$Pb^{2+}, \: S$ and

$NO$ .

What is the oxidation number of sulphur in

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

0%
+7
0%
+5
0%
+6
0%
This compound does not exist

Explanation

${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }+2{ H }_{ 2 }O\longrightarrow 2{ H }_{ 2 }{ SO }_{ 4 }+{ H }_{ 2 }{ O }_{ 2 }$
The structure of

${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$ has been given below:
We can see that it has a peroxide linkage in which each oxygen atom has oxidation number of

$-1$ .
Let

$x$ be the oxidation state of S in

${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$ .
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,

$+2+2x- 2(4)-2 -4 =0$
or,

$x=+6$

$\left[ Cr({ O }_{ 2 }){ \left( { NH }_{ 3 } \right)}_{ 4 }{ (H }_{ 2 }O) \right] { Cl }_{ 2 }$ , oxidation number of

$Cr$ is

$+3$ then,

${O}_{2}$ will be in the form of :

0%
dioxo
0%
peroxo
0%
superoxo
0%
oxo

Explanation

Let the oxidation number of

$O_2$ be

$x$ .
Since the overall charge on the complex is

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

$0$ .

$\left[ \underset { +3 }{ Cr } \underset { +x }{ ({ O }_{ 2 }) } \underset { +0 }{ { \left( { NH }_{ 3 } \right) }_{ 4 } } \underset { +0 }{ \left( { H }_{ 2 }O \right) } \right] \underset { -1 }{ { Cl }_{ 2 } }$
Therefore,

$x+3+0+0-2 =0; x=-1$
So,

$({O}_{2})$ exists as

${ O }_{ 2 }^{ - }$ (superoxide ion).

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

${ H }_{ 2 }S{ O }_{ 5 }$ , both have

$+6$ oxidation state of sulphur. It is due to the :

0%
presence of peroxy group
0%
presence of superoxo group
0%
presence of neutral ${O}_{2}$
0%
presence of ozone

Explanation

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

${ H }_{ 2 }S{ O }_{ 5 }$ , both have

$+6$ oxidation state of sulphur. It is due to the presence of peroxy group. Both peroxomonosulphuric acid and peroxodisulphuric acid contains 1 peroxy group each.

The ratio of oxygen atom having

$-2$ and

$-1$ oxidation numbers in

$S_2O^{2-}_8$ is _____.

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

Explanation

The ratio of oxygen atom having

$-2$ and

$-1$ oxidation numbers in

$S_2O^{2-}_8$ is three as only one peroxy linkage is present.
So, we can see from the structure below that the number of oxygen atoms having

$-2$ oxidation state is

$6$ while those having

$-1$ oxidation state is

$2$ . Hence, the ratio is

$\frac{6}{2} = 3$ .

Prussian blue has two types of iron with oxidation number as shown below:

$\overset { III }{ Fe } \left[ \overset { II }{ Fe } { \left( CN \right) }_{ 6 } \right]$
What is the net charge on prussian blue?

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

Explanation

According to given conditions,

$\overset { III }{ Fe } \left[ \overset { II }{ Fe } { \left( CN \right) }_{ 6 } \right]$
Sum of the charges =

$+3+2-6\times 1 = -1$ .
Therefore, net charge on compound is

$-1$ .

Oxidation number of chromium in

$Cr{ O }_{ 2 }^{ 2+ }$ is ____.

0%
-2
0%
+4
0%
+6
0%
-6

Explanation

Let

$x$ be the oxidation number of Cr in

$Cr{ O }_{ 2 }^{ 2+ }$ .
Since the overall

charge on the complex is

$+2$ , the sum of oxidation states of all elements in

it should be equal to

$+2$ .
Therefore,

$x+2(2-) = +2$
or,

$x = +6$
Thus, the oxidation number of Cr in

$Cr{ O }_{ 2 }^{ 2+ }$ is

$+6$ .

Oxidation number of

$Cl$ atoms in

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

0%
zero on each
0%
$-1$ on $\overset { \ast }{ C } l$ and $+1$ on $\overset { \ast }{ C } \overset { \ast }{ l }$
0%
$+1$ on ${ Cl }^{ \ast }$ and $-1$ on ${ Cl }^{ \ast \ast }$
0%
$+1$ on each

Explanation

$\displaystyle CaOCl_2$ is the mixed salt of HCl and HOCl with

$\displaystyle Ca(OH)_2$ and can be represented as

$\displaystyle \underset {\displaystyle \text { -1}}{Cl}-Ca- \underset {\displaystyle \text { +1}}{OCl}$ .
The oxidation number of chlorine in HCl is

$\displaystyle -1$ .
The oxidation number of chlorine in HOCl is

$\displaystyle +1$ .

Oxidation number of

$Mn$ in

$Mn{ O }_{ 3 }Cl$ is:

0%
+5
0%
+7
0%
+6
0%
+4

Explanation

Let

$x$ be the oxidation number of

$Mn$ in

$Mn{ O }_{ 3 }Cl$ .
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,

$x+3(-2)+(-1)=0$
or,

$x=7$ .

Statement: The oxidation number of carbon in

$NaCNS$ is

$+4$

State whether the given statement is true or false.

0%
True
0%
False
0%

Anomalous
0%
None of these

Explanation

Let

$x$ be the oxidation number of C in

$NaCNS$ .
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,

$x+1+(-2) +(-3) = 0$
or,

$x=+4$
Hence, the oxidation number of carbon in

$NaCNS$ is

$+4$ .

Hence, the given statement is

$\text{true}$

Example(s) of a compound with chlorine in

$+7$ oxidation state is/are :

0%
$KClO_4$
0%
$Cl_2O_7$
0%
$Cl_2O_6$
0%
all of these

Explanation

Examples of compound where oxidation number of chlorine are +

$7$ is

$KClO_4$ and

$Cl_2O_7$ .
(a) For

$KCIO_4$ ,
Let the oxidation number of Cl be

$x$ .
Since the overall charge on the complex is

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

$0$ .

$+1+x+4(-2)=0$
or,

$x=+7$
(b) For

$Cl_2O_7$ ,
Let the oxidation number of Cl be

$x$ .

$2x+7(-2)=0$
or,

$x=+7$

Oxidation number of carbon in

$(CN)_2$ is _____.

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

Explanation

Let

$x$ be the oxidation number of C in

$(CN)_2$ .
Since the overall

charge on the compound is

$0$ , the sum of oxidation states of all elements in

it should be equal to

$0$ .
Therefore,

$2(x+(-3)) =0$
or,

$x=+3$
Hence, the oxidation number of C in

$(CN)_2$ is

$+3$ .

The _________ process involves the complete transfer of electrons in the formation of ionic bonds or partial transfer or shift of electrons in the formation of covalent bonds.

0%
redox
0%
oxidation
0%
reduction
0%
none

Explanation

The redox process involves the complete transfer of electrons in the formation in ionic bonds or partial transfer or shift of electrons in the formation of covalent bonds.
An example of the first type is the reaction between mercuric chloride and stannous chloride.

$2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$
In this reaction,

$SnCl_2$ is oxidized into

$SnCl_4$ and

$HgCl_2$ is reduced into

$Hg_2Cl_2$ .

Corrosion of iron is :

0%
redox process
0%
neutrilization process
0%
precipitation process
0%
none of the above

Explanation

Corrosion of iron is redox process.
At anode, iron is oxidized.

$\displaystyle Fe \rightarrow Fe^+ + 2e^- \: \:\ \ \ \ \ \ \ \ \ \ \ \ E^0_{oxid} = 0.44 V$
Reduction occurs at cathode.

$\displaystyle 2H^+ + 1/2 O_2 +2e^- \rightarrow H_2O \: \: \ \ \ \ \ \ \ \ E^0_{red} = 1.23 V$

Perxenate

$(XeO^{4-}_6$ ) and perchromate (

$CrO_5$ ) both have peroxide bonds.

0%
True
0%
False
0%
Ambiguous
0%
None of these

Explanation

$\text { Therefore, the given statement is False }$
1990372_183569_ans_1110a7cba3174a748e3d130bf5d18a81.PNG

The sum of oxidation number of all the atoms in a neutral molecule must be zero.

0%
True
0%
False
0%
Ambiguous
0%
None of these

Explanation

The sum of oxidation number of all the atoms in a neutral molecule must be zero.
For example, neutral molecules such as

$O_2, P_4, O_3, S_8$ and

$KMnO_4$ have the sum of oxidation number of all the atoms equal to zero.
For an ion, the sum of oxidation number of all the atoms is equal to the charge on the ion.
For example, in cyanide ion

$(CN^-)$ , the sum of oxidation number of all the atoms is equal to

$-1$ .
In ammonium ion

$(NH_4)^+$ , the sum of oxidation number of all the atoms is equal to

$+1$ .

The lowest possible oxidation state of nitrogen is

$-3$ as in

$N^{3-}$ .

0%
True
0%
False
0%
Ambiguous
0%
None of these

Explanation

The lowest possible oxidation state of nitrogen is

$-3$ as in

$N^{3-}$ .

Nitrogen can form compounds in which oxidation state ranges from

$-3$ to

$+5$ .

Ammonia,

$NH_3$ and magnesium nitride,

$Mg_3N_2$ have N in

$-3$ oxidation state.

N has

$5$ valence electrons. It accepts

$3$ electrons to complete its octet.

Thus, it shows

$-3$ oxidation state.

What is the oxidation number and valency of carbon in methanal

$\left(HCHO\right)$ ?

0%
0, 4
0%
1,4
0%
0,2
0%
None of these

Explanation

For compounds containing C atoms, the oxidation number of C atom

$= \frac {2n_O-n_H}{n_C}$
Here,

$n_O, n_H and n_C$ represents the number of oxygen, hydrogen and carbon atoms respectively.
For methanal

$CH_2O$ , the oxidation number of C atom

$\displaystyle = \frac {2n_O-n_H}{n_C}=\frac {2 \times 1-2}{1}=0$
The valency of C in methanal is 4 as it is attached to 2 H atoms by two single bonds and one oxygen atom by a double bond.

$\displaystyle H- \overset {\overset {\displaystyle O}{||}}{C}-H$

In a reaction,

$4$ mol of electrons are transferred to one mol of

$HNO_3$ when it acts as an oxidant. The possible reduction product is :

0%
$0.5\:mol\:N_2$
0%
$0.5\:mol\:N_2O$
0%
$1\:mol\:NO_2$
0%
$1\:mol\:NH_3$

Explanation

As it accepts

$4$ mol of electrons, oxidation state of N in product is

$5-4=1$ . Therefore, product is

$N_2O$ , where oxidation state of N is

$+1$ .

Which of the following is a wrong statement about redox reaction?

0%
Share of electrons
0%
Transfer of electrons
0%
Gain of oxygen
0%
Loss of Hydrogen

Which of the following are primary standard?

0%
Oxalic acid
0%
Potassium permanganate
0%
Potassium dichromate
0%
Sodium hydroxide

Determine the oxidation number of underlined atoms in following :

$8K\underline{Cl}O_3 + 24HCl \rightarrow + 8K\underline {Cl}+ 12 H_2O + 9\underline{Cl}_2+ 6\underline{Cl}O_2$

0%
$+5,\: -1,\: 0,\: +4$
0%
$+5,\: -1,\: 0,\: +2$
0%
$+4,\: -1,\: 0,\: +5$
0%
None of these

Explanation

Let X be the oxidation number of Cl in

$K{Cl}O_3$

$\displaystyle 1+X+3(-2) = 0$

$\displaystyle X = +5$

Let X be the oxidation number of Cl in

$K{Cl}$

$\displaystyle 1+X = 0$

$\displaystyle X= -1$

Let X be the oxidation number of Cl in

${Cl}_2$

$\displaystyle 2X=0$

$\displaystyle X=0$

Let X be the oxidation number of Cl in

${Cl}O_2$

$\displaystyle X+2(-2) = 0$

$\displaystyle X = +4$

The oxidation number of underlined atoms in

$8K\underline{Cl}O_3 + 24HCl \rightarrow + 8K\underline {Cl}+ 12 H_2O + 9\underline{Cl}_2+ 6\underline{Cl}O_2$ are

$+5,\: -1,\: 0,\: +4$ respectively.

Tailing of mercury is ___________ redox change.

0%
Intramolecular
0%
Intermolecular
0%
Disproportion
0%
None

Explanation

When ozone is passed through mercury, mercurous oxide

$(Hg_2O)$ is formed.
Due to this, mercury loses its meniscus and starts sticking to the glass.
This phenomenon is known as Tailing of mercury.

$2Hg + O_3 \rightarrow Hg_2O + O_2$ .
In this reaction, the oxidation number of mercury changes from 0 to +1. Thus, it is oxidized.
The oxidation number of oxygen changes from 0 to -2. Thus, it is reduced.
Hence, tailing of mercury is intermolecular redox change.

Select the primary standard(s).

0%
$Ce\left (NO_3\right)_4$
0%
$As_2O_3$
0%
$Na_2C_2O_4$
0%
$Na_2CO_3$

Explanation

A primary standard is a reagent that is extremely pure, stable, has no water of hydration and has a high molecular weight.

$As_2O_3$ ,

$Na_2C_2O_4$ act as primary standard reagents.

$Ce(NO_3)_4$ ,

$Na_2CO_3$ are not primary standard reagents because they have less molecular weight and they can be hydrated.
Hence, options B and C are correct.

The change in the oxidation number of the underlined nitrogen atom in the following chemical reactions are :
a. Nitrous acid

$\left( H\underline { N } { O }_{ 2 } \right)$ reduces

$Mn{ O }_{ 4 }^{ \ominus }$ in acid solution.
b. Nitrous acid

$\left( H\underline { N } { O }_{ 2 } \right)$ oxidises

${ I }^{ \ominus }$ to

${ I }_{ 2 }$ in acid solution.

0%
(a) +3 to +5
(b) +3 to +2
0%
(a) +3 to +6
(b) +3 to +2
0%
(a) +3 to +5
(b) +3 to +1
0%
(a) +3 to +2
(b) +3 to +5

Explanation

The balanced reactions are as follows:
i.

$2Mn{ O }_{ 4 }^{ \ominus } + 6{ H }^{ + } + 5N{ O }_{ 2 }^{ \ominus } \longrightarrow 5N{ O }_{ 3 }^{ \ominus } + 3{ H }_{ 2 }O + 2{ Mn }^{ 2+ }$
ii.

$2HN{ O }_{ 2 } + 2{ H }^{ \oplus } + 2{ I }^{ \ominus } \longrightarrow 2NO + 2{ H }_{ 2 }O + { I }_{ 2 }$
Oxidation number of

$N$ changes from +3 to +5 in (i).
Oxidation number of

$N$ changes from +3 to +2 in (ii).

One mole of

${N}_{2}{H}_{4}$ loses

$10$ moles of electrons to form a new compound

$A$ . Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in

$A$ ?

[There is no change in the oxidation state of hydrogen]

0%
$+1$
0%
$-3$
0%
$+3$
0%
$+5$

Explanation

The oxidation state of N in hydrazine is -2.

1 mole of hydrazine contains 2 moles of N and loses 10 moles of electrons. Hence, 1 N atom will lose 5 electrons. Hence, its oxidation number will increase by 5.

Hence, the oxidation number of N in compound A will be

$-2+5=+3$ .

Hence, the correct option is

$C$

The oxidation state of

$C$ in diamond is :

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

Explanation

Carbon in diamond is in elemental state, so the oxidation state of

$C$ in diamond is zero.
The structure of diamond has been shown below:

Assign oxidation numbers to the elements in the following ionic compounds.
a.

$NaBr$ b.

$MgO$ c.

$Al{F}_{3}$

0%
a. Na = $+1$ , Br = $-1$ ; b. Mg = $+2$ , O = $-2$ ; c. Al = $+3$ , F = $-1$
0%
a. Na = $+2$ , Br = $-2$ ; b. Mg = $+2$ , O = $-2$ ; c. Al = $+3$ , F = $-1$
0%
a. Na= $+1$ , Br = $-1$ ; b. Mg = $+1$ , O = $-1$ ; c. Al = $+3$ , F = $-1$
0%
None of these

Explanation

In sodium bromide (

$NaBr$ ), the oxidation number of Na is

$+1$ and that of Br is

$-1$ .
In magnesium oxide (

$MgO$ ), the oxidation number of Mg is

$+2$ and that of O is

$-2$ .
In Aluminium fluoride (

$AlF_3$ ), the oxidation number of Al is

$+3$ and that of F is

$-1$ .

Which of the following represent redox reactions?

0%
${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+2\overset { \ominus }{ O } H\longrightarrow 2Cr{ O }_{ 4 }^{ 2- }+{ H }_{ 2 }O$
0%
$S{ O }_{ 3 }^{ 2- }+{ H }_{ 2 }O+{ I }_{ 2 }\longrightarrow S{ O }_{ 4 }^{ 2- }+2{ I }^{ \ominus }+2{ H }^{ \oplus }$
0%
$Ca{ \left( OH \right) }_{ 2 }+{ Cl }_{ 2 }\longrightarrow Ca{ \left( OCl \right) }_{ 2 }+Ca{ Cl }_{ 2 }$
0%
$P{ Cl }_{ 5 }\longrightarrow P{ Cl }_{ 3 }+{ Cl }_{ 2 }$

Explanation

${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+2\overset { \ominus }{ O } H\longrightarrow 2Cr{ O }_{ 4 }^{ 2- }+{ H }_{ 2 }O$
+6 +6

$S{ O }_{ 3 }^{ 2- }+{ H }_{ 2 }O+{ I }_{ 2 }\longrightarrow S{ O }_{ 4 }^{ 2- }+2{ I }^{ \ominus }+2{ H }^{ \oplus }$
+4 0 +6 -1

$Ca{ \left( OH \right) }_{ 2 }+{ Cl }_{ 2 }\longrightarrow Ca{ \left( OCl \right) }_{ 2 }+Ca{ Cl }_{ 2 }$
0 +1 -1

$P{ Cl }_{ 5 }\longrightarrow P{ Cl }_{ 3 }+{ Cl }_{ 2 }$
+5 +3
In (A), there in no change in the oxidation state (+6) of Cr.
Therefore, it is not a redox reaction.

Which molecules represented by the underlined atoms are in their highest oxidation state?

0%
${H}_{2}{\underline S}_{2}{O}_{8}$
0%
${\underline P}_{4}{O}_{10}$
0%
${F}_{2}\underline O_2$
0%
${\underline {Mn}}_{2}{O}_{7}$

Explanation

${Mn}_{2}{O}_{7}$
+7

$F_2 \underline O_2$
+1

${P}_{4}{O}_{10}$
+5

${H}_{2}{S}_{2}{O}_{8}$
+6

Which molecules represents the unerlined atoms in their lowest oxidation state?

0%
${F}_{2}\underline O$
0%
${H}_{2}\underline S$
0%
$\underline P{H}_{3}$
0%
$\underline {N}_{2}{H}_{4}$

Explanation

${H}_{2}\underline S$ and

$\underline P{H}_{3}$ represents the unerlined atoms in their lowest oxidation state.
The oxidation state of S in

${H}_{2}\underline S$ is -2 and the oxidation state of P in

$\underline P{H}_{3}$ is -3.

Calculate the oxidation number of the underlined elements in the following compounds.
a.

$K\underline { Mn } { O }_{ 4 }$ b.

$\underline { Cr } { O }_{ 2 }{ Cl }_{ 2 }$ c.

$Na\underline { I } { O }_{ 3 }$

0%
a. $+7$ b. $+6$ c. $+5$
0%
a. $+9$ b. $+6$ c. $+5$
0%
a. $+7$ b. $0$ c. $+5$
0%
None of these

Explanation

a) Let

$x$ be the oxidation number of Mn in potassium permanganate

$KMnO_4$ .
Since, the overall charge on the complex is

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

$0$ .

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

$x=+7$
b) Let

$x$ be the oxidation number of Cr in chromyl chloride

$CrO_2Cl_2$ .

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

$x=+6$
c) Let

$x$ be the oxidation number of I in nitric acid

$NaIO_3$ .

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

$x=+5$

Which of the following statements is/are correct about

$C{H}_{2} = C{Cl}_{2}$ ?

0%
Both carbons are in $+2$ oxidation state
0%
Both carbons are in $-2$ oxidation state
0%
The first carbon has $+2$ and the second has $-2$ oxidation states
0%
The average oxidation number of carbon is zero

Explanation

$C{H}_{2} = C{Cl}_{2}$

-2 +2
So, the first carbon has

$+2$ and the second has

$-2$ oxidation states as shown above.
Therefore, the average oxidation number of carbon is

$\dfrac{(-2)+(2)}{2} =0$

Which of the following reactions does not involve oxidation-reduction?

0%
$2Rb+2{ H }_{ 2 }O\longrightarrow 2RbOH+{ H }_{ 2 }$
0%
$2Cu{ I }_{ 2 }\longrightarrow 2CuI+{ I }_{ 2 }$
0%
$N{ H }_{ 4 }Cl+NaOH\longrightarrow NaCl+N{ H }_{ 3 }+{ H }_{ 2 }O$
0%
$4KCN+Fe{ \left( CN \right) }_{ 2 }\longrightarrow { K }_{ 4 }\left[ Fe{ \left( CN \right) }_{ 6 } \right]$

Explanation

The reactions are as follows:

$2Rb+2{ H }_{ 2 }O\longrightarrow 2RbOH+{ H }_{ 2 }$
0 +1 +2 0

$2Cu{ I }_{ 2 }\longrightarrow 2CuI+{ I }_{ 2 }$
+2 +1 0

$N{ H }_{ 4 }Cl+NaOH\longrightarrow NaCl+N{ H }_{ 3 }+{ H }_{ 2 }O$

$4KCN+Fe{ \left( CN \right) }_{ 2 }\longrightarrow { K }_{ 4 }\left[ Fe{ \left( CN \right) }_{ 6 } \right]$
As we can see that in (C) and (D), the oxidation numbers of the various elements do not change.
Therefore, they are not redox reactions.

Calculate the oxidation number of the underlined elements.
a.

$\underline { P } { H }_{ 3 }$ b.

$\underline { S } { O }_{ 2 }$ c.

$H\underline { N } { O }_{ 3 }$ d.

${ H }_{ 3 }\underline { P } { O }_{ 4 }$

0%
a. $-3$ ; b. $+4$ ; c. $+5$ ; d. $+5$
0%
a. $-3$ ; b. $+2$ ; c. $+5$ ; d. $+5$
0%
a. $-3$ ; b. $+4$ ; c. $+5$ ; d. $+4$
0%
None of these

Explanation

a) Let

$x$ be the oxidation number of P in phosphine

$PH_3$ .
Since, the overall charge on the complex is

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

$0$ .

$x+3(+1)=0$

$x=+3$ .
Similarly,
b) Let

$x$ be the oxidation number of S in sulphur dioxide

$SO_2$ .

$x+2(-2)=0$

$x=+4$ .
c) Let

$x$ be the oxidation number of N in nitric acid

$HNO_3$ .

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

$x=+5$ .
Let

$x$ be the oxidation number of P in phosphoric acid

$H_3PO_4$ .

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

$x=+5$ .

Which of the following reactions do not involve oxidation-reduction?
I.

$2Cs+2{ H }_{ 2 }O\longrightarrow 2CsOH+{ H }_{ 2 }$
II.

$2Cu{ I }_{ 2 }\longrightarrow 2CuI+{ I }_{ 2 }$
III.

$N{ H }_{ 4 }Br+KOH\longrightarrow KBr+N{ H }_{ 3 }+{ H }_{ 2 }O$
IV.

$4KCN+Fe{ \left( CN \right) }_{ 2 }\longrightarrow { K }_{ 4 }\left[ Fe{ \left( CN \right) }_{ 6 } \right]$

0%
I, II
0%
I, III
0%
I, III, IV
0%
III, IV

Explanation

$Cs\longrightarrow { Cs }^{ 1+ }+{ e }^{ - }$ (Oxidation)

${ H }_{ 2 }O+{ e }^{ - }\longrightarrow \overset { \ominus }{ O } H+\frac { 1 }{ 2 } { H }_{ 2 }$ (Reduction)
Hence, it is a redox reaction.
ii.

${ e }^{ - }+{ Cu }^{ 2+ }\longrightarrow { Cu }^{ 1+ }$ (Reduction)

$2{ I }^{ \ominus }\longrightarrow { I }_{ 2 }+2{ e }^{ - }$ (Oxidation)
Hence, it is a redox reaction.
iii. No change in oxidation number.
Hence, the above reaction is not redox.
iv. No change in the oxidation number of either

${ Fe }^{ 2+ }$ or

${ CN }^{ \ominus }$ in both reactant and product occurs.
Hence, it is not a redox reaction.

Which of the following represents a redox reaction?

0%
$NaOH+HCl\longrightarrow NaCl+{ H }_{ 2 }O$
0%
$Ba{ Cl }_{ 2 }+{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow BaS{ O }_{ 4 }+2HCl$
0%
$CuS{ O }_{ 4 }+2{ H }_{ 2 }O\longrightarrow Cu{ \left( OH \right) }_{ 2 }+{ H }_{ 2 }S{ O }_{ 4 }$
0%
$Zn+2HCl\longrightarrow Zn{ Cl }_{ 2 }+{ H }_{ 2 }$

Explanation

$NaOH+HCl\longrightarrow NaCl+{ H }_{ 2 }O$

$Ba{ Cl }_{ 2 }+{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow BaS{ O }_{ 4 }+2HCl$
The above two reactions are neutralisation reactions as they are acid-base reactions.

$CuS{ O }_{ 4 }+2{ H }_{ 2 }O\longrightarrow Cu{ \left( OH \right) }_{ 2 }+{ H }_{ 2 }S{ O }_{ 4 }$
The oxidation state of

$Cu$ is

$+2$ in both reactant and product and that of

$S{O}_{4}^{2-}$ ion does not change. So, it is none.

$Zn+2HCl\longrightarrow Zn{ Cl }_{ 2 }+{ H }_{ 2 }$
It is a redox reaction.
The half cell reactions are given below:

$Zn\longrightarrow Zn^{ 2+ }+2{ e }^{ - }$

$2{ H }^{ \oplus }+2{ e }^{ - }\longrightarrow { H }_{ 2 }$

The oxidation state of chromium in

$Cr{ \left( CO \right)}_{ 6 }$ is :

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

Explanation

Let

$x$ be the oxidation state of

$Cr$ in

$Cr(CO)_6$ .
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,

$x + 6(0) = 0$
or,

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

$Cr$ in

$Cr(CO)_6$ is

$0$ .

The oxidation number of carbon in

$C{H}_{2}{Cl}_{2}$ is

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

Explanation

Let

$x$ be the oxidation state of

$C$ in

$CH_2Cl_2$ .

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,

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

or,

$x = 0$ .

Hence, the oxidation state of

$C$ in

$CH_2Cl_2$ is

$0$ .

So, the correct option is

$A$ .

The oxidation state of

$Fe$ in

$Fe{ \left( CO \right) }_{ 5 }$ is :

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

Explanation

Let

$x$ be the oxidation state of

$Fe$ in

$Fe(CO)_5$ .
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,

$x + 5(0) = 0$
or,

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

$Fe$ in

$Fe(CO)_5$ is

$0$ .

Which of the following is a redox reaction?

0%
${H}_{2}S{O}_{4}$ with $NaOH$
0%
In atmosphere, ${O}_{3}$ from ${O}_{2}$ by lightning
0%
Nitrogen oxides form nitrogen and oxygen by lightning
0%
Evaporation of ${H}_{2}O$

Explanation

$H_2SO_4+2NaOH \rightarrow Na_2SO_4+2H_2O$
This is a neutralisation reaction.

B. In atmosphere, formation of

${O}_{3}$ from

${O}_{2}$ by lightning is the change of an element in one form (Oxidation state = 0) to another form.

C.

$N_2+O_2 \rightarrow N_2O$
It is a redox reaction.

D. Evaporation of

${H}_{2}O$ .
It is just a physical change.

In which of the following pairs is there the greatest difference in the oxidation numbers of the underlined elements?

0%
$\underline { N } { O }_{ 2 }$ and $\underline { { N }_{ 2 } } { O }_{ 4 }$
0%
$\underline { { P }_{ 2 } } { O }_{ 5 }$ and $\underline { { P }_{ 4 } } { O }_{ 10 }$
0%
$\underline { { N }_{ 2 } } O$ and $\underline { N } O$
0%
$\underline { S } { O }_{ 2 }$ and $\underline { S } { O }_{ 3 }$

Explanation

The oxidation number of elements have been mentioned in brackets as shown below:
a.

$N{ O }_{ 2 }\:( x=4)$ and

${ N }_{ 2 }{ O }_{ 4 }\:( x=4)$
Difference in oxidation state of

$N=0$ .
b.

${ P }_{ 2 }{ O }_{ 5 }\:(x=+5)$ and

${ P }_{ 4 }{ O }_{ 10 }\:(x=+5)$
Difference in oxidation state of

$P$ is zero.
c.

${ N }_{ 2 }O\:( x=1)$ and

$NO\:( x=2)$
Difference in oxidation state of

$N=1$ .
d.

$S{ O }_{ 2 }\:( x=4)$ and

$S{ O }_{ 3 } \:( x=6)$
Difference in oxidation state of

$S=2$ .
Hence, the difference in oxidation states of underlined elements is greatest in

$SO_2$ .

The oxidation number of oxygen in

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

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

Explanation

Hint: Sum of the oxidation state of all the elements present in any atom is equal to zero.

Concept used:

1. Total Oxidation number of any atom is always equal to zero, i.e. the sum of the oxidation state of all the elements present in any atom is equal to zero.

2. In

$O F_{2}$ there is one Oxygen molecule and two fluorine molecules.

3. Oxidation state of Fluorine is always

$-1$ .

4. In spite of being Oxygen as an electronegative element, here, the oxidation number of Oxygen in

$O F_{2}$ will be positive, as Fluorine is more electronegative than Oxygen.

STEP-1:

As we know that,

Fluorine is more electronegative than Oxygen,

As, one Fluorine atom have always

$-1$ Oxidation state,

So, due to 2 molecules of Fluorine,

Total oxidation number of Difluoride will be

$=(2 \times-1)=-2$

We Know that,

Total oxidation state of the element is always zero.

Let the oxidation number of Oxygen in

$O F_{2}$ is

${ }^{\text {" }} \mathrm{x}$ .

STEP-2

So,

As per question,

Total Oxidation state of

$O F_{2}$ will be equal to zero.

Hence,

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

$\quad \Rightarrow x-2=0$

$\Rightarrow x=0-(-2)$

$\quad \Rightarrow x=2$

Conclusion:

Hence, in

$O F_{2}$ the oxidation number of Oxygen will be equal to

$+2$ .

The oxidation state of nitrogen is correctly given for :

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

Explanation

(A) oxidation number of

$N$ is 0

$\left[C O\left(N H_{3}\right)_{5} \text { Cl }\right] \text{Cl}_{2}$

we know ,

$Cl$ has -1 state,

$N H_{3}$ is neutral

i.e O charge,

$H^{+}$ has a +1 charge.

so net charge should be 0 .

Let

$x$ be the charge on

$N$ atom

$x \times 1+3\times1=0$

so oxidation state of Nitrogen is -3.

hence

$A$ is wrong

(B)

$\mathrm{NH}_{2} \mathrm{OH}$ oxidation number of

$N$ is -2

we know net charge on

$H^{+}$ is +1 , net charge on

$OH$ is -1

since compound should have neutral charge

Let

$x$ bo charge on N atom

$x \times 1+1\times2-1\times1=0$

$x=-1$ so oxidation state of Nitregen is -1

hence

$B$ is wrong

( C)

$\left(N_{2} H_{5}\right)_{2} SO_{4}$ oxidation number of

$N$ is +2

$[2 \times x+5 \times 1] \times 2- 2\times 1=0$

$\Rightarrow x=-2$

$C$ is wrong

(D)

$M g_{3} N_{2} \quad O \cdot S \Rightarrow-3$

we know oxidation state of

$Mg$ is +2

Let

$x$ bo charge on N atom

$3 \times 2+2 \times x=0$

$x=-3$ so oxidation state of Nitregen is -3

hence

$D$ is correct

Which of the following is not a redox reaction?

0%
$CaC{ O }_{ 3 }\longrightarrow CaO+C{ O }_{ 2 }$
0%
${ O }_{ 2 }+2{ H }_{ 2 }\longrightarrow 2{ H }_{ 2 }O$
0%
$Na+{ H }_{ 2 }O\longrightarrow NaOH+\frac { 1 }{ 2 } { H }_{ 2 }$
0%
$Mn{ Cl }_{ 3 }\longrightarrow Mn{ Cl }_{ 2 }+\frac { 1 }{ 2 } { Cl }_{ 2 }$

Explanation

a. Oxidation states of

$Ca$ and

$C$ are +2 and +4 respectively in both the reactant and product.
Hence, it is not a redox reaction.
b.

$4{ e }^{ - }+{ O }_{ 2 }\longrightarrow 2{ O }^{ 2- }$ (Reduction)

${ H }_{ 2 }\longrightarrow 2{ H }^{ \oplus }+2{ e }^{ - }$ (Oxidation)
Hence, it is a redox reaction.
c.

$Na\longrightarrow { Na }^{ \oplus }+{ e }^{ - }$ (Oxidation)

$2{ H }^{ \oplus }+2{ e }^{ - }\longrightarrow { H }_{ 2 }$ (Reduction)
Hence, it is a redox reaction.
d.

${ e }^{ - }+{ Mn }^{ 3+ }\longrightarrow { Mn }^{ 2+ }$ (Reduction)

$2{ Cl }^{ \ominus }\longrightarrow { Cl }_{ 2 }+2{ e }^{ - }$ (Oxidation)
Hence, it is a redox reaction.

Which of the following statement is not correct?

0%
The oxidation number of $S$ in ${ \left( N{ H }_{ 4 } \right) }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$ is $+6$
0%
The oxidation number of $Os$ in $Os{ O }_{ 4 }$ is $+8$
0%
The oxidation number of $S$ in ${ H }_{ 2 }S{ O }_{ 5 }$ is $+8$
0%
The oxidation number of $O$ in $K{ O }_{ 2 }$ is $-\frac { 1 }{ 2 }$

Explanation

${ \left( \overset { +1 }{ { NH }_{ 4 } } \right) }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }^{ 2- }$

$\therefore$ Oxidation state of

$S = +6$ . Since,

${ S }_{ 2 }{ O }_{ 8 }^{ 2- }$ has one peroxide bond.
B. Oxidation state of

$Os$ in

$OsO_4$ is

$+8$ .
C. Oxidation state of

$S$ in

${ H }_{ 2 }S{ O }_{ 5 }=+6$ (Since it has one peroxide bond).
D.

${ K }^{ 1+ }{ O }_{ 2 }^{ 1- }$ , oxidation state of

$O= - \dfrac{ 1 }{ 2 }$ .
Hence, only option C is incorrect.

The oxidation number of Phosphorus in

$Mg_2P_2O_7$ is:

0%
+ 3
0%
+ 2
0%
+ 5
0%
- 3

The oxidation number of

$Pr$ in

${Pr}_{6}{O}_{11}$ is :

0%
$\displaystyle\frac{22}{6}$
0%
$\displaystyle\frac{20}{6}$
0%
$3$
0%
$4$

Explanation

Let

$x$ be the
oxidation state of

$Pr$ in

$Pr_6O_{11}$ .
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,

$$6x + 11(-2) = 0$$

or,

$x = + \dfrac{22}{6}$
Hence,
the oxidation state of

$Pr$ in

$Pr_6O_{11}$ is

$+\dfrac{22}{6}$ .

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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