MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 1

The oxidation number of sulphur is -4 in

0%
${{\text{H}}_{\text{2}}}{\text{S}}\,\,\,$
0%
$\,{\text{C}}{{\text{S}}_{\text{2}}}\,\,\,$
0%
$\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\,$
0%
${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}$

What are the oxidation numbers of '

$N$ ' in

${ NH }_{ 4 }N{ O }_{ 3 }$ ?

0%
+3 , -5
0%
-3, +5
0%
+3, +6
0%
-2, +2

Is the compounds is according to decreasing order of the oxidation state of nitrogen?

$HNO_3$ >

$NO$ >

$N_2$ >

$NH_4Cl$

0%
True
0%
False

In the reaction :

$xBrO_{3}^{-}+yCr^{3+}+zH_{2}O\rightarrow Br_{2}+CrO_{4}^{2-}+H^{+}$
The coefficient

$x,\ y$ and

$z$ are:

0%
$6,10,11$
0%
$6,10,20$
0%
$6,8,22$
0%
$6,10,22$

$CrO^{2-}_4 \overset{pH=x}{\rightarrow}Cr_2O^{2-}_7 \overset{pH=y}{\rightarrow}CrO^{2-}_4$ .

x and y can be

0%
3 and 5
0%
3 and 8
0%
8 and 3
0%
8 and 11

Oxygen exhibit

$-1$ oxidation state in:

0%
$OF_{2}$
0%
$H_{2}O_{2}$
0%
$HC1O$
0%
$H_{2}O$

Explanation

${ O }^{ -1 }$ exist where there is peroxide linkage. Hence

${ H }_{ 2 }O_{ 2 }$ (Hydrogen peroxide).

$H-O-O-H$

Which is not redox change ?

0%
$CaCO_{3} \rightarrow CaO+CO_{2}$
0%
$2H_{2}+O_{2} \rightarrow 2H_{2}O$
0%
$Na+H_{2}O \rightarrow NaOH+\dfrac{1}{2} H_{2}$
0%
$MnCI_{3} \rightarrow MnCI_{2}+\dfrac{1}{2} CI_{2}$

Explanation

In the Reaction

$CaC{O}_{3} \rightarrow CaO + {C}{O}_{2}$ , there is no change in the oxidation states of Ca , O and C.

No Redox change has happened

(i)

$Br_2+2NaOH[cold] \rightarrow NaBr+(X)+H_2O$
(ii)

$2Br_2+6NaOH[hot]\rightarrow NaBr+(Y)+H_2O$

What is the oxidation number of

$Br$ in

$(X)$ and

$(Y)$ respectively?

0%
$+1,+7$
0%
$+1,+5$
0%
$+4,+5$
0%
None of these

Identify the element that can have only two oxidation states.

0%
Hydrogen
0%
Fluorine
0%
Carbon
0%
Chromium

The highest oxidation state of Mn is shown by

0%
${{\rm{K}}_{\rm{2}}}{\rm{Mn}}{{\rm{O}}_4}$
0%
${\rm{KMn}}{{\rm{O}}_4}$
0%
${\rm{Mn}}{{\rm{O}}_2}$
0%
${\rm{M}}{{\rm{n}}_2}{{\rm{O}}_3}$

Which of the following has lowest oxidation state?

0%
${{\text{K}}_{\text{2}}}{\text{Fe}}{{\text{O}}_{\text{4}}}$
0%
${{\text{K}}_{\text{4}}}{\text{Fe}}{\left( {{\text{CN}}} \right)_{\text{6}}}$
0%
${\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
0%
${\text{Fe}}{\left( {{\text{CO}}} \right)_{\text{5}}}$

The oxidation state of cobalt in

0%
2
0%
3
0%
4
0%
6

The oxidation state of tungsten in

$N{a_2}{W_4}{O_{13}}.10{H_2}O$ is

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

The oxidation state of the underlined element in the given compound is:

$\underline{Ba}Cl_2$

0%
+2
0%
-2
0%
0
0%
none of these

Explanation

$BaCl_2$

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

$\Rightarrow x=2$

As chlorine needs only one electron to get octet.

The correct statement for the molecule

$CsI_{3}$ is that:

0%
it contains $Cs^{3+}$ and $I^{-}$ ions
0%
it contains $Cs^{+}, I^{-}$ and lattice $I_{2}$ molecule
0%
it is a covalent molecule
0%
it contains $Cs^{+}$ and $I_{3}{^-}$ ions

Explanation

$CsI_{3} \rightarrow Cs^{+}+{I_3}^{-}$

$\Rightarrow Cs$ cannot show

$+3$ oxidation state.

$\Rightarrow I_{2}$ molecules are too large to be accommodated in the lattice.

Hence, the correct option is

$D$

The pair of compounds having metals in their highest oxidation state is:

0%
$[Fe(CN)_6]^{3-}$ and $[Cu(CN)_4]^{2-}$
0%
$MnO_2$ and $CrO_2Cl_2$
0%
$[FeCl_4]^-$ and $Co_2O_3$
0%
$[NiCl_4]^{2-}$ and $[CoCl_4]^{2-}$

The correct order of acid strength is.

0%
$HClO < HClO_2 < HClO_3 < HClO_4$
0%
$HClO_4 < HClO < HClO_2 < HClO_3$
0%
$HClO_2 < HClO_3 < HClO_4 < HClO$
0%
$HClO_4 < HClO_3 < HClO_2 < HClO$

Explanation

$Explanation:$

The acid strength is directly proportional to the oxidation state of chlroine. The oxidation state of Chlorine in given acids are:

$Compound$	$Oxidation\ of\ Cl$
$HClO$	$+1$
$HClO_2$	$+3$
$HClO_3$	$+5$
$HClO_4$	$+7$

Therefore, the correct order of acidic strength above compounds are:

$HClO<HClO_2<HClO_3<HClO_4$

$Final\ Answer:$ Option

$A$ .

Identify Z in the reaction given below:

$2Mn{ O}^-_4+5H_2O_2+6H^+\rightarrow 2Z+5O_2+8H_2O$

0%
$Mn^{2+}$
0%
$Mn^{4+}$
0%
$Mn$
0%
$MnO_2$

Explanation

$\displaystyle 2Mn{ O }^-_4+5H_2O_2+6H^+\rightarrow 2Mn^{2+}+5O_2+8H_2O$

Hydrogen peroxide reduces

$Mn(VII)$ to

$Mn(II)$ in acidic medium.

Thus

$Z$ is

$Mn^{2+}$ .

Hence, option A is correct.

The characteristic oxidation number of atoms in free metals is:

0%
$-1$
0%
$+1$
0%
any number
0%
0

The brown ring complex compound is formulated as

$[Fe(H_2O)_5(NO)]SO_4$ . The oxidation state of iron is:

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

Explanation

When brown ring complex

$[Fe{ ({ H }_{ 2 }O) }_{ 5 }(NO)]{ SO }_{ 4 }$ dissolves in water, it dissociates to give

${ [Fe{ ({ H }_{ 2 }O) }_{ 5 }(NO)] }^{ 2+ }$ and

${ S{ O }_{ 4 } }^{ 2- }$ .

${ [Fe{ ({ H }_{ 2 }O) }_{ 5 }(NO)] }^{ 2+ }$ ,

$NO$ is in +1 oxidation state and

${ H }_{ 2 }O$ is neutral.

So, oxidation state of

$Fe$ is

$x+5(0)+1=2$ i.e, oxidation state of

$Fe$ is

$+1$ .

What is the oxidation number of lithium in

$LiCl$ ?

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

Explanation

Oxidation number of

$Li$ in

$LiCl:x-1=0$

$\Rightarrow x=+1$

What is the oxidation number of

$O$ in a diatomic molecule

$({O}_{2})$ ?

0%
$+2$
0%
$\pm 2$
0%
$+8$
0%
$0$

Oxidation state of iodine in

$I Cl_{2}^{-}$ is :

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

Explanation

Cl is in -1 oxidation state since it has lower atomic number than iodine. Let the oxidation state of iodine be x.

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

$\therefore x=+1$

A redox reaction is one in which:

0%
both the substance are reduced.
0%
both the substance are oxidised.
0%
one substance is reduced and other is oxidised.
0%
all of the above.

Explanation

In a redox reaction both oxidation and reduction is happening together.

Hence, the correct option is $C$ .

The oxidation states of sulphur in the anions

$SO_{3}^{2-}, S_{2}O_{4}^{2-}$ and

$S_{2}O_{6}^{2-}$ follow the order -

0%
$S_{2}O_{4}^{2-} < SO_{3}^{2-} < S_{2}O_{6}^{2-}$
0%
$SO_{3}^{2-} < S_{2}O_{4}^{2-} < S_{2}O_{6}^{2-}$
0%
$S_{2}O_{4}^{2-} < S_{2}O_{6}^{2-} < SO_{3}^{2-}$
0%
$S_{2}O_{6}^{2-} < S_{2}O_{4}^{2-} < SO_{3}^{2-}$

$2Al_2O_3\rightarrow Al_2O_3+2H_2O$ is a balanced chemical equation.

Is the above reaction correct?

0%
True
0%
False

Explanation

$2Al_2O_3\longrightarrow Al_2O_3+2H_2O$

In balanaced equation as:

$Al_2O_3$ : On LHS= 2 moles.

: On RHS= 1 mole.

And hydrogen, oxygen atom on both the side are not balanced.

Oxidation state of Fe in

$Fe_{0.94} O_1$ is?

0%
200
0%
$\dfrac{200}{94}$
0%
$\dfrac{94}{200}$
0%
none of these

The oxidation state of the underlined element in the given compound is:

$\underline{I}F_7$

0%
+5
0%
+7
0%
+6
0%
none of these

Explanation

$IF_7$

Fluorine is the most electronegative element and it has

$-1$ .

$\Rightarrow x-7 \times 1=0$

$\Rightarrow x=7$ .

What is the oxidation number of sulfur in

$H_2S$ ?

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

Explanation

In a neutral compounds, sum of oxidation state of all the atoms present in the compounds is zero.

$H_2S$ molecule, Oxidation number of hydrogen atom is

$+1$ .

Suppose, oxidation number of sulphur atom in

$H_2S$ be

$x$ .

Now,

$2+x=0$

$\Rightarrow x=-2$

Hence, oxidation number of sulfur in

$H_2S$ is

$-2$ .

The brown ring complex compound is formulated as

$[Fe(H_{2}O)_{5}(NO)]SO_{4}$ . The oxidation state of iron in this complex is:

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

$Cu(S) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq)+2Ag(S)$

In the given reaction, the reduction half cell reaction is:

0%
$Cu$ $^{2+}$ $+$ $2e$ $^{-}$ $\rightarrow$ $Cu$
0%
$Cu$ $-$ $2e$ $^{-}$ $\rightarrow$ $Cu$ $^{2+}$
0%
$Ag$ $^{+}$ $+$ e $^{-}$ $\rightarrow$ $Ag$
0%
$Ag$ $-$ $e$ $^{-}$ $\rightarrow$ $Ag$ $^{+}$

Explanation

$Cu$ is getting oxidised and

$Ag^+$ is getting reduced

$\therefore$ The reduction half-cell is

$Ag^+ + e^-\rightarrow Ag$

Hence, the correct option is

$\text{C}$

Oxidation number of

$Fe$ in

$Na_2[Fe(CN)_{5}(NO)].2H_{2}O$ is____________.

0%
$-2$
0%
$+1$
0%
$+3$
0%
$+5$

Explanation

Let

$x$ be the oxidation number of

$Fe$ .
The oxidation numbers of

$Na^+,\ CN^-$ ion and

$NO$ group are

$+1, \:-1$ and

$0$ respectively. The sum of the oxidation numbers is zero as the compound is neutral.
Hence,

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

$x = +3$ .
The oxidation state of iron in

$Na_2[Fe(CN)_5(NO)].2H_2O$ is

$+3$ .

In which of the following, iron is present in its lowest oxidation state?

0%
$FeSO_{4}.7H_{2}O$
0%
$Fe(CO)_{5}$
0%
$Fe_{0.94}O$
0%
$Fe_{3}O_{4}$

Explanation

The oxidation state of iron in iron pentacarbonyl is

$0$ .

In all other compounds, iron has a positive oxidation state.

Thus,

$Fe(CO)_5$ has the lowest oxidation state of iron.

The oxidation state of phosphorus is maximum in_________.

0%
phospine $(PH_{3})$
0%
diphosphine $(P_{2}H_{4})$
0%
metaphosphoric acid $(HPO_{3})$
0%
phosphorus acid $(H_{3}PO_{3})$

Explanation

The oxidation states of phosphorus in phosphine, diphosphine, metaphosphoric acid and phosphorus acid are

$-3,-2,+5$ and

$+3$ respectively. Hence, it is maximum in metaphosphoric acid.

A half cell reaction is one which:

0%
involves only half a mole of electrolyte
0%
goes only half way to completion
0%
takes place at one electrode
0%
consumes half a unit of electricity

Oxidation state of nitrogen is not an integer in:

0%
hydroxyl amine $(NH_{2}OH)$
0%
ammonia $(NH_{3})$
0%
hydrazine $(N_{2}H_{4})$
0%
hydrazoic acid $(N_{3}H)$

Explanation

$N_3H$ , the oxidation state of

$N$ is

$-\dfrac{1}{3}$ . Hence, it is not an integer.

The oxidation state of

$N$ in

$NH_2OH$ ,

$NH_3$ and

$N_2H_4$ are

$-1,\:-3$ and

$-2$ respectively.

The oxidation state of oxygen is maximum in_________.

0%
bleaching powder $(CaOCl_{2})$
0%
oxygen difluoride $(OF_{2})$
0%
dioxygen difluoride $(O_{2}F_{2})$
0%
hydrogen peroxide $(H_{2}O_{2})$

Explanation

The oxidation states of oxygen in bleaching powder, oxygen difluoride, dioxygen difluoride and hydrogen peroxide are

$-2,\: +2,\: +1$ and

$-1$ respectively. Hence, it is maximum in

$OF_2$ .

The oxidation state of the metal atom is zero in:

0%
ferrous sulphate
0%
ferric sulphate
0%
rust
0%
iron carbonyl

The oxidation number of chlorine is maximum in __________.

0%
$HOCl$
0%
$Cl_{2}O_{6}$
0%
$KClO_{4}$
0%
$NaClO_{3}$

Explanation

The oxidation numbers of chlorine in

$HOCl$ ,

$Cl_2O_6, KClO_4$ and

$NaClO_3$ are

$+1,\: +6,\: +7$ and

$+5$ respectively. Hence, the oxidation number of chlorine is maximum in

$KClO_4$ .

Oxidation number of sulphur in marshall's acid

$(H_{2}S_{2}O_{8})$ is

0%
$+5$
0%
$+8$
0%
$+6$
0%
$+7$

Explanation

Let

$x$ be the oxidation number of sulphur in marshall's acid

$(H_2S_2O_8)$ . It contains a peroxide linkage. Thus, the oxidation number of two O atoms is

$-1$ and that of the remaining O atoms is

$-2$ . The oxidation number of H is

$+1$ .
For the neutral molecule, the sum of the oxidation numbers is zero.
Therefore,

$2(+1)+2x+2(-1)+6(-2)=0$ or,

$x=+6$
The structure is given below:

The oxidation number of manganese in

$KMnO_{4}$ is:

0%
$+3$
0%
$+5$
0%
$+7$
0%
$+9$

Explanation

$KMnO_4$ : Potassium permanganate

Let the oxidation state of

$Mn$ be

$x$ .

Since the complex is neutral, the sum of all oxidation states will be equal to zero.

Hence,

$+1+x -8=0$ or,

$x =+7$

Oxidation number of

$Cr$ in

$KCrO_{4}^-$ is:

0%
$+4$
0%
$+6$
0%
$+2$
0%
$+8$

Explanation

Let x be the oxidation number of Cr in

$\displaystyle KCrO_{4}^-$

$\displaystyle +1+X+4(-2) = -1$

$\displaystyle +1+X-8=-1$

$\displaystyle X-7=-1$

$\displaystyle X=+6$

Thus, the oxidation number of Cr in

$\displaystyle KCrO_{4}^-$ is +6.

Option B is correct.

The oxidation number of manganese in potassium manganate is:

0%
$+7$
0%
$+6$
0%
$+4$
0%
$+2$

Explanation

$K_2MnO_4$ : Potassium manganate.

For 2

$K^{+}$ , the oxidation state is

$+2$ .

For

$MnO_{4}^{ 2-}$ ;

$Mn + 4\times -2 = -2$ or,

$x = +6$ .

therefore,

$Mn$ has the oxidation state of

$+6$ .

Option B is correct.

In which of the following compounds, sulphur atom has different oxidation number?

0%
$H_{2}SO_{4}$
0%
$HSO_{3}^-$
0%
$SO_{2}Cl_{2}$
0%
$SO_{3}$

Explanation

The oxidation state of sulphur in

$HSO_3^-$ is

$+4$ .
In

$H_2SO_4, SO_2Cl_2$ and

$SO_3$ , the oxidation state of sulfur is

$+6$ which is the same.

In calcium hypophosphite, oxidation number of phosphorus is

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

Explanation

The charge on the hypophosphite,

$H_2PO_2^-$ ion is

$-1$ .
The oxidation numbers of H and O are

$+1$ and

$-2$ respectively.
Let

$x$ be the oxidation number of phosphorus, then the sum of the oxidation numbers will be equal to

$-1$ , which is the charge on the ion. Hence,

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

$x= +1$ . So, the oxidation state of phosphorus is +1.

In which of the following substances, sulphur has the lowest oxidation number?

0%
$H_{2}SO_{4}$
0%
$SO_{2}$
0%
$H_{2}SO_{3}$
0%
$H_{2}S$

Explanation

$H_2S$ , the oxidation state of sulfur is

$-2$ .

In other given compounds, sulfur has positive oxidation states.

The oxidation states of sulfur in

$H_2SO_4$ ,

$SO_2$ and

$H_2SO_3$ are

$+6$ ,

$+4$ and

$+4$ respectively.

Option D is correct.

Oxidation number of sulphur in Caros acid

$(H_{2}SO_{5})$ is:

0%
$+5$
0%
$+8$
0%
$+6$
0%
$+7$

Explanation

$H_2SO_5$ , one peroxide bond is present.
So, the oxidation number of

$2$ oxygen atoms is

$-2$ . (Compare hydrogen peroxide,

$H_2O_2$ ,

$2$ atoms of oxygen will get

$-1$ oxidation number each).
The remaining

$3$ atoms will have

$-2$ oxidation number each. This is a special case. Oxidation number of each hydrogen atom is

$+1$ only.
So, finally the oxidation number of sulphur is given by,

$x -2 -6 +2 = 0$ (equal to zero as the compound is neutral)
or,

$x= +6$
The structure of caros acid is as follows:

Oxidation number of carbon in

$SCN^{-}$ ion is

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

Explanation

In thiocyanate ion (

$SCN^-$ ), carbon atom is attached to

$N$ via single bond and

$S$ via triple bond. The oxidation numbers of S and N are

$-2$ and

$-3$ respectively.
Let,

$x$ be the oxidation number of C. The total charge on ion is

$-1$ . Hence,

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

$x=+4$ .
The structure of

$SCN^-$ is given below:

The sum of the oxidation number of the carbon atoms in

$CH_{3} CHO$ is:

0%
$+2$
0%
$-4$
0%
$-1$
0%
$-2$

Explanation

Let

$x$ be the sum of the oxidation states of carbon atoms in acetaldehyde

$(CH_3CHO)$ .

The oxidation numbers of hydrogen and oxygen are

$+1$ and

$-2$ respectively.

Let, $$x$ be the sum of the oxidation state of C.

For a neutral molecule, the sum of the oxidation numbers of all the molecules is zero.

Hence,

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

$x=-2$ .

Hence, the correct option is

$A$

Which of the following change is neither oxidation nor reduction?

0%
$K_{2}Cr_{2}O_{7}$ to $K_{2}CrO_{4}$
0%
$Na_{2}SO_{4}$ to $NaHSO_{4}$
0%
$CuCl_{2}$ to $CuSO_{4}$
0%
All the above

Explanation

$K_{2}Cr_{2}^{+6}O_{7}\rightarrow K_{2}Cr^{+6}O_{4}$

$Na^{1+}$

$\rightarrow$

$Na^{1+}$

$Cu^{2+}$

$\rightarrow$

$Cu^{2+}$
Therefore, all the above are not redox reactions.

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 1 - 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 1 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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