MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 2

In the reaction between acidified

$KMnO_{4}$ and hot oxalic acid, the species that gains electrons is:

0%
$K^{+}$
0%
$MnO_{4}^{-}$
0%
$C_{2}O_{4}^{2-}$
0%
$CO_{2}$

Explanation

The reaction between potassium permanganate and oxalic acid is a redox reaction.

During the reaction, potassium permanganate (which is a strong oxidizing agent) is reduced and oxalic acid is oxidized.

Thus,

$MnO_4^-$ ion gains electrons.

Option B is correct.

$H_{2}O_{2}+O_{3}\rightarrow H_{2}O+2O_{2}$ , in this reaction

0%
$H_{2} O_{2}$ is bleached
0%
$H_{2} O_{2}$ is oxidised
0%
$H_{2} O_{2}$ is dehydrated
0%
$H_{2} O_{2}$ is neither oxidised nor reduced

Explanation

Oxidation half reactions is,

$H_{2}O_{2}\rightarrow 2H^{+}+O_{2}+2e^{-}$
There is a loss of electrons. Therefore,

$H_2O_2$ is oxidized.

Which of the following statement is wrong?

0%
Oxidation number of oxygen is $+1$ in peroxides.
0%
Oxidation number of oxygen is $+2$ in oxygen difluoride.
0%
Oxidation number of oxygen is $-\frac{1}{2}$ in super oxides.
0%
Oxidation number of oxygen is $-2$ in most of its compounds.

Explanation

In peroxides (

$H_2O_2$ ), oxygen has an oxidation state of

$-1$ .

$OF_2$ , oxidation state of oxygen is equal to

$+2$ .

$KO_2$ , oxidation state of oxygen is equal to

$-\cfrac {1} {2}$

Oxidation number of oxygen is

$-2$ in most of its compounds.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

$1$ mole of

$N_{2}H_{4}$ lost

$10$ moles of electrons and formed a new compound,

$Y$ , in which all hydrogens are present without any change in the oxidation state. The oxidation number of nitrogen in the new compound,

$Y$ , is:

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

Explanation

The oxidation state of two nitrogen in hydrazine is

$-4$ as four hydrogen atoms have

$=1$ each.

Therefore,

$1$ nitrogen has oxidation state of

$-2$ .

When hydrazine containing two nitrogen atoms loses

$10$ electrons, the oxidation state of two nitrogen changes from

$-4$ to

$+6$ .

Therefore, the oxidation state changes from

$-2 +5 = +3$ per nitrogen atom.

Option C is correct.

Which is not a redox reaction?

0%
$CaCO_{3}\rightarrow CaO+CO_{2}$
0%
$2H_{2}+O_{2}\rightarrow 2H_{2}O$
0%
$Na+H_{2}O\rightarrow NaOH+ \frac{1}{2}H_{2}$
0%
$MnCl_{3}\rightarrow MnCl_{2}+ \frac{1}{2}Cl_{2}$

Explanation

$CaCO_{3}\rightarrow CaO+CO_{2}$

Since, the reaction does not involve oxidation and reduction, it is not a redox reaction.

In the decomposition of

$CaCO_{3}$ , there occurs no change in the oxidation state of elements.

Which of the following is a redox reaction?

0%
$NaCl+KNO_{3}\rightarrow NaNO_{3}+KCl$
0%
$CaC_{2}O_{4}+2HCl\rightarrow CaCl_{2}+H_{2}C_{2}O_{4}$
0%
$Mg(OH)_{2}+2NH_{4}Cl\rightarrow MgCl_{2}+2NH_{4}OH$
0%
$Zn+2AgCN\rightarrow 2Ag+Zn(CN)_{2}$

Explanation

$Zn^{0}+2\mathrm{A}g^{+1}CN\rightarrow 2\mathrm{A}g^{0}+Zn^{+2}(CN)_{2}$

Oxidation of Zn while reduction of Ag occurs.

Hence, it is a redox reaction.

Iron shows least oxidation state in which of the following?

0%
$K_{4}[Fe(CN)_{6}]$
0%
$K_{3}[Fe(CN)_{6}]$
0%
$FeSO_{4}(NH_{4})_{2}SO_{4}(6H_{2}O)$
0%
$Fe_{2}(CO)_{9}$

Explanation

The oxidation states of iron in

$K_4[Fe(CN)_6], \ K_3[Fe(CN)_6], \ FeSO_4(NH_4)_2SO_4(6H_2O) \ and \ Fe_2(CO)_9$ are

$+2, \:+3,\: +2$ and

$0$ respectively.
Hence, the least oxidation state of Fe is in

$Fe_2(CO)_9$ .

Oxidation state of

$N$ is given correct in :

0%
$0$ in $[Co(NH_{3})_{5}Cl]Cl$
0%
$-1$ in $NH_{4}OH$
0%
$-3$ in $(N_{2}H_{5})_{2}SO_{4}$
0%
$-3$ in $Mg_{3}N_{2}$

Explanation

The oxidation numbers of nitrogen in

$[Co(NH_{3})_{5}Cl]Cl, NH_{4}OH, (N_{2}H_{5})_{2}SO_{4}$ and

$Mg_{3}N_{2}$ are

$-3, \:-3,\: -2$ and

$-3$ respectively.
Thus, the oxidation state of

$N$ is correctly given in

$Mg_{3}N_{2}$ .

In which one of the following compounds does oxygen have an oxidation number of

$+2$ ?

0%
$F_{2}O$
0%
$Cl_{2}O$
0%
$Na_{2}O_{2}$
0%
$Na_{2}O$

Explanation

Flourine is more electronegative than oxygen which in turn is more electronegative than Cl and Na.
The oxidation number of oxygen in

$F_2O, Cl_2O, Na_2O_2$ and

$Na_2O$ are

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

$-2$ respectively.

In which of the following compounds, is the oxidation number of sulphur is the least?

0%
$SO_{2}$
0%
$SO_{3}$
0%
$Na_{2}S_{4}O_{8}$
0%
$H_{2}SO_{4}$

Explanation

The oxidation number of sulphur in

$SO_2, \: SO_3,\: Na_2SO_4$ and

$H_2SO_4$ are

$+4,\: +6,\: +2.5$ and

$+6$ respectively. Hence, it is least in

$Na_2SO_4$ .

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

Explanation

Sulphur in

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

$S_{2}O_{6}^{-2}$ has

$+4$ ,

$+3$ and

$+5$ oxidation state respectively.
Hence, the order is

$S_{2}O_{4}^{2-}$ <

$SO_{3}^{2-}$ <

$S_{2}O_{6}^{2-}$ .

Which of the following does not involve the change of valency?

0%
$SO_{2}+2H_{2}S\rightarrow 2H_{2}O+2S$
0%
$2Na+O_{2}\rightarrow Na_{2}O_{2}$
0%
$Na_{2}O_{2}+H_{2}SO_{4}\rightarrow Na_{2}SO_{4}+H_{2}O_{2}$
0%
$Zn+CuSO_{4}\rightarrow ZnSO_{4}+Cu$

The oxidation state of the most electronegative atom in each of the product in the following reaction is

$H_{2}+O_{2}\rightarrow H_{2}O_{2}+H_{2}O$

0%
$-2,\: -2$
0%
$+1,\:-2$
0%
$+2,\:-1$
0%
$-1,\:-2$

Explanation

In hydrogen peroxide, the oxidation number of O is

$-1$ as it is a peroxide (O-O) while in water it is

$-2$ as it is an oxide (O

$^{2-}$ ).

Oxidation state of central atom in

$Ni(CN)_{4}^{2-}$ is :

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

Explanation

Let

$x$ be the oxidation state of nickel in

$Ni(CN)^{-2}_4$ .
The oxidation state of cyanide group is

$-1$ .
The sum of the oxidation states is equal to

$-2$ which is the charge on the ion.
Therefore

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

$x=+2$
Hence, the oxidation state is

$+2$ .

What is the oxidation number of chlorine in

$ClO_{3}^{-}$ ?

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

Explanation

Let

$x$ be the oxidation number of chlorine in

$ClO_3^-$ .
The oxidation number of oxygen is

$-2$ .
The sum of the oxidation numbers of chlorine and oxygen is

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

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

$x=+5$ .

The oxidation number of sulphur in

$S_8$ ,

$S_2F_2$ ,

$H_2S$ respectively are:

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

The number of moles of

$K_{2}Cr_{2}O_{7}$ that will be needed to react with one mole of

$H_{2}S$ in acidic medium is :

0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{5}$
0%
$\dfrac{2}{3}$

Explanation

The balanced chemical equation for the reaction between potassium dichromate and hydrogen sulphide is as given below:

$K_2Cr_2O_7 +4H_2SO_4+3H_2S \rightarrow K_2SO_4+Cr_2(SO_4)_3+7H_2O+3S$

$1$ mol of potassium dichromate reacts with

$3$ moles of hydrogen sulphide.

Hence, the number of moles of

$K_2Cr_2O_7$ that will be needed to react one mole of

$H_2S$ in acidic medium is

$\dfrac{1}{3}$ .

Option B is correct.

Oxidation number of

$P$ in

$PO^{3-}_4$ , of

$S$ in

$SO^{2-}_4$ and that of

$Cr$ in

$Cr_2O^{2-}_7$ are respectively:

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

Explanation

$PO_4^{3-}$ , Oxidation No. of

$P= +5$

$SO_4^{2-}$ , Oxidation No. of

$S=+6$

$Cr_2O_7^{2-}$ , Oxidation No. of

$Cr =+6$

Oxidation states of P in

$H_{4}P_{2}O_{5},H_{4}P_{2}O_{6},H_{4}P_{2}O_{7},$ are respectively:

0%
+3, +5, + 4
0%
+5, +3, + 4
0%
+5, +4, + 3
0%
+3, +4, + 5

Explanation

$H_4P_2O_5$

$H_4P_2O_6$

$H_4P_2O_7$

$4+2x -10=0$

$4+2x -12=0$

$4+2x -14=0$

$2x=6$

$2x=8$

$2x=10$

$x=+3$

$x=+4$

$x=+5$

Hence, the correct option is

$C$

Oxidation state of Nitrogen in Nitrogen dioxide is:

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

Explanation

$_2$ (Nitrogen Dioxide)

x-4=0

x=+4

The oxidation state of Nitrogen in NO

$_2$ is (+4)

The stability of

$+1$ oxidation state increases in the sequence:

0%
$Ga < In < Al < Tl$
0%
$Al < Ga < In < Tl$
0%
$Tl < In < Ga < Al$
0%
$In < Tl < Ga < Al$

Explanation

Hint: As we go down the group the common oxidation state decrease by $2$ due to the inert pair effect.

Explanation:

As we moving down the group, due to the poor shielding of

$d,f$ orbitals leads to the increase in effective nuclear charge on the

$s$ electrons in the outermost shell.

Therefore these two

$s$ electrons don't participate as often, and hence the stability of the

$n-2$ oxidation state decreases on moving down the group.

Since the order of moving down the group is

$Al \rightarrow Ga \rightarrow In \rightarrow Tl$ .

The increasing order of atomic size is as;

$Tl > In > Ga > Al$

The stability of

$+1$ oxidation state will be the reverse of this order,

i.e.

$Al<Ga<In<Tl$ .

Correct Option: $B$

Carbon is in lowest oxidation state in :

0%
$HCOOH$
0%
$HCHO$
0%
$CF_4$
0%
$CO_2$

Among the following, the number of elements showing only one non-zero oxidation state is O, Cl, F, N, P, Sn, Tl, Na, Ti

0%
2
0%
8
0%
5
0%
9

Among the following, the species in which the oxidation number of an element is

$+6$ is :

0%
$MnO_4^-$
0%
$Cr(CN)_6^{3-}$
0%
$NiF_6^{2-}$
0%
$CrO_2Cl_2$

Explanation

The oxidation number of manganese in

$MnO_4^-$ is

$7$ .
The oxidation number of chromium in

$Cr(CN)_6^{3-}$ is

$+3$ .
The oxidation number of nickel in

$NiF_6^{2-}$ is

$+4$ .
The oxidation number of chromium in

$CrO_2Cl_2$ is

$+6$ .
Hence, the species in which an element has oxidation number

$+6$ is

$CrO_2Cl_2$ .

For the redox reaction:

$MnO_4^-+C_2O_4^{2-}+H^+\rightarrow Mn^{2+}+CO_2+H_2O$
The correct coefficients of the reactants

$MnO_4^-, C_2O_4^{2-}$ and

$H^+$ for the balanced reaction are ____, ____ and ____ respectively.

0%
$2,\: 5,\: 16$
0%
$16,\: 5,\: 2$
0%
$5,\: 16,\: 2$
0%
$2,\: 16,\: 5$

Metals exhibit ..... oxidation states in their compounds.

0%
Positive
0%
Negative
0%
Neutral
0%
none of these

Explanation

As metals have great tendency in donating electrons it will have positive oxidation states.

For example Na is metal has atomic number 11

So electronic configuration is 2,8,1

So after donating one electron its octet get completed and become stable and becomes

$Na^+$

Option A is correct.

The number of moles of

$KMnO_4$ that will be needed to react completely with one mole of ferrous oxalate in acidic medium is:

0%
$\dfrac {2}{5}$
0%
$\dfrac {3}{5}$
0%
$\dfrac {4}{5}$
0%
$1$

Explanation

The balanced chemical reaction is as shown below:

$3MnO_4^-+5FeC_2O_4+24H^+\rightarrow 3Mn^{2+}+5Fe^{3+}+10CO_2+12H_2O$

$3$ moles of

$KMnO_4$ react with

$5$ moles of

$FeC_2O_4$ .

Hence, the number of moles of

$KMnO_4$ that will be needed to react completely with one mole of ferrous oxalate in acidic medium is

$\dfrac {3}{5}$ .

Hence, the correct option is

$B$

The number of moles of

$KMnO_4$ that will be needed to react with one mole of sulphite ion in acidic solution is :

0%
$\dfrac {2}{5}$
0%
$\dfrac {3}{5}$
0%
$\dfrac {4}{5}$
0%
$1$

Explanation

The balanced chemical reaction is as given below,

$2MnO_4^-+5SO_3^{2-}+6H^+\rightarrow 2Mn^{2+}+5SO_4^{2-}+3H_2O$

$5$ moles of

$SO_3^{2-}$ reacts with

$2$ moles of

$KMnO_4$ .

Hence,

$1$ mole of

$SO_3^{2-}$ will react with

$\dfrac{2}{5}$ moles of

$KMnO_4$ .

The oxidation number of manganese in

$KMnO_4$ is +7.

Is the above statement true?

0%
True
0%
False

Explanation

$KMnO_4=1+x+(-8)=0$

$\Rightarrow x=8-1$

$\Rightarrow x=+7$ .

The oxidation state of

$Mn$ in

$K_2MnO_4$ is same as that in

$MnO_2$ .

Is the above statement true?

0%
True
0%
False

Explanation

$K_2MnO_4$ = Oxidation number of

$Mn$ :

$2+x+(-8)=0$

$\Rightarrow x=8-2$

$\Rightarrow x=+6$ .

$MnO_2$ =Oxidation number of

$Mn: x+(-4)=0$

$\Rightarrow x=+4$ .

The oxidation number of

$Ni$ in

$K_4[Ni(CN)_4]$ is :

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

Explanation

Let the oxidation state of

$Ni$ be

$x$ in

$K_4[Ni(CN)_4]$ .
Since the overall charge on the complex is

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

$0$ .
Therefore,

$4+x-4=0$
or,

$x=0$

What are the types of the following reactions?

$3CrO_3^-+2 H^+ \rightarrow 2 Cr O_4^{-2} + Cr^{3+}+H_2O$

0%
Disproportionation
0%
Combustion
0%
Redox
0%
Double displacement

Explanation

Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives off heat. The original substance is called the fuel, and the source of oxygen is called the oxidizer. The fuel can be a solid, liquid, or gas, although for airplane propulsion the fuel is usually a liquid

Redox reactions include all chemical reactions in which atoms have their oxidation state changed; in general, redox reactions involve the transfer of electrons between species.

Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atoms or ions.

Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.

Double displacement reactions may be defined as the chemical reactions in which one component each of both the reacting molecules is exchanged to form the products. During this reaction, the cations and anions of two different compounds switch places, forming two entirely different compounds.

Disproportionation reaction is the reaction in which a substance is simultaneously oxidized and reduced, giving two different products.

${CrO_3}^{-}(V)$ is oxidised to

$CrO^{2-}_4(VI)$ and reduced to

$Cr^{3+}(III)$ .
So in the above reaction both oxidation and reduction was done, so it is redox and Disproportionation reaction.

Hence options B & D are correct.

Which of the following is a spectator ion in the reaction?

$2Na+2HCl \rightarrow 2NaCl + H_2$

0%
$Na^+$
0%
$Cl^-$
0%
$H^+$
0%
None of these

Explanation

Spectator ions are the ions which donot take part in the reaction.

$2Na+2HCl \rightarrow 2NaCl + H_2$
0 +1 -1 +1 -1 0
So, ionic reaction is,

$2Na +2H^+ \rightarrow 2Na^+ + H_2$
Only for chloride ions, oxidation state doesnt change, so they are spectator here.

Which process is not involved in the reaction?

$KO_2+H_2O+CO_2\rightarrow KHCO_3+O_2$

0%
Hydrolysis
0%
Auto-redox
0%
Acid-base reaction
0%
Non redox change

Explanation

$KO_2+H_2O+CO_2\rightarrow KHCO_3+O_2$

$-\frac{1}{2}$

$-2$

$0$
So, its a auto-redox reaction.
Also, its acid-base and hydrolysis reaction.
However, non redox change doesn't happen here.

In the reaction

$2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2 NaI$ , the oxidation state of sulphur is :

0%
decreased
0%
increased
0%
unchanged
0%
none of these

Explanation

Given the reaction:

$2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2 NaI$
+2 +2.5
We can see that the oxidation state of S increases.

When

$SO_2$ is passed in a solution of potassium iodate, the oxidation state of iodine changes from :

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

Explanation

The reaction between

$SO_2$ and

$KIO_3$ is given below:

$5SO_2 + 4H_2O + 2KIO_3 \rightarrow 4H_2SO_4 + I_2 + K_2SO_4$
+4 +5 +6 0
Thus, we can see that oxidation state of iodine changes fron

$+5$ in

$KIO_3$ to

$0$ in

$I_2$ .

${ Fe }^{ 2+ }$ is oxidised by

$Mn{ O }_{ 4 }^{ - }$ or

${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$ in acidic medium. What is new oxidation number of Fe?

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

Explanation

$5{ Fe }^{ 2 }+\underset { pinkish\\ violet }{ Mn{ O }_{ 4 }^{ - } } +8{ H }^{ + }\longrightarrow \underset { colourless }{ { Mn }^{ 2+ } } +5{ Fe }^{ 3+ }+4{ H }_{ 2 }O\quad$ ;

$\quad 6{ Fe }^{ 2+ }+\underset { orange }{ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- } } +14{ H }^{ + }\longrightarrow 6{ Fe }^{ 3+ }+2{ Cr }^{ 3+ }+7{ H }_{ 2 }O$
New oxidation state of iron is

$+3$ .

The oxidation number of phosphorus in

Ca(H2PO2)2Ca(H2PO2)2 is:

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

Among the following, identify the species with an atom in

$+4$ oxidation state.

0%
$MnO^-_4$
0%
$Cr(CN)_6^{3-}$
0%
$NiF^{2-}_6$
0%
$CrO_2Cl_2$

Which of the following statements are correct?

0%
Oxidation of a substance is followed by reduction of another.
0%
Reduction of a substance is followed by oxidation of another.
0%
Oxidation and reduction are complementary reactions.
0%
It is not necessary that both oxidation and reduction should take place in the same reaction.

Which is the anhydride of

${HNO}_{3}$ out of

${N}_{2}O,NO,{N}_{2}{O}_{5}$ ?

0%
${N}_{2}{O}_{5}$
0%
$NO$
0%
${N}_{2}O$
0%
${HNO}_{3}$

Explanation

The anhydride of

$HNO_3$ is

${N}_{2}{O}_{5}$ (oxidation number of the element

$(N)$ in oxyacid and its anhydride remains same).

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

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 a total of

$5$ atoms.
Let the oxidation state of

$C$ in

$C_3O_2$ be

$x$ .
The oxidation state of

$C$ is

$3x-4=0$ or,

$x = \frac{4}{3}$ (since the compound is neutral, sum of oxidation states is equated to zero).

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

0%
$HNO_3, NO, NH_4Cl, N_2$
0%
$HNO_3, NO, N_2, NH_4Cl$
0%
$HNO_3, NH_4Cl, NO, N_2$
0%
$NO, HNO_3, NH_4Cl, N_2$

Explanation

The oxidation state of nitrogen in different compounds are as follows:
Compound Oxidation state

$HNO_3$ +5

$NO$ +2

$N_2$ 0

$NH_4Cl$ -3
Hence, the correct ordering according to the decreasing order of the oxidation state of nitrogen is

$HNO_3,\ NO,\ N_2,\ NH_4Cl$ .

Choose the correct option regarding change in oxidation number of the underlined atom in the following conversion.

$2NaOH+\underline { Br_{ 2 } } \rightarrow NaBr+NaBrO+H_{ 2 }O$

0%
Decreases to $-1$ w.r.t $NaBr$ and increases to $-3$ w.r.t $NaBrO$
0%
Decreases to $-1$ w.r.t $NaBr$ and increases to $+2$ w.r.t $NaBrO$
0%
Decreases to $-1$ w.r.t $NaBr$ and increases to $+1$ w.r.t $NaBrO$
0%
Decreases to $-2$ w.r.t $NaBr$ and increases to $+1$ w.r.t $NaBrO$

Explanation

The oxidation number of bromine in

$Br_2, NaBr$ and

$NaBrO$ is

$0,-1$ and

$+1$ respectively.
Hence, during the reaction, the oxidation number of bromine in

$Br_2$ decreases to

$-1$ w.r.t

$NaBr$ and increases to

$+1$ w.r.t

$NaBrO$ .
Thus, the correct option will be C.

Which one of the following 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.
The oxidation number of N in

$(CN)_2, HCN$ and

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

Oxidation number of oxygen in superoxide ion,

${O}_{2}^{-}$ is :

0%
$-2$
0%
$-\cfrac{1}{2}$
0%
$-\cfrac{3}{2}$
0%
none of these

Explanation

Given superoxide ion,

${ O }_{ 2 }^{-}$ .
If oxidation number if oxygen

$=x$
then,

$2x=-1$

$\therefore x=-\cfrac{1}{2}$

Determine the oxidation state of the underlined element in

${ K }_{ 4 }\underline { { P }_{ 2 } } { O }_{ 7 }$ .

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

Explanation

Let

$x$ be the oxidation number of

$P$ in

$K_4P_2O_7$ .
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,

$4(+1) + 2(x) +7(-2)=0$ .

$4+2x-14=0$

$2x =10$

$x = +5$
Hence, the oxidation number of P in

$K_4P_2O_7$ is

$+5$ .

${ K }_{ 3 }Cr{ \left( { O }_{ 2 } \right) }_{ x }$ has peroxy linkage with oxidation number of

$Cr$ as

$+5$ . The value of

$x$ is :

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

Explanation

${ K }_{ 3 }Cr{ \left( { O }_{ 2 } \right) }_{ x }$ has

$Cr{ \left( { O }_{ 2 } \right) }_{ x }^{ 3- }$ ion.
Therefore,

$+5-2x=-3$
or,

$-2x=-8$
or,

$x=4$ .
It is

${ K }_{ 3 }Cr{ \left( { O }_{ 2 } \right) }_{ 4 }$ .

When

${Cl}^{-}$ is converted into

${Cl}_{2}$ , change in oxidation number is :

0%
$3$ units
0%
$1$ units
0%
$2$ units
0%
none of these

Explanation

The reaction is

$\underset { -1 }{ { Cl }^{ - } } \longrightarrow \cfrac { 1 }{ 2 } \underset { 0 }{ { Cl }_{ 2 } }$
Change in oxidation number

$=1$ unit.

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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