MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 10

Oxidation number of carbon in chloroform

$(CHCl_3)$ is :

0%
$+2$
0%
$\frac{+4}{3}$
0%
$+3$
0%
$+4$

In the following redox equation,

$xUO^{+2}+Cr_2O_7^{2-}\rightarrow aUO_2 ^{2+}+zCR^{+3}+bH_2O$
The values of coefficients x, y and z respectively are:

0%
3,8,2
0%
3,8,7
0%
3,2,4
0%
None of the above

Explanation

The balanced chemical equation may be written as

$3UO^{2+}+{Cr_2O_7}^{2-}+8H^+\rightarrow 3{UO_2}^{2+}+2Cr^{+3}+4H_2O$

On comparison,

$x=3$ ,

$y=8$ ,

$z=2$

Hence, A is correct

Oxidation number of osmium (Os) in

$OsO_4$ is

0%
+8
0%
+6
0%
+7
0%
+4

In which of the following reactions, there is no change valency ?

0%
${ 4KClO }_{ 3 }\rightarrow { 3KClO }_{ 4 }+KCl$
0%
${ SO }_{ 2 }+{ 2H }_{ 2 }S\rightarrow { 2H }_{ 2 }O+3S$
0%
${ BaO }_{ 2 }+{ H }_{ 2 }{ SO }_{ 4 }\rightarrow { BaSO }_{ 4 }+H_{2}O_{2}$
0%
$2BaO +{ O }_{ 2 }\rightarrow { 2BaO }_{ 2 }$

In the reaction,

$Fe_2O_3 + 3C \rightarrow 2Fe + 3CO$ , oxidation number of

$C$ :

0%
increases by 3 per atom
0%
decreases by 3 per atom
0%
increases by 2 per atom
0%
no change

Explanation

Initial oxidation number of

$C=0$

Final oxidation number of

$C$ in

$CO=+2$

Hence, the oxidation number of

$C$ increases by

$2$ per atom

To a 25 ml

$H_2O_2$ solution excess acidified solution of

$KI$ was added. The iodine liberated required 20 ml of 0.3 N sodium thiosulphate solution. Use these data to choose the correct statement from the following

0%
The weight of $H_2O_2$ present in 25 ml solution is 0.102 g.
0%
The molarity of $H_2O_2$ solution is 0.12 M.
0%
The weight of $H_2O_2$ present in 1 L of the solution is 0.816 g.
0%
The volume strength of $H_2O_2$ is 1.344 L.

Oxidation number of

$Cl$ in

$CaOC{l_2}$ (bleaching powder is)

0%
Zero , since it contains $C{l_2}$
0%
-1 , since it contains $C{l^ - }$
0%
+1 , since it contains $Cl{O^ - }$
0%
+1 and -1 since it contains $Cl{O^ - }$ and $C{l^ - }$

In the reaction

$MnO_{4}^{-}+SO_{3}^{-2}+H^{+}\longrightarrow SO_{4}^{2+}+Mn^{2+}+H_{2}O$

0%
$MnO_{4}^{-}$ and $H^{+}$ both are reduced.
0%
$MnO_{4}^{-}$ is reduced and $H^{+}$ is oxidised.
0%
$MnO_{4}^{-}$ is reduced and $SO_{3}^{2-}$ is oxidised.
0%
$MnO_{4}^{-}$ is oxidised and $SO_{3}^{2-}$ is reduced.

In the fluorite structure, the oxidation number of

$Ca$ is:

0%
+2
0%
+6
0%
+8
0%
+3

The oxidation number of oxygen in

$O_2F_2$ is:

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

If the following half cells have the

$E^o$ values as

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

$E^o= +0.77V$ and

${Fe}^{+2}+2 {e}^{-} \longrightarrow {Fe}$ ;

$E^o= -0.44 V$ The

$E^o$ of the half cell is

$Fe^{+3} + 3e^- \rightarrow Fe$ will be:

0%
$0.33$ V
0%
$1.21$ V
0%
$0.04$ V
0%
$0.605$ V

Which compound contains two different elements with identical oxidation states?

0%
HClO
0%
$Mg(OH)_2$
0%
$Na_2SO_4$
0%
$NH_4Cl$

Explanation

$HClO$ , the oxidation state of

$H$ and

$Cl$ is same, i.e.,

$+1$ .

In the reaction,

$MnO^{-}_{4(aq)}+Br^{-}_{(aq)}\rightarrow MnO_{2(s)}+BrO^{-}_{3(aq)}$ , the correct change in oxidation number of the species involved is:

0%
$Br^{5+}$ to $Br^{-}$
0%
$Mn^{7+}$ to $Mn^{2+}$
0%
$Mn^{7+}$ to $Mn^{3+}$
0%
$Br^{-}$ to $Br^{5+}$

Explanation

From the oxidation state assigned to each species, we can see clearly that the oxidation state of

$Br$ is changed from

$-1$ to

$+5$ .

Hence, option

$D$ is correct.
1407863_1676498_ans_36d09d8e7095494ea57e58220e4982f2.PNG

Balance the following redox equations in basic medium by oxidation number method :

0%
$\mathrm { Zn } _ { ( \mathrm { s } ) } + \mathrm { NO } _ { 3 ( \mathrm { aq } ) } ^ { - } \longrightarrow \mathrm { NH } _ { 3 ( \mathrm { aq } ) } + \mathrm { Zn } ( \mathrm { OH } ) _ { 6 ( \mathrm { aq } ) } ^ { 2 - }$
0%
$\mathrm { MnCl } _ { 2 ( \mathrm { aq } ) } + \mathrm { HO } _ { 2 ( \mathrm { aq } ) } ^ { - } \longrightarrow \mathrm { Mn } ( \mathrm { OH } ) _ { 3 ( \mathrm { s } ) } + \mathrm { Cl } _ { ( \mathrm { aq } ) } ^ { - }$
0%
$\mathrm { Mn } _ { ( \mathrm { aq } ) } ^ { 2 + } + \mathrm { H } _ { 2 } \mathrm { O } _ { 2 ( \mathrm { aq } ) } \longrightarrow \mathrm { MnO } _ { 2 ( \mathrm { s } ) } + \mathrm { H } _ { 2 } \mathrm { O } _ { ( l ) }$
0%
$\mathrm { Cu } ( \mathrm { OH } ) _ { 2 ( \mathrm { s } ) } + \mathrm { N } _ { 2 } \mathrm { H } _ { 4 } ( \mathrm { aq } ) \longrightarrow \mathrm { Cu } _ { ( \mathrm { s } ) } + \mathrm { N } _ { 2 ( \mathrm { g } ) }$

In the compounds

$KMnO_4$ and

$K_2Cr_2O_7$ , the lowest oxidation state is of:

0%
potassium
0%
chromium
0%
oxygen
0%
manganese

The compound of Xe and F is found to have 53.5% Xe. What is oxidation number of Xe in this compound? (Xe = 131)

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

Explanation

Let the mass of given compound is = 100 g

Mass of

$Xe= 53.5g$

Mass of

$F= 100-53.5= 46.5g$

Moles of

$Xe=\dfrac{53.5}{131}= 0.41$ moles

$\Rightarrow \dfrac{0.41}{0.41}=1$

Moles of

$F=\dfrac{46.5}{19}= 2.45$ moles

$\Rightarrow \dfrac{2.45}{0.41}=6$

Molecular formula of given compound =

$XeF_{6}$

Oxidation state of

$Xe= +6$

Which reaction is not a redox reaction?

0%
$Mg+2HNO_3\rightarrow Mg(NO_3)_2+H_2$
0%
$2Mg(NO_3)_2\rightarrow 2MgO+4NO_2+O_2$
0%
$SO_2+NO_2\rightarrow SO_3+NO$
0%
$SO_3+H_2O\rightarrow H_2SO_4$

Oxidation number of sodium in sodium amalgam is :

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

Explanation

The oxidation state of sodium in

$(Na)Hg$ is zero because it is a fused state of sodium in mercury which is a free state there in a free state the oxidation state of any element always zero.

The order of the oxidation state of the phosphorus atom in

$H_{3}PO_{2}$ ,

$H_{3}PO_{4}$ ,

$H_{3}PO_{3}$ and

$H_{4}P_{2}O_{6}$

0%
$H_{3}PO_{4} > H_{3}PO_{2}> H_{3}PO_{3}>H_{4}P_{2}O_{6}$
0%
$H_{3}PO_{4} >H_{4}P_{2}O_{6}> H_{3}PO_{3}> H_{3}PO_{2}$
0%
$H_{3}PO_{2}> H_{3}PO_{3}> H_{2}P_{2}O_{6}> H_{3}PO_{4}$
0%
$H_{3}PO_{3}>H_{3}PO_{2} > H_{3}PO_{4}> H_{2}P_{2}O_{6}$

Explanation

Explanation:

Oxidation state of Phosphorus in

$H_3PO_4=(+5),\ H_4P_2O_6=(+4),\ H_3PO_3=(+3),\ H_3PO_2=(+1)$

So option B is correct

The oxidation states of Cr in

$\left [Cr(H_{2}O_{6}) \right ] Cl_{3}$ ,

$\left [ Cr(C_{6}H_{6})_{2} \right ]$ and

$K_{2}\left [ Cr(CN)_{2}(O_{2})(O_{2})(NH_{3}) \right ]$ respectively are

0%
+3,0 and +4
0%
+3,+4 and +6
0%
+3,+2 and +4
0%
+3,+0 and +6

Explanation

The oxidation states of Cr in

$\left [Cr(H_{2}O_{6}) \right ] Cl_{3}$ ,

$\left [ Cr(C_{6}H_{6})_{2} \right ]$ and

$K_{2}\left [ Cr(CN)_{2}(O_{2})(O_{2})(NH_{3}) \right ]$ respectively are +3,+0 and +6.

Hence Option "D" is the correct answer.

Which of the following reaction is an example of redox reactions

0%
$XeF_{4}+O_{2}F_{2}\rightarrow XF_{6} +O_{2}$
0%
$XeF_{2}+PF_{5}\rightarrow \left [ XeF \right ]^{+}PF_{6}^{-}$
0%
$XeF_{6}+H_{2}O\rightarrow XeoF_{4}+2HF$
0%
$XeF_{6}+2H_{2}O\rightarrow Xeo_{2}F_{2}+4HF$

The oxidation numbers of

$S$ in

$S_{8},\ S_{2}F_{2}$ and

$H_{2}S$ respectively are

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

Explanation

Since sulphur exists in elemental form in

$S_8$ , its oxidation number is 0.

The oxidation number of sulphur in

$S_2F_2$ is +1 and the oxidation number of fluorine is -1.

The oxidation number of sulphur in

$H_2S$ is -2 and oxidation number of hydrogen is +1. Option A is correct.

Oxidation number of

$N$ in

${({NH}_{4})}_{2}{SO}_{4}$ is

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

Explanation

${({NH}_{4})}_{2}{SO}_{4}\rightleftharpoons 2{NH}_{4}^{+}+{SO}_{4}^{2-}$

$\overset { \ast }{ N } {H}_{4}^{+}$

Let,

$x$ be the oxidation state of

$N.$

$x+4\times 1=+1;\ \Rightarrow x=-3$

$H{NO}_{3}$ changes into

${N}_{2}O$ , the oxidation number is changed by

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

Explanation

$H{NO}_{3}\rightleftharpoons \overset { \ast }{ {N}_{2} } O$

$1+x-6=0;2x'-2=0$

$x=+5$ ;

$x'=+1$

Hence the oxidation no. is changed by +4.

The oxidation number of

$C$ in

${CO}_{2}$ is

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

Explanation

$\overset { \ast }{ C } { O }_{ 2 }$

Let,

$x$ be the oxidation number of

$C$ in

${CO}_{2}.$

Now,

$x+2\times (-2)=0;\ \Rightarrow x-4=0;\ \Rightarrow x=+4.$

Oxidation number of osmiium

$(Os)$ in

$Os{O}_{4}$ is

0%
$+4$
0%
$+6$
0%
$+7$
0%
$+8$

Explanation

$\overset { \ast }{ O } s{O}_{4}$

$x+4(-2)=0;x=+8$

The oxidation states of the most electronegative element in the products of the reaction of

$Ba{O}_{2}$ with dilute

${H}_{2}{SO}_{4}$ are

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

Explanation

${H}_{2}{O}_{2}$ oxygen shows

$=-1$ (peroxide) oxidation state and in

$Ba{SO}_{4}$ oxygen shows

$=-2$ oxidation state

The oxidation number of

$Fe$ and

$S$ in iron pyrites are

0%
$4$ , $-2$
0%
$2$ , $-1$
0%
$3$ , $-1.5$
0%
$3$ , $-1$

Explanation

$\overset { \ast }{ F }e{S}_{2}$

$Fe\overset { \ast }{ {S}_{2} }$

$x-4=0;4+2x=0$

$x=\cfrac{-4}{2}=-2$

The oxidation number of

$Ba$ in barium peroxide is

0%
$+6$
0%
$+2$
0%
$1$
0%
$+4$

Explanation

Barium peroxide is

$BaO_2.$

The oxidation number of

$Ba$ in barium peroxide is

$+2$ as

$Ba$ is second group element.

${Sn}^{2+}$ loses two electrons in a reaction. What will be the oxidation number of tin after the reaction?

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

Explanation

${Sn}^{2+}\rightarrow {Sn}^{4+}+2{e}^{-}$

${Sn}^{2+}$ loses two electrons in the reaction to form

$Sn^{4+}$ and its oxidation number changes from

$+2$ to

$+4.$

The sum of the oxidation numbers of all the carbons in

${C}_{6}{H}_{5}CHO$ is

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

Explanation

Let,

$x$ be the oxidation numbers of the carbon in

${C}_{6}{H}_{5}CHO.$

Now,

$7\times x+6\times(+1)+(-2)=0;\ \Rightarrow x=-\dfrac{4}{7}$

So, the sum of the oxidation numbers of all the carbons in

${C}_{6}{H}_{5}CHO$ is

$-4.$

In which of the following compounds the oxidation number of carbon is maximum?

0%
$HCHO$
0%
$CH{Cl}_{3}$
0%
${CH}_{3}OH$
0%
${C}_{12}{H}_{22}{O}_{11}$

Explanation

The oxidation state of

$C$ in the following compounds is as follows:

$HCHO\rightarrow{0}$
$CHCl_3\rightarrow{+2}$
$CH_3OH\rightarrow{-2}$
$C_{12}H_{22}O_{11}\rightarrow{0}$

Hence option

$(B)$ is correct.

In which of the following pair of compounds, oxidation number of chromium is same?

0%
$K_2CrO_4$ and $KCrO_2$
0%
$KCrO_2$ and $Cr(CO)_6$
0%
$K_2Cr_2O_7$ and $Cr(CO)_4$
0%
$K_2Cr_2O_7$ and $K_2CrO_4$

Explanation

$KCrO_2$ and

$Cr(CO)_6$ .

Balance the following equation and choose the quantity which is the sum of the coefficients of reactants and products:
.........

$PtCl_4$ + ...........

$XeF_4 \rightarrow PtF_{6}$ + ....

$ClF$ + ....

$Xe$

0%
$16$
0%
$13$
0%
$18$
0%
$12$

Explanation

Balanced reaction is

$PtCl_4 + 5XeF_4 \rightarrow PtF_{6} + 4ClF + 5Xe$

$\therefore$ Sum of stoichiometric coefficients

$= 1 + 5 + 1 + 4 + 5 = 16$

The value of coefficients to balance the following reaction are

$Cr(OH)_3+ClO^-+OH^- \rightarrow CrO_4^{2-}+Cl^-+H_2O$

0%
$Cr(OH)_3\ 2,\ ClO^-\ 3,\ CrO_4^{2-}\ 3,\ Cl^-\ 3$
0%
$Cr(OH)_3\ 2,\ ClO^-\ 4,\ CrO_4^{2-}\ 3,\ Cl^-\ 2$
0%
$Cr(OH)_3\ 2,\ ClO^-\ 4,\ CrO_4^{2-}\ 4,\ Cl^-\ 2$
0%
$Cr(OH)_3\ 2,\ ClO^-\ 3,\ CrO_4^{2-}\ 2,\ Cl^-\ 3$

The oxidation number of

$\mathrm{N}$ in

$\mathrm{NF}_{3}$ is

$+3 .$

0%
True
0%
False

The oxidation number of sulphur in

$\mathrm{S}_{8}, \mathrm{S}_{2} \mathrm{F}_{2}, \mathrm{H}_{2} \mathrm{S}$ respectively, are

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

Explanation

$0,+1$ and -2

Which of the following statements is correct?

0%
Hydrogen has the same ionization potential as alkali metals
0%
$H^-$ has more electronegativity as halogens
0%
$H^-$ has oxidation number of $-1$
0%
$H^-$ will Dot be liberated at the anode.

Oxidation state of Osmium (Os) in

$\mathrm{OsO}_{4}$ is

0%
$+7$
0%
$+6$
0%
$+4$
0%
$+8$

Explanation

$+8$

The reaction,

$3 \mathrm{ClO}^{-}(a q) \rightarrow$

$\mathrm{ClO}_{3}^{-}(a q)+2 \mathrm{Cl}^{-}(a q)$ is an example of

0%
Oxidation reaction
0%
Reduction reaction
0%
Disproportionation reaction
0%
Decomposition reaction.

A metal ion

$M^{3+}$ loses 3 electrons, its oxidation number will be

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

Explanation

$\mathrm{M}^{3+} \rightarrow \mathrm{M}^{6+}+3 e^{-}$

The oxidation number of carbon in

$\mathrm{CH}_{2} \mathrm{Cl}_{2}$ is

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

Oxidation states of the metal in the minerals haematite and magnetite, respectively are:

0%
II, III in haematite and III in magnetite
0%
II, III in haematite and II in magnetite
0%
II in haematite and II, III in magnetite
0%
III in haematite and II, III in magnetite

Explanation

Correct option: $D$

Explanation:

Step: 1 Calculation of oxidation state of Magnetite:
Magnetite is one of the main iron ores found in the earth's crust. Chemical formula is $Fe_3O_4$ . It is the stoichiometric composition of $FeO$ and $Fe_2O_3$ .
For $FeO$ , let oxidation number assumed be $X$

So,

$X+\left ( -2 \right )=0$

$\Rightarrow X=+2$

For $Fe_2O_3$ , let the oxidation number assumed be $X$

So, $2X+3\left ( -2 \right )=0$ $\Rightarrow X=+3$

Step: 2

Hematite is an ore containing oxide of iron with the molecular formula

$Fe_2O_3$ .

For $Fe_2O_3$ , let the oxidation number assumed be $X$

So, $2X+3\left ( -2 \right )=0$ $\Rightarrow X=+3$

Final answer:

In haematite, the oxidation state of

$Fe$ is

$+3$ and in magnetite, the oxidation state of

$Fe$ is

$+2$ and

$+3$ both.

The number of moles of

$KMnO_{4}$ that will be needed to react completely with one mole of ferrour oxalate

$Fe(C_{2}O_{4})$ in acidic solution is :

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

Explanation

The number of moles of

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

$\dfrac{3}{5}$ .
Potassium permangante is a very strong oxidizing agent. It oxidizes iron(II) ion as well as the oxalate ion.
Oxalate oxidation:

$2MnO_4- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2$
Iron(II) oxidation:

$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$

Hence, moles of

$KMnO_{4}$ that will be needed to react completely with one mole of ferrour oxalate

$Fe(C_{2}O_{4})$ is

$\dfrac{3}{5}$ (2 moles for oxalate and 1 mole for iron (II)).

What is the oxidation state of Fe in the product formed when acidified potassium ferrocyanide

$(K_{4}[Fe(CN)_{6}])$ is treated with hydrogen peroxide?

0%
$+2$
0%
$+3$
0%
$+1$
0%
$+6$

Explanation

When acidified potassium ferrocyanide

$(K_4[Fe(CN)_6])$ reacts with hydrogen peroxide, it yields potassium ferricyanide

$(K_3[Fe(CN)_6])$ .

It is a redox reaction, which is described as follows:

$K_4[Fe{(CN)}_6] + H_2SO_4 + H_2O_2 \rightarrow K_3[Fe{(CN)}_6] + K_2SO_4 + H_2O$

In this redox reaction,

${\text 2 Fe(II) - 2 e^-} \rightarrow {\text 2 Fe(III)}$ (oxidation)

${\text 2 O^- (I) + 2 e^-} \rightarrow {\text 2 O^- (II)}$ (reduction)

$K_4[Fe(CN)_6]$ acts as a reducing agent, whereas hydrogen peroxide is a very strong oxidizing agent that oxidizes iron.

The oxidation state of Fe changes from +2 to +3.

One gas bleaches the colour of flowers by reduction while the other gas by oxidation. The two gases are, respectively:

0%
$CO$ & $Cl_{2}$
0%
$H_{2}S$ & $Br_{2}$
0%
$SO_{2}$ & $Cl_{2}$
0%
$NH_{3}$ & $SO_{3}$

Explanation

Sulphur dioxide and chlorine gas are the two gases that bleach the colour of flower by oxidation and reduction.

Chlorine reacts with water and produces a single oxygen atom also called nascent oxygen. This nascent oxygen when combines with any colour make it colourless.

$Cl_{2}+H_{2}O\rightarrow 2HCl+(O)$ (O is oxidizing)

Sulphur dioxide when reacts with coloured substances it releases oxygen from the substance resulting in loss of colour.

$SO_{2}+2H_{2}O\rightarrow H_{2}SO_{4}+2(H)$ (H is reducing)

Hence, the correct option is

$C$

When

$K_{2}Cr_{2}O_{7}$ is mixed with

$H_{2}SO_{4}$ and thoroughly shaken with

$H_{2}O_{2}$ in presence of ether, then a floated blue coloured complex

$X$ is formed. The change in oxidation state and the percentage of

$Cr$ in the complex

$X$ is:

0%
$+6, 49.4$
0%
$+4, 39.4$
0%
$0, 39.4$
0%
$+4, 59.4$

Explanation

When

$K_2Cr_2O_7$ reacts with

$H_2O_2$ , a blue colored compound is formed which is

$CrO_5$ .

The oxidation number of

$Cr$ in

$CrO_5$ is

$+6$ , because four of oxygen atoms are involved in peroxide linkage.

So, change in oxidation state from

$K_2Cr_2O_7$ to

$CrO_5$ is

$0$ .

The percentage of chromium in the complex is

$\dfrac { 52 }{ 132 } \times 100=39.4$ %.

The structure of

$CrO_5$ is shown above.

The oxidation number of

$Cr$ in

$CrO_{5}$ is:

0%
$+10$
0%
$+8$
0%
$+6$
0%
$+4$

Explanation

$CrO_5$ , one oxygen is attached to Cr through the double bond. It has oxidation number

$-2$ . Four oxygen atoms are attached to Cr atom through a single bond. Each of these oxygen has an oxidation number

$-1$ .

Let

$x$ be the oxidation number of chromium. The sum of all the oxidation numbers is zero.

Hence,

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

$x = 6$ .

Thus, the oxidation number of chromium in

$CrO_5$ is

$+6$ .

The structure of

$CrO_5$ is shown above.

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

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

Explanation

The oxidation states of nitrogen are given below:

$HNO_3\rightarrow N^{+5}$

$NO\rightarrow N^{+2}$

$NH_4Cl\rightarrow N^{-3}$

$N_2\rightarrow N^o$

Hence, the descending order of the oxidation state of nitrogen is

$HNO_3, NO, N_2, NH_4Cl$ .

$1$ mol of ferric oxalate is oxidised completely by

$x$ mol of

$MnO_{4}^{-}$ in acidic medium. Then,

$x$ is :

0%
$1.2$
0%
$1.6$
0%
$1.8$
0%
$1.5$

Explanation

The reaction involved is shown below:

$Fe_{2}(C_{2}O_{4})_{3}\rightarrow 2Fe^{3+}+3C_{2}O_{4}^{2-}$

$3 (C^{3+})_{2}\rightarrow 6C^{4+}+6e^{-}$

$Mn^{7+}+5e^{-}\rightarrow Mn^{2+}$

$6$ mol of

$Mn^{7+}$ needs

$15$ mol of

$C_{2}O_{4}^{2-}$ or

$5$ mol of

$Fe_{2}(C_{2}O_{4})_{3}$ .
Therefore,

$1$ mol will require

$\dfrac{6}{5}$

$=$

$1.2$ .

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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