MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 7

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

$Na_2[\ \underline{Fe}(CN)_5NO\ ]$

0%
-2
0%
+2
0%
+3
0%
none of these

Explanation

Given compound

$Na_2[Fe(CN)_5NO]$

On dividing into ions we get

$[Fe(CN)_5NO]^{-3}$ and

$3Na^{+1}$

$\Rightarrow x+(-5)+0=-3$

$\Rightarrow x= -3+5$

$\Rightarrow x=+2$ .

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

$Ba_2\underline{Xe}O_2$

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

Explanation

$Ba_2 Xe O_2$

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

$\Rightarrow x=0$

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

$\underline{Ru}O_4$

0%
+8
0%
+6
0%
+4
0%
none of these

Explanation

$RuO_4$ oxidation state be

$x$

Then,

$x + 4 (-2) = 0$

$x - 8 = 0$

$x = +8$

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

$K_2\underline{Ta}F_7$

0%
+7
0%
+4
0%
+5
0%
+1

Explanation

$K_2T_aF_1$ Let oxidation state be

$x$

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

$x = +5$

Which of the following has been arranged in order of increasing oxidation number of nitrogen?

0%
$NH_3 < N_2O_5 < NO < N_2$
0%
$NO^+_2 < NO^-_3 < NO^-_2 < N^-_3$
0%
$NH^+_4 < N_2H_4 < NH_2OH < N_2O$
0%
$NO_2 < NaN_3 < NH^+_4 < N_2O$

Explanation

Oxidation state of

$H$

$=+1$

Oxidation state of

$O$

$=-2$

For option A oxidation state of

$N$ in

$NH_3$ is -3,in

$N_2O_5$ is +5,in

$NO$ is +2. So this is not increasing order of oxidation state of nitrogen.

For option B oxidation state of

$N$ in

$NO_2^{+}$ is +5,in

$NO_3^{-}$ is +5. So this is not increasing order of oxidation state of nitrogen.

For option C oxidation state of

$N$ in

$NH_4^{+}$ is -3,in

$N_2H_4$ is -2,in

$NH_2OH$ is -1 and in

$N_2O$ is +1. So this is correct answer as compounds are arranged in increasing order of oxidation number of nitrogen.

For option A oxidation state of

$N$ in

$NO_2$ is -4,in

$NaN_3$ is -1/3,in

$NH_4^{+}$ is -3. So this is not increasing order of oxidation state of nitrogen.

The oxidation number of nitrogen atoms in

$NH_4NO_3$ are:

0%
$+3, +3$
0%
$+3, -3$
0%
$-3, +5$
0%
$-5, +3$

Explanation

$NH_4NO_3\rightleftharpoons NH^+_4+NO^-_3$

$NH^+_4x+4=+1$

$x=-3$

$NO^-_3x-6=-1$

$x=+5$ .

The oxidation states of sulphur in Caro's and Marshall's acid are:

0%
$+6, +6$
0%
$+4, +6$
0%
$+6, -6$
0%
$+6, +4$

Explanation

$H_2O_5$ Caro's Acid

$H-O-\overset{\displaystyle \overset{\displaystyle O}{||}}{\underset{\displaystyle \underset{\displaystyle O}{||}}{\displaystyle S}}-O-O-H$ .

$\underset{State}{(+6)}$

$H_2S_2O_8$ Marshall's Acid

$H-O-\overset{\displaystyle \overset{\displaystyle O}{||}}{\underset{\displaystyle \underset{\displaystyle O}{||}}{S}}-O-O-\overset{\displaystyle\overset{\displaystyle O}{||}}{\underset{\displaystyle \underset{\displaystyle O}{||}}{S}}-O-H$

$\underset{State}{(+6)}$
Both these acids have peroxy link.

Oxidation state(s) of chlorine in

$CaOCl_2$ (bleaching powder) is/are?

0%
$+1$ and $-1$
0%
$+1$ only
0%
$-1$ only
0%
None of these

Explanation

$CaOCl_2$ , two

$Cl$ are in different oxidation states because one

$Cl$ is attached to oxygen atom exists as

$OCl^{-}$ and other is attached directly to

$Ca$ exists as

$Cl^{-}$ .One attached directly to

$Ca$ has -1 oxidation state and one attached to oxygen has +1 oxidation state. So

$Cl$ exists in two different oxidation states -1 and +1.

Oxidation states of carbon atoms in diamond and graphite are.

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

Oxidation number of

$C$ in

$HCCOH$ is________.

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

Oxidation number of sulphur in

$S_8, S_2F_2$ and

$H_2S$ are:

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

Explanation

Oxidation number of sulphur in

$S_8$ is 0 as in

$S_8$ , sulphur exists in elemental form so it has 0 oxidation state.

Oxidation number of sulphur in

$S_2F_2$ is +1 as oxidation number of

$F$ is -1.

Oxidation number of sulphur in

$H_2S$ is -2 as oxidation number of hydrogen is +1.

In the ethylene molecule the two carbon atoms have the oxidation numbers.

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

Explanation

Formula of ethylene

$=$

$C_2H_4$

Let oxidation number of

$C$ is

$x$ and we know oxidation number of

$H$ is

$+1$ .

So,

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

$x=-2$

So oxidation number of each carbon is

$-2$ .

Oxalic acid dihydrate,

$H_{2}C_{2}O_{4}\cdot 2H_{2}O(s)$ is often used as a primary reagent to standardise sodium hydroxide solution. Which of these facts are reasons to choose this substance as a primary standard?
I. It is diprotic.
II. It is a stable compound that can be weighed directly in air.
III. It is available in pure form.

0%
III only
0%
I and II only
0%
II and III only
0%
I, II and III

Explanation

$\begin{array}{l} { H_{ 2 } }{ C_{ 2 } }{ O_{ 4 } }\cdot 2{ H_{ 2 } }O\left( s \right) \to oxadic\, acid\, dihydrate\, \\ Since,\, it\, is\, very\, stable\, and\, also\, available\, in\, pressure \\ from\, in\, the\, laboratory\, i.e.\, it's\, always\, used\, as\, primary\, s\tan dard\, in\, the\, lab. \\ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$

Select the species having zero oxidation state at the underlined elements.

0%
$(CH_3)_2\underset{-}{S}O$
0%
$\underline{C}_{12}H_{22}O_{11}$
0%
$H_2\underset{-}{S}O_3$
0%
$\underline{N }_2H_4$

Explanation

$(CH_3)_2SO$

$\Rightarrow 2(+4-3)+x-2=0$

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

$\Rightarrow x=0$

$C_{12}H_{22}O_{11}$

$\Rightarrow 12x+22-22=0$

$\Rightarrow 12x=0$

$\Rightarrow x=0$

$H_2SO_3$

$S \Rightarrow 2+x-6=0$

$\Rightarrow x=4$

$N_2H_4$

$\Rightarrow 2x+4=0$

$\Rightarrow x=-1$

Which of the following have been arrange in order of decreasing oxidation number of sulphur?

0%
$Na_{2}S_{4}O_{6} > H_{2}S_{2}O_{7} > Na_2S_2O_3 > S_8$
0%
$SO_{2}^{+} > SO^{2-}_4 > SO^{2-}_3 > HSO^-_4$
0%
$H_2SO_5 > H_2SO_3 > SCl_2 > H_2S$
0%
$H_2SO_4 > SO_2 > H_2S > H_2S_2O_8$

Explanation

1.Oxidation number of

$S$ in

$H_2S_2O_7$ is +6,in

$Na_2S_4O_6$ +2.5,in

$Na_2S_2O_3$ is +2 and in

$S_8$ is 0.

Oxidation number of sulphur in decreasing order:

$H_2S_2O_7$ >

$Na_2S_4O_6$ >

$Na_2S_2O_3$ >

$S_8$ .

2.Oxidation number of

$S$ in

$SO_2^{+}$ is +5,in

$SO_4^{2-}$ 6,in

$SO_3^{2-}$ is +4 and in

$HSO_4^{-}$ is +6.

Oxidation number of sulphur in decreasing order:

$HSO_4^{-}$ =

$SO_4^{2-}$ >

$SO_2^{+}$ >

$SO_3^{2-}$ .

3.Oxidation number of

$S$ in

$H_2SO_5$ is +8,in

$H_2SO_3$ +4,in

$SCl_2$ is +2 and in

$H_2S$ is -2.

Oxidation number of sulphur in decreasing order:

$H_2SO_5$ >

$H_2SO_3$ >

$SCl_2$ >

$H_2S$ .

4.Oxidation number of

$S$ in

$H_2SO_4$ is +6,in

$SO_2$ +4,in

$H_2S$ is -2 and in

$H_2S_2O_8$ is +7.

Oxidation number of sulphur in decreasing order:

$H_2S_2O_8$ >

$H_2SO_4$ >

$SO_2$ >

$H_2S$ .

So A and C are correct answer.

The oxidation number of Cr is

$+6$ in_______________.

0%
$FeCr_2O_4$
0%
$KCrO_3Cl$
0%
$CrO_5$
0%
$[Cr(OH)_4]^-$

Explanation

1. In

$FeCr_2O_4$ ,

$Fe$ is in +2 oxidation state so oxidation state of

$Cr$ is

$2+2(x)+4(-2)=0,x=+3$ .

2. In

$KCrO_3Cl$ ,

$K$ is in +1 oxidation state so oxidation state of

$Cr$ is

$1+x+3(-2)+(-1)=0,x=+6$ .

3. In

$CrO_5$ ,four

$O$ are attached through peroxide linkage so two oxygen are in -1 oxidation state and one oxygen is in -2 oxidation state so oxidation so oxidation state of

$Cr$ is +6.

4. In

$[Cr(OH)_4]^{-}$ ,oxidation state of

$Cr$ is +3.

Which among the following are auto redox reactions?

0%
$P_4+OH^-\rightarrow H_2PO^-_4+PH_3$
0%
$S_2O^{2-}_3\rightarrow SO^{2-}_4+S$
0%
$H_2O_2\rightarrow H_2O+O_2$
0%
$AgCl+NH_3\rightarrow [Ag(NH_3)_2]Cl$

Explanation

Auto redox reaction is the reaction in which same substance itself undergoes both oxidation as well reduction.

$P$ is oxidised from 0 to +5 and reduced from 0 to -3 so this is an auto redox reaction.

$S$ is oxidised from +2 to +6 and reduced from +2 to 0 so this is an auto redox reaction.

$O$ is oxidised from -1 to 0 and reduced from -1 to -2 so this is an auto redox reaction.

4.Oxidation state of

$Ag$ does not change during reaction so this is not auto redox reaction.

The sum of oxidation number of nitrogen in

$NH_4NO_3$ is:

0%
+1
0%
+2
0%
+4
0%
none of these

Explanation

Given

$NH_4NO_3$ in the ionic state is present in the form of

$NH_4^++{NO_3}^-$ .

$\Rightarrow x+4=+1$

$\Rightarrow x-6=-1$

$\Rightarrow x= -3$

$\Rightarrow x=+5$

Sum of two oxidation states of

$N$ gives

$+2$ .

${MnO_{4}}^{-}$ ions are reduced in acidic condition to

$Mn^{2+}$ ions whereas they are reduced in neutral condition to

$MnO_{2}$ . The oxidation of

$25\ mL$ of a solution

$X$ containing

$Fe^{2+}$ ions required in acidic condition,

$20\ mL$ of a solution

$Y$ containing

${MnO_{4}}^{-}$ ions. What volume of solution

$Y$ would be required to oxidise

$25\ mL$ of solution

$X$ containing

$Fe^{2+}$ ions in neutral condition?

0%
$11.4\ mL$
0%
$12\ mL$
0%
$33.3\ mL$
0%
$35\ mL$

Explanation

Acid medium

$MnO_{4}^{-} + 5Fe^{2+} + 8H^{+}\rightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O$

$\dfrac {M_{1}V_{1}}{1} = \dfrac {M_{2}V_{2}}{5}$

$\dfrac {M_{1}\times 20}{1} = \dfrac {M_{2} \times 25}{5}$

$M_{1} = \dfrac {M_{2}}{4} .... (i)$
Neutral medium

$MnO_{4}^{-} + 3Fe^{2+} + 4H^{+} \rightarrow MnO_{2} + 3Fe^{3+} + 2H_{2}O$

$\dfrac {M_{1}V_{1}}{1} = \dfrac {M_{2}V_{2}}{3}$

$\left (\dfrac {M_{2}}{4}\right )\times V_{1} = \dfrac {M_{2}\times 25}{3}$

$V_{1} = 33.3\ mL$ .

The oxidation number of carbon is zero in_____________.

0%
HCHO
0%
$CH_2Cl_2$
0%
$C_6H_{12}O_6$
0%
$C_{12}H_{22}O_{11}$

Explanation

Oxidation number of

$H$

$=+1$

Oxidation number of

$Cl$

$=-1$

Oxidation number of

$O$

$=-2$

Let oxidation number of

$C$ is

$x$

$HCHO$

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

$x=0$

$CH_2Cl_2$

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

$x=0$

$C_6H_{12}O_6$

$x+6(-2)+12(+1)=0$

$x=0$

$C_{12}H_{22}O_{11}$

$12(x)+22(+1)+11(-2)=0$

$x=0$

Oxidation state of carbon is zero in all compounds.

The value of n in the molecular formula

$Be_nAl_2Si_6O_{18}$ is:

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

Of the following reactions, only one is a redox reaction. Identify this reaction.

0%
$Ca{ \left( OH \right) }_{ 2 }+2HCl\longrightarrow Ca{ Cl }_{ 2 }+2{ H }_{ 2 }O$
0%
$2{ S }_{ 2 }{ O }_{ 7 }^{ 2- }+2{ H }_{ 2 }O\longrightarrow 2S{ O }_{ 4 }^{ 2- }+4{ H }^{ + }$
0%
$Ba{ Cl }_{ 2 }+MgS{ O }_{ 4 }\longrightarrow BaS{ O }_{ 4 }+Mg{ Cl }_{ 2 }$
0%
${ Cu }_{ 2 }S+2FeO\longrightarrow 2Cu+2Fe+S{ O }_{ 2 }$

Explanation

Redox reaction means oxidation and reduction reaction must occur simultaneously.

$Cu_2S+2FeO\longrightarrow 2Cu+2Fe+SO_2$

The oxidation number of cobalt in

$K[Co(CO)_4]$ is:

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

Explanation

The oxidation number of colbalt in

$K[Co(CO)_4]$ is:

$+1+x+0\times 4=0$

$\therefore x=-1$ .

Hence, option

$A$ is correct.

Which of the statement is correct?

0%
The oxidation state of iron in $Na_2[Fe(CN)_5NO]$ is +2
0%
$[Ag(NH_3)_2]^+$ linear in shape
0%
In $[Fe(H_2O)_6]^{3+}$ , Fe is $sp^2d^2$ hybridised
0%
In $[Co(H_2O)_6]^{3+}$ , Co is $d^2sp^3$ hybridised.

Which one of the following leads to redox reaction?

0%
$AgN{ O }_{ 3 }+HCl$
0%
$KOH+HCl$
0%
$KI+{ Cl }_{ 2 }$
0%
${ NH }_{ 3 }+HCl\quad$

Explanation

Redox reactions are characterized by the transfer of electrons between chemical species, most often with one species undergoing oxidation while another species undergoes reduction.

$2\overset { -1 }{ KI } +\overset { 0 }{ { Cl }_{ 2 } } \longrightarrow 2\overset { 1 }{ KCl } +\overset { 0 }{ { I }_{ 2 } }$

It is an example of a redox reaction.

Hence, option

$C$ is correct.

The oxidation number of sulphur in

$H_2S_2O_8$ is:

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

Explanation

$H-O-\overset{\overset{\displaystyle O}{\displaystyle ||}}{\underset{\underset{\displaystyle O}{\displaystyle ||}}{\displaystyle S}}-O-O-\overset{\overset{\displaystyle O}{\displaystyle ||}}{\underset{\underset{\displaystyle O}{\displaystyle ||}}{\displaystyle S}}-O-H$

$\because$

$2\ O$ -atoms are in peroxide linkage

$\therefore$ Their oxidation state

$=-1$
and, oxidation state of remaining

$6\ O$ -atoms

$=-2$
Oxidation state of

$H-$ atoms

$=+1$

Let oxidation state of

$S$ atom

$=x$

$S_2$

$H_2$

$O_6$

$O_2$

$\therefore 2x+2(+1)+6(-2)+2(-1)=0$

$2x+2-12-2=0$

$\therefore x=+6$ .

The oxidation state of central atom in the anion of compound

$NaH_2PO_2$ will be:

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

Explanation

$Na(H_2)PO_2$

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

$x -1= 0$
or

$x= 1$

The oxidation states of P atom in

$POCl_3, H_2PO_3$ and

$H_4P_2O_6$ , respectively are:

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

Explanation

Oxygen has an oxidation state of

$-2$ and hydrogen has

$+1$ .

$POCl_3=x+(-2)+3(-1)=0,x=+5$

$H_2PO_3=2(+1)+x+3(-2)=0,x=+4$

$H_4P_2O_6=4(+1)+2(x)+6(-2)=0,x=+4$

Hence, option A is correct.

Oxidation number of

$'N'$ in

$N_3H$ (hydrazoic acid) is:

0%
$-\cfrac{1}{3}$
0%
$+3$
0%
$0$
0%
$-3$

Explanation

The oxidation state of

$N$ in

$N_3H$ (hydrazoic acid) is:

$\Rightarrow 3x+1=0$

$\Rightarrow 3x=-1$

$\Rightarrow x=- \dfrac 13$

Hence, option

$A$ is correct.

In Redox reaction:

$Fe + HNO_{3}\rightarrow Fe(NO_{3})_{2} + NH_{4}NO_{3} + H_{2}O$ the coefficient of

$HNO_{3}, Fe(NO_{3})_{2}, NH_{4}NO_{3}$ is:

0%
$1 : 10 : 4$
0%
$10 : 4 : 1$
0%
$4:10:4$
0%
$4:1:4$

Explanation

After balancing the equation, we get

$4Fe + 10HNO_3 \rightarrow 4Fe(NO_3)_2 + NH_4NO_3 + 3H_2O$

Therefore, Ratio of

$HNO_3,\ Fe(NO_3)_2\ and\ NH_4NO_3$ is :

10:4:1.

Hence, Option B is correct.

The oxidation number of iron in potassium ferricyanide

$[K_{3}Fe(CN)_{6}]$ is :

0%
Two
0%
Six
0%
Three
0%
None

Explanation

To calculate: Oxidation state of iron in potassium ferro cyanide.

Solution :

Oxidation state of Potassium , K = +1

Oxidation state of Cyanide ion =-1

Let oxidation state of Iron be x

We know that sum of all the oxidation states in the compound is equal to overall charge of the compound.

Therefore, (+1)3 +[ x +6(-1)] = 0

$\Rightarrow 3+ [ x-6] =0$

$\Rightarrow x= +6-3 =+3$

Hence, the correct option is C.

When benzaldehyde is oxidised to give benzoic acid then the oxidation state of carbon of aldehydic group is changed from:

0%
$+2$ to $+3$
0%
$+1$ to $+3$
0%
Zero to $+2$
0%
No change

Explanation

Bonds between same atoms donot affect the oxidation state. Therefore only functional group affects the oxidation state of

$C$

In benzaldehyde:

$=O$ has

$-2$ and

$H$ has

$+1$ , therefore in

$CHO$ O.S. of

$C$ is:

$x-2+1=0$

$C$ is

$+1$

In Benzoic acid

$=O$ has

$-2$ and

$OH$ has

$-1$ , therefore in

$COOH$ O.S. of

$C$ in

$CO-OH$ is:

$x-2-1=0$

$C$ is

$+3$

Thus O.S. of

$C$ changes from

$+1$ to

$+3$ . Option B is correct

The reaction which proceeds in the forward direction is:

0%
$Fe_3O_4 + 6HCl \rightarrow2 FeCl_3 + 3H_2O$
0%
$NH_3 + H_2O + NaCl\rightarrow NH_4Cl + NaOH$
0%
$SnCl_4 +Hg_2Cl_2\rightarrow SnCl_2 + 2HgCl_2$
0%
$2CuI + I_2 + 4 K^+ \rightarrow 2 Cu^{2+} + 4 KI$

Explanation

$FeCl_3$ and

$H_2O$ do not react to give back

$Fe_3O_4$ and

$HCl$ .

Oxidation number of P in

$Ba\, (H_2PO_2)_2$ is:

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

Explanation

$Ba(H_{2}PO_{2})_{2}$ is

$Ba(H_{2}PO_{2})_{2}\Rightarrow (H_{2}PO_{2})_{2}^{2-}$

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

$+2+x-4=-1$

$x=-1+4-2$

$\boxed{x=+1}$

So oxidation number of

$P=+1$

In which of the following, the oxidation number of oxygen has been arranged in increasing order?

0%
$BaO_{2} < KO_{2} < O_{3} < OF_{2}$
0%
$OF_{2} < KO_{2} < BaO_{2} < O_{3}$
0%
$BaO_{2} < O_{3} < OF_{2} < KO_{2}$
0%
$KO_{2} < OF_{2} < O_{3} < BaO_{2}$

Explanation

Hint: The oxidation state of oxygen differs from compound to compound.

Step 1: Oxidation state calculation.

A. In $BaO_2$ , oxygen is in a peroxide state. therefore has a charge of $-1$ .

B. In $KO_2$ , oxygen is in a superoxide state. therefore has charge of $\frac {-1}{2}$ .

C. In $O_3$ , as there is $3$ oxygen. First one has charged $-1$ , second one has a charge of $0$ and third one has therefore had a charge of $+1$ .

Hence, overall charge is $0$ .

In $OF_2$ , oxygen is less electronegative than fluorine. Therefore has a charge of $-2$ .

Step 2: Oxidation state of given compounds.

As shown above, the calculated oxidation state of compounds is as follows.

A. $BaO_2$ $\rightarrow$ $-1$

B. $KO_2$ $\rightarrow$ $\frac {-1}{2}$

C. $O_3$ $\rightarrow$ $0$

D. $OF_2$ $\rightarrow$ $+2$

Final step: From the above-calculated value, the increasing order of oxidation state is in option A.

Hence A is the correct option.

Which of the following elements never show positive oxidation number?

0%
$O$
0%
$Fe$
0%
$Ga$
0%
$F$

The sum of the oxidation states of all the carbon atoms present in the compound

$C_{6}H_{5}CHO$ is:

0%
$-4$
0%
$3$
0%
$+5$
0%
$-4/7$

Explanation

As we know that sum of oxidation state of all atoms in a compound is equal to the total charge on that compound.

$\therefore \text{ In } {C}_{6}{H}_{5}CHO$ -

Let the oxidation state of carbon in

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

$\therefore \; 6 \times x + 5 \times 1 + x + 1 + \left( -2 \right) = 0$

$\Rightarrow \; 7x = -5 - 1 + 2 = -4$

As there are seven carbon atoms are present in the compound, the sum of oxidation state of all the carbon atoms will be

$7x$ , i.e.

$-4$ .

Minerals associated with redox reactions are

0%
Na, Cu
0%
N, Cu
0%
Fe, Cu
0%
Ca, Fe

The ratio of number of moles of

$KMnO_4$ and

$K_2Cr_2O_7$ required to oxidise

$0.1$ mol

$Sn^{+4}$ in acidic medium :

0%
$6 : 5$
0%
$5 : 6$
0%
$1 : 2$
0%
$2 : 1$

Explanation

$\overset { +7 }{ KMnO_4 } \quad \underrightarrow { acidic \quad medium }\quad \overset { +2 }{ Mn^{+2} }$

$\overset { +6 }{ K_2Cr_2O_7 } \quad \underrightarrow { acidic \quad medium }\quad \overset { +3 }{ Cr^{+3} }$

$\overset { +4 }{Sn^{+4} } \quad \underrightarrow { reducing \quad agent}\quad \overset { +2 }{ Sn^{+2} }$

Number of equivalents

$= moles \times nf$

$n(Sn^{+4})=0.1$

$nf(Sn^{+4})=2$

Equating number of equivalents of

$KMn{O}_{4}$ and

${Sn^{+4}}$

${ n }_{ 1 }{ nf }_{ 1 }={ n }_{ 2 }{ nf }_{ 2 }$

$0.1\times 2=5n_{ 2 }\\ n_{ 21 }=0.04$

Equating number of equivalents

$K_2$

$Cr_2$

$O_7$ and

${Sn^{+4}}$

$n_1f_1=n_22f_22$

$0.1 \times2=6n_{22}$

$n_{22} =\cfrac{0.1} {3}$

Ratio of moles of

$KMnO_3$ and

$K_2 Cr_2 O_7=\cfrac{0.04} {0.1}=\cfrac{6} {5}$

The pair of compounds in which the metals are in their highest oxidation state is:

0%
$MnO_2, FeCl_3$
0%
$Mn{ O }_{ 4 }^{ - }, CrO_2Cl_2$
0%
$MnCl_2 , CrCl_3$
0%
$[NiCl_4]^{2-}, [ CoCl_4]^-$

Explanation

The charges on oxygen and chlorine atoms are

$-2$ and

$-1$ respectively. Therefore, the oxidation states of the central atoms in the given compounds are calculated as follows:

$MnO_2 , FeCl_3 \rightarrow +4 , +3$

$MnO_{4}^{-} , CrO_2Cl_2 \rightarrow +7 , +6$

$MnCl_2 , CrCl_3 \rightarrow +2 , +3$

$[NiCl_4]^{-2} , [CoCl_4]^- \rightarrow +2 , +3$

The element that does not show positive oxidation state is:

0%
$O$
0%
$N$
0%
$Cl$
0%
$F$

Which of the following shows highest oxidation number in combined state?

0%
$Os$
0%
$Ru$
0%
Both (1) & (2)
0%
None

Explanation

The possible oxidation states of

$Os$ and

$Ru$ are :-

$Os \Rightarrow -4,-2,-1,0,+1,+2,+3,+4,+5,+6,+7,+8$

$Ru \Rightarrow -4,-2,0,+1,+2,+3,+4,+5,+6,+7,+8$

Therefore, both

$Os$ and

$Ru$ show highest oxidation number of

$+8$ .

Therefore,

$(C)$ is the correct answer.

Carbon is in the lowest oxidation state in:

0%
$CH_4$
0%
$CCl_4$
0%
$CF_4$
0%
$CO_2$

Explanation

Let the oxidation number of carbon be

$x$ .

Oxidation state of

$H=+1$ ,

$Cl=F=-1$ and of

$O=-2$ .

(A)

$CH_4$ :

$x+4 \times 1=0$

$x+4=0$ or

$x=-4$

(B)

$CCl_4$ :

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

$x-4=0$ or

$x=+4$

(C)

$CF_4$ :

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

$x-4=0$ or

$x=+4$

(D)

$CO_2$ :

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

$x-4=0$ or

$x=+4$

Lowest oxidation of carbon is

$-4$ in

$CH_4$ .

Which compound amongst the following has the highest oxidation number of

$Mn$ ?

0%
$KMnO_4$
0%
$K_2MnO_4$
0%
$MnO_2$
0%
$Mn_2O_3$

Explanation

Let the oxidation number of

$Mn$ be

$x$ .

Oxidation state of

$K=+1$ ,

$O=-2$ .

(A)

$KMnO_4$ :

$1+x+4 \times (−2)=0$

$x=+7$

(B)

$K_2MnO_4$ :

$2 \times 1+x+4 \times (−2)=0$

$x=+6$

(C)

$MnO_2$ :

$x+2 \times (−2)=0$

$x=+4$

(D)

$Mn_2O_3$ :

$2x+3 \times (−2)=0$

$x=+3$

therefore highest oxidation number of

$Mn$ is +7 in

$KMnO_4$

Which of the following is a decreasing order of oxidation states of the central atoms?

0%
$PCl_5 , HIO_4, Cl_2{ O }_{ 7 }^{ 2- },Cl_2O$
0%
$Cr_2{ O }_{ 7 }^{ 2- }, Cl_2O, HIO_4, PCl_5$
0%
$HIO_4, Cr_2{ O }_{ 7 }^{ 2- }, PCI_5,Cl_2O$
0%
$Cr_2{ O }_{ 7 }^{ 2- }, HIO_4,Cl_2O,PCl_5$

Explanation

Let the oxidation number of the central atom be

$x$ .

Oxidation state of side atoms be:

$H=+1$ ,

$Cl=I=-1$ and of

$O=-2$ .

$PCl_5$ :

$x+5 \times (-1)=0$

$x=+5$

$HIO_4$ :

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

$x=+7$

$Cl_2O_7^{-2}$ :

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

$x=+6$

$Cl_2O$ :

$2x+(-2)=0$

$x=+1$

$Cr_2O_7^{-2}$ :

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

$x=+6$

Thus the decreasing order of the oxidation state of central atom is as:

$HIO_4>Cr_2O_7^{-2}=Cl_2O_7^{-2}>PCl_5>Cl_2O$

Various oxidation states of few elements are mentioned, Which of the options is not correctly matched?

0%
Phosphorus : +3 to +5
0%
Nitrogen : +1 to +5
0%
Iodine : -1 to +7
0%
Chromium : -3 to +6

Choose the redox reaction from the following.

0%
$Cu+2H_2SO_4 \rightarrow CuSO_4+SO_2+2H_2O$
0%
$BaCl_2+H_2SO_4 \rightarrow BaSO_4+2HCl$
0%
$2NaOH+H_2SO_4 \rightarrow Na_2SO_4+2H_2O$
0%
$KNO_2+H_2SO_4 \rightarrow 2HNO_2+K_2SO_4$

Explanation

A redox reaction is the one in which oxidation and reduction take place simultaneously.

In option A,

$\cfrac { Cu }{ O } +\cfrac { 2{ H }_{ 2 }{ SO }_{ 4 } }{ +6 } \longrightarrow \cfrac { CuS{ O }_{ 4 } }{ +2 } +\cfrac { { SO }_{ 2 } }{ +4 } +2H_{ 2 }O$

Oxidation of

$Cu$ from

$"0"$ to

$"+2"$ and reduction of S from

$"+6"$ to

$"+4"$ occurs in the same reaction and hence it is a redox reaction.

The oxidation number of nitrogen in

$( N_2H_5)^+$ is:

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

Explanation

Let the oxidation number of nitrogen be

$x$ .

Oxidation number of

$H=+1$

$(N_2H_5)^+$ :

$2 \times x+ 5 \times 1=+1$

$2x+5=1$ or

$x=-2$

Thus Oxidation number of

$N$ in

$(N_2H_5)^+$ is +2

Which is not true about the oxidation state of the following elements?

0%
Sulphur +6 to -2
0%
Carbon +4 to -4
0%
Chlorine +7 to -1
0%
Nitrogen +3 to -1

Which compound among the following has lowest oxidation number of chlorine?

0%
$HClO_4$
0%
$HClO_3$
0%
$HCl$
0%
$HOCl$

Explanation

Let the oxidation number of chlorine be

$x$ .

Oxidation number of

$H=+1$ and

$O=-2$ .

(A)

$HClO_4$ :

$1+x+4 \times (−2)=0$

$x-7=0$ or

$x=7$

(B)

$HClO_3$ :

$1+x+3 \times (−2)=0$

$x-5=0$ or

$x=5$

(C)

$HCl$ :

$1+x=0$

$x=-1$

(D)

$HOCl$ :

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

$x-1=0$ or

$x=+1$

Therefore lowest oxidation number of chlorine is -1 in

$HCl$ .

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 7 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 7 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.