Match the column I with column II and mark the appropriate choice. Column I ( Compound ) Coloumn II (Oxidation state of $$Fe$$) ( A ) $$K_3[(Fe(OH) _6]$$ (i) +8/3 ( B ) $$ K_2[FeO_4]$$ (ii) +2 ( C ) $$FeSO_4, (NH_4)_2SO_46H_2O$$ (iii) +3 ( D) $$Fe_3O4$$ (iv) +6

$$( A ) \rightarrow ( iii ) , ( B ) \rightarrow ( iv ) , ( C ) \rightarrow ( ii ) ,( D ) \rightarrow ( i )$$

$$( A ) \rightarrow ( iii ) , ( B ) \rightarrow ( i ) , ( C ) \rightarrow ( ii ) ,( D ) \rightarrow ( iv )$$

$$( A ) \rightarrow ( i ) , ( B ) \rightarrow ( iii ) , ( C ) \rightarrow ( ii ) ,( D ) \rightarrow ( iv )$$

$$( A ) \rightarrow ( iv ) , ( B ) \rightarrow ( ii ) , ( C ) \rightarrow ( i ) ,( D ) \rightarrow ( iii )$$

MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 8

Examples of few compounds in a particular oxidation state are given . Mark the example which is not correct.

0%
Phosphorus in +1 oxidation state - $H_3PO_2$
0%
Chlorine in +7 oxidation state - $HClO$
0%
Chromium in +6 oxidation state - $CrO_2Cl_2$
0%
Carbon in 0 oxidation state - $C_{12}H_{22}O_{11}$

Explanation

Let the oxidation number of a desired atom be

$x$ .

Oxidation number of

$H=+1$ ,

$Cl=-1$ and of

$O=-2$ .

(A) Phosphorus in

$H_3PO_2$ :

$3 \times 1+x+2 \times (−2)=0$

$x-1=0$ or

$x=+1$

Thus, phosphorus in +1 oxidation state in

$H_3PO_2$

(B) Chorine in

$HClO$ :

$1+x+(−2)=0$

$x-1=0$ or

$x=+1$

Chlorine in +1 oxidation state - HClO

$CrO_2Cl_2$ :

$x+ 2 \times (−2)+2 \times (−1)=0$

$x-6=0$ or

$x=+6$

Chromium in +6 oxidation state -

$CrO_2Cl_2$

(D) Chromium in

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

$12x+22 \times 1+11 \times (−2)=0$

$12x+22-22=0$ or

$x=0$

Carbon in 0 oxidation state -

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

Which of the following oxidation numbers is not correctly matched?

0%
$P$ in $NaH_2PO_4 = +5$
0%
$Ni$ in $[ Ni(CN)_6]^{4-} = +2$
0%
$P$ in $Mg_2P_2O_7 = +6$
0%
$Cr$ in $(NH_4)_2Cr_2O_7 =+6$

Explanation

Let the oxidation number of a desired atom be

$x$ .

Oxidation number of

$H=Na=+1$ ,

$CN=-1$ ,

$O=-2$ ,

$Mg=+2$ ,

$NH_4=+1$

(A) P in

$NaH_2PO_4$ :

$1+2 \times 1+x+4 \times (−2)=0$

$x-5=0$ or

$x=+5$

(B)Ni in

$[Ni(CN)_6]^{4−}$ :

$x+ 6 \times (-1)=-4$

$x-6=-4$ or

$x=+2$

$Mg_2P_2O_7$ :

$2 \times 2+2 \times x+7 \times (−2)=0$

$2x-10=0$ or

$x=+5$

(D) Cr in

$(NH_4)_2Cr_2O_7$ :

$2 \times 1+2 \times x +7 \times (−2)=0$

$2x-12=0$ or

$x=+6$

Oxidation number of sulphur in peroxomonosulphuric acid

$(H_2SO_5)$ is:

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

Explanation

$H_2SO_5$ has the following structure:

$H-O-\overset { \overset { O }{ \parallel } }{ \underset { \underset { O }{ \parallel } }{ S } } -O-O-H$

the oxygens in peroxide linkage i.e.

$O-O$ has oxidation number =-1 each

and in oxide linkage i.e

$=O$ and

$H-O-S$ it is -2

Let the oxidation number of sulphur be

$x$ .

Oxidation number of

$H=+1$

$H_2SO_5$ :

$2 \times 1+x+3 \times (−2)+2 \times (−1)=0$

$x+2-6-2=0$ or

$x=+6$

thus Oxidation number of sulphur in peroxomonosulphuric acid (

$H_2SO_5$ ) is

$+6$ .

In which of the following compounds oxidation state of chlorine has two different values?

0%
$CaCl_2$
0%
$NaCl$
0%
$CaOCl_2$
0%
$CCl_4$

Explanation

Ionic compounds can exist in equillibrium with the constituting ions as:

$CaCl_2 \rightarrow Ca^{2+}+2Cl^-$

Oxidation state of

$Cl$ is -1

$NaCl \rightarrow Na^{+}+Cl^-$

Oxidation state of

$Cl$ is -1

$CaOCl_2 \rightarrow Ca^{2+}+OCl^-+Cl^-$

Oxidation state of

$Cl$ is -1 in

$Cl^-$ and in

$OCl^-$ it is:

$-2+x=-1\ or\ x=+1$

$CCl_4$ is covalent and thus oxidation of all

$Cl$ is same.

Match the column I with column II and mark the appropriate choice.

	Column I ( Compound )		Coloumn II (Oxidation state of $Fe$ )
( A )	$K_3[(Fe(OH) _6]$	(i)	+8/3
( B )	$K_2[FeO_4]$	(ii)	+2
( C )	$FeSO_4, (NH_4)_2SO_46H_2O$	(iii)	+3
( D)	$Fe_3O4$	(iv)	+6

0%
$( A ) \rightarrow ( iii ) , ( B ) \rightarrow ( i ) , ( C ) \rightarrow ( ii ) ,( D ) \rightarrow ( iv )$
0%
$( A ) \rightarrow ( iii ) , ( B ) \rightarrow ( iv ) , ( C ) \rightarrow ( ii ) ,( D ) \rightarrow ( i )$
0%
$( A ) \rightarrow ( i ) , ( B ) \rightarrow ( iii ) , ( C ) \rightarrow ( ii ) ,( D ) \rightarrow ( iv )$
0%
$( A ) \rightarrow ( iv ) , ( B ) \rightarrow ( ii ) , ( C ) \rightarrow ( i ) ,( D ) \rightarrow ( iii )$

Explanation

Let the oxidation number of

$Fe$ be

$x$ .

Oxidation number of the ligands are:

$H=K=+1$ ,

$O=-2,OH=-2+1=-1,NH_4=+1,SO_4=-2,H_2O=0$

$(A)$

$K_3[Fe(OH)_6]$ :

$3 \times 1 +x+6 \times (-1)=0$

$x+3-6=0$ or

$x=+3$ =(iii)

$(B)$

$K_2[FeO_4]$ :

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

$x+2-8=0$ or

$x=+6$ =(iv)

$(C)$

$FeSO_4.(NH_4)_2SO_4.6H_2O$ :

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

$x-2+2-2+0=0$ or

$x=+2$

$(D)$

$Fe_3O_4$ :

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

$3x-8=0$ or

$x=+8/3$ $$=(i)$$

$(A)-(iii), (B)-(iv), (C)-(ii), (D)- (i )$

Mark the correct statement from the following.

0%
Copper metal can be oxidised by $Zn^{2+}$ ions
0%
Oxidation number of phosphorus in $P_4$ is 4
0%
An element in the highest oxidation state acts only as a reducing agent.
0%
The element which shows highest oxidation number of +8 is $OsO_4$

Explanation

A. Copper metal cannot be oxidised by

$Zn^{2+}$ ion because the reduction potential of copper ions is higher than zinc ions. Thus will get reduced in the presence of zinc metal as:

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

B. Oxidation number of an atom in its elemental form is zero. Thus oxidation number of phosphorus is zero in

$P_4$

C. An element in its highest oxidation state can not loose more electrons and thus can not be oxidised further. Therefore will get reduced only and will act as an oxidising agent or oxidant.

D. Osmium is the element which shows highest oxidation number of +8 in

$OsO_4$ as

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

$x=+8$ .

The oxidation state of

$Fe$ in

$K_4[Fe(CN)_6]$ is:

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

Explanation

Let the oxidation number of

$Fe$ be

$x$ .

Oxidation number of

$CN=-1$ ,

$K=+1$ .

$K_4[Fe(CN)_6]$ :

$4 \times 1+x+6 \times (-1)=0$

$4+x-6=0$ or

$x=+2$

The oxidation state of Fe in

$K_4[Fe(CN)_6]$ is +2.

In which of the following compounds carbon is in highest oxidation state?

0%
$CH_3Cl$
0%
$CCl_4$
0%
$CHCl_3$
0%
$CH_2Cl_2$

Explanation

Let the oxidation number of carbon be

$x$ .

Oxidation number of

$H=+1$ ,

$Cl=-1$

(A)

$CH_3Cl$ :

$x+3 \times 1+(−1)=0$

$x+3-1=0$ or

$x=-2$

(B)

$CCl_4$ :

$x+4 \times (−1)=0$

$x-4=0$ or

$x=+4$

(C)

$CHCl_3$ :

$x+1+3 \times (−1)=0$

$x+1-3=0$ or

$x=+2$

(D)

$CH_2Cl_2$ :

$x+2 \times 1+2 \times (−1)=0$

$x+2-2=0$ or

$x=0$

the highest oxidation state of carbon is +4 in

$CCl_4$

What is the oxidation state of

$P$ in

$Ba(H_2PO_2)_2$ ?

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

Explanation

Let the oxidation number of

$P$ be

$x$ .

Oxidation number of

$H=+1$ ,

$Ba=+2$ and of

$O=-2$ .

$Ba(H_2PO_2)_2$ :

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

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

$2x=2$

$\therefore x=+1$

The oxidation state of P in

$Ba(H_2PO_2)_2$ is +1.

What are the oxidation states of phosphorus in the following compounds?

$H_3PO_2,H_3PO_4,Mg_2P_2O_7,PH_3,HPO_3$

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

Explanation

Let the oxidation number of phosphorous be

$x$ .

Oxidation number of

$H=+1$ ,

$Mg=+2$ and of

$O=-2$ .

$H_3PO_2$ :

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

$3+x-4=0$ or

$x=+1$

$H_3PO_4$ :

$3 \times 1+x +4 \times (-2)=0$

$3+x-8=0$ or

$x=+5$

$Mg_2P_2O_7$ :

$2 \times 2+2x +7 \times (-2)=0$

$4+2x-14=0$ or

$x=+5$

$PH_3$ :

$x+3 \times 1=0$

$x=−3$

$HPO_3$ :

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

$1+x-6=0$ or

$x=+5$

Thus the oxidation states of phosphorus in:

$H_3PO_2, H_3PO_4, Mg_2P_2O_7, PH_3,HPO_3$ is +1,+5,+5,-3,+5, respectively.

The oxidation state of molybdenum in its oxo complex species

$[Mo_2O_4(C_2H_4)_2(H_2O_2)]^{2-}]$ .

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

Explanation

Let the oxidation number of molybdenum

$(Mo)$ be

$x$ .

Oxidation number of

$H_2O_2=C_2H_4=0$ ,

$O=-2$ .

$[Mo_2O_4(C_2H_4)_2(H_2O_2)]^{2-}$

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

$2x - 8 = -2$ or

$x = +3$

The oxidation state of molybdenum in its oxo complex species

$[Mo_2O_4 (C_2H_4)_2(H_2O_2)]^{2−}$ is +3

Which of the following reaction shows reduction of water?

0%
$2H_2O+2Na\rightarrow 2NaOH+H_2$
0%
$6CO_2+12H_2O\rightarrow C_6H_12O_6+6H_2O+6O_2$
0%
$2F_2+2H_2O\rightarrow 4H^++4F^-+O_2$
0%
$P_4O_10+6H_2O\rightarrow 4H_3PO_4$

The sum of oxidation states of sulphur in

$H_2S_2O_8$ is?

0%
$+2$
0%
$+6$
0%
$+7$
0%
$+12$

Explanation

Structure of

$H_2S_2O_8$ is:

$HO-\overset { \overset { O }{ || } }{ \underset { \overset { || }{ O } }{ S } } -O-O-\overset { \overset { O }{ || } }{ \underset { \overset { || }{ O } }{ S } } -OH$

there is one peroxy linkage

$S-O-O-S$ . Hence two of the eight oxygen atoms have -1 oxidation state and rest 6 have -2 oxidation state. Let the total oxidation state of S be x

then

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

$2+x-12-2=0$

$x=+12$

Sum of oxidation state of both S is x i.e. 12 in

$H_2S_2O_8$ .

The correct sequence of the oxidation state of Fe, Ta, P and N elements in the given compounds are respectively :

$Na_2[Fe(CN)_5NO],\ K_2TaF_7,\ Mg_2P_2O_7,\ Na_2S_4O_6,\ N_3H$

0%
$+3, +5 ,+5 .+2.5,-\frac { 1 }{ 3 }$
0%
$+5,+3,+5,+3,+\frac { 1 }{ 3 }$
0%
$+3, +3 ,+5 ,+5 ,-\frac { 1 }{ 3 }$
0%
$+5 , +5 ,+3 ,+2.5, +\frac { 1 }{ 3 }$

Explanation

Let the oxidation number of the desired atom be

$x$ .

Oxidation number of

$H=Na=+1,Mg=+2,CN=F=-1,$ and of

$O=-2$ .

$Na_2[Fe(CN)_5NO]$ :

$2 \times 1+x+5 \times (-1) + 0 = 0$

$x+2-5=0$ "or"

$x=+3$

$K_2TaF_7$ :

$2 \times 1+x+7 \times (-1) = 0$

$x+2-7=0$ "or"

$x=+5$

$Mg_2P_2O_7$ :

$2 \times 2+2x+7 \times (-2) = 0$

$2x+4-14=0$ "or"

$x=+5$

$Na_2S_4O_6$ :

$2 \times 1+4x+6 \times (-2) = 0$

$4x+2-12=0$ "or"

$x=+2.5$

$N_3H$ :

$3 \times x+1 = 0$

"or"

$x=-1/3$

Thus oxidation number of

$Fe$ in

$Na_2[Fe(CN)_5NO]$ ,

$Ta$ in

$K_2TaF_7$ , P in

$Mg_2P_2O_7$ , S in

$Na_2S_4O_6$ and

$N$ in

$N_3H$ is: +3,+5,+5,+2.5,-1/3, respectively.

Which of the following arrangements represent increasing oxidation number of the central atom ?

0%
$CrO_{2}^{-}$ , $ClO_{3}^{-}$ , $CrO_{4}^{2-}$ , $MnO_{4}^{-}$ ,
0%
$ClO_{3}^{-}$ , $CrO_{4}^{2-}$ , $MnO_{4}^{-}$ , $CrO_{2}^{-}$ ,
0%
$CrO_{2}^{-}$ , $ClO_{3}^{-}$ , $MnO_{4}^{-}$ , $CrO_{4}^{2-}$ ,
0%
$CrO_{4}^{2-}$ , $MnO_{4}^{-}$ , $CrO_{2}^{-}$ , $ClO_{3}^{-}$ ,

Explanation

: In the oxides, the oxidation number of oxygen is $-2$

Correct Answer: Option A

Explanation:

Let the oxidation number of the central atom be $x$
$CrO^{-}_{2}$ :

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

$\therefore$ $x=+3$

$CrO_{4}^{2-}$ :

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

$\therefore$ $x=+6$

$ClO^{-}_{3}$ :

$x\ +\ (3\times (-2))=-1$

$\therefore$ $x = +5$

$MnO_{4}^{-}$ :

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

$\therefore$ $x = +7$

Hence, correct answer is $CrO^{-}_{2}$ , $ClO^{-}_{3}$ , $CrO_{4}^{2-}$ , $MnO_{4}^{-}$

The largest oxidation number exhibited by an element depends on its outer
electronic configuration. With which of the following outer electronic configurations
the element will exhibit the largest oxidation number?

0%
$3d^{1}4s^{2}$
0%
$3d^{3}4s^{2}$
0%
$3d^{5}4s^{1}$
0%
$3d^{5}4s^{2}$

Explanation

Hint: Count the total number of electrons in the valence shell.

Correct Answer: Option D

Explanation:

$3d^{1}4s^{2}$ , there are

$3$ electrons in the outermost shell. Hence

$+3$ is the maximum possible oxidation state.

$3d^{3}4s^{2}$ , there are

$5$ electrons in the outermost shell. Hence

$+5$ is the maximum possible oxidation state.

$3d^{5}4s^{1}$ , there are

$6$ electrons in the outermost shell. Hence

$+6$ is the maximum possible oxidation state.

$3d^{5}4s^{2}$ , there are

$7$ electrons in the outermost shell. Hence

$+7$ is the maximum possible oxidation state.

Final Answer: Hence option $D$ is the correct answer.

In the reaction :

$I_2+2S_2{ O }_{ 3 }^{ 2- }\rightarrow 2I^- +S_4{ O }_{ 6 }^{ 2- }$ ?

0%
$I_2$ is reducing agent
0%
$I_2$ is oxidising agent and $S_2{ O }_{ 3 }^{ 2- }$ is reducing agent
0%
$S_2{ O }_{ 3 }^{ 2- }$ is oxidising agent
0%
$I_2$ is reducing agent and $S_2{ O }_{ 3 }^{ 2- }$ is oxidising agent

Explanation

$\overset { 0 }{ I }_2 \rightarrow 2\overset { 0 }{ I }^-$

$I_2$ undergoes reduction hence acts as oxidising agents.

${ S }_{ 2 }^{ +2 }{ O}_{ 6 }^{ 2- } \rightarrow { S }_{ 4}^{ +2.5 }{ O }_{ 6 }^{ 2- }$
It undergoes oxidation hence acts as a reducing agent.

Calculate the oxidation number of underlined element in the given compound.

$K[\underline{Co}(C_2O_4)_2(NH_3)_2]$

0%
+5
0%
+6
0%
+2
0%
+3

Explanation

So, ionic form is

${ { [Co({ C }_{ 2 }{ O }_{ 4 }) }_{ 2 }{ { (NH }_{ 3 }) }_{ 2 }] }^{ - }$ .

Since,

${ NH }_{ 3 }$ is a neutral ligand. So, charge on

${ NH }_{ 3 }$ = 0

${ C }_{ 2 }{ O }_{ 4 }$ is a negative ligand. Change on

${ C }_{ 2 }{ O }_{ 4 }$ = -2

Now, suppose oxidation state of

$Co$ is

$x$ .

So,

$x+2\times (-2)+2\times 0\quad =\quad -1\\ x-4\quad =\quad -1\\ x\quad =\quad +3$ .

So, correct answer is option D.

Which of the following arrangements represent increasing oxidation number of the central atom?

0%
${CrO}_{2}^{-},{ClO}_{3}^{-},{CrO}_{4}^{2-},{MnO}_{4}^{-}$
0%
${ClO}_{3}^{-},{CrO}_{4}^{2-},{MnO}_{4}^{-},{CrO}_{2}^{-}$
0%
${CrO}_{2}^{-},{ClO}_{3}^{-},{MnO}_{4}^{-},{CrO}_{4}^{2-}$
0%
${CrO}_{4}^{2-},{MnO}_{4}^{-},{CrO}_{2}^{-},{ClO}_{3}^{-}$

Explanation

Hint

The total number of electrons that an atom acquires or loses in order to form a chemical connection with another atom is known as the oxidation number.

Explanation

Let the central oxidation number is $x$ .

The oxidation number of $O$ is $2$ .

Step 1: the oxidation state of $CrO_{2}^{-}$

Thus, the oxidation state if $Cr$ in $CrO_{2}^{-}$ .

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

$x-4=-1$

$x=-1+4$

$x=3$

Step 2 : the oxidation state of $CrO_{4}^{2-}$ ,

$x+4\times \left( -2 \right)=-2$

$x-8=-2$

$x=8-2$

$x=6$

Step 3 : the oxidation state of $ClO_{3}^{-}$ ,

$x+3\times \left( -2 \right)=-1$

$x-6=-1$

$x=6-1=5$

Step 4: the oxidation state of $MnO_{4}^{-}$ ,

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

$x-8=-1$

$x=8-1=+7$

Hence, the increasing order of the central atom is $CrO_{2}^{-}<ClO_{3}^{-}<CrO_{4}^{2-}<MnO_{4}^{-}$ .

Final answer

The correct answer is option (A).

$H^+$ ions are present in the balanced chemical equation of reaction in a basic medium.

Is the above statement true or false?

0%
True
0%
False

Explanation

${ H }^{ + }$ ions and

${ OH }^{ - }$ ions are used to balance number of oxygen and hydrogen atoms in acidic and basic medium respectively. Hence, the given redox reaction must be in basic medium.

$2OH^- + 2OCN^- + 3OCl^- \rightarrow 2CO_3{2^-} + N_2 + 3Cl^- + H_2O$
This is the balanced chemical equation of the reaction in which medium?

0%
Acidic
0%
Basic
0%
Neutral
0%
Any of the above

Explanation

The medium is basic clearly seen in the reaction with

$OH^-$

In the reaction

$4Fe+3O_2\rightarrow 4Fe^{3+}+6O^{2-}$ which of the following statements is incorrect?

0%
It is a redox reaction
0%
Metallic iron is a reducing agent
0%
$Fe^{3+}$ is an oxidising agent
0%
Metallic iron is reduced to $Fe^{3+}$

Explanation

$\text { Given, } 4 \mathrm{Fe}+3 \mathrm{O}_{2} \longrightarrow 4 \mathrm{Fe}^{+3}+6 \mathrm{O}^{2} \text { . }$

clearly, given is a redox reaction in Which, Fe gets oxidised to

$\mathrm{Fe}^{+3}$ , ithive

$0_{2}$ gets reduced to

$0^{2-}$ .

Thus motallic iron (Fe) is a reducing agent and

$\mathrm{Fe}^{+3}$ is an oxidising agent.

So, option (D) is correct.

Oxidation number of

$Cr$ in

$CrO_5$ is:

0%
$+10$
0%
$+6$
0%
$+4$
0%
$+5$

Explanation

$\displaystyle CrO_5$ has one

$\displaystyle Cr=O$ double bond. The oxygen atom of this double bond has oxidation number of

$\displaystyle -2$ . The remaining 4 oxygen atoms are part of peroxide linkages. The oxygen atom of peroxide linkage has oxidation number of

$\displaystyle -1$ . Let, X be the oxidation number of

$\displaystyle Cr$

$\displaystyle X+(-2)+4(-1)=0$

$\displaystyle X-2-4=0$

$\displaystyle X=+6$
Hence, the oxidation number of

$\displaystyle Cr=+6$

$\displaystyle$

957593_1021185_ans_f21f3bf95cbd4d4891769b1fdea91b95.png

In which of the following conditions is the potential for the following half-cell reaction maximum?

$2H^+ + 2e^{-} \longrightarrow H_2$

0%
$1.0 \,M \,HCl$
0%
A solution having $pH = 4$
0%
Pure water
0%
$1.0 \,M \,NaOH$

Explanation

In the given options,

(A). Concentration of

${ H }^{ + }$ in 1.0 M HCl =

$1M$

(B). Since,

$pH$ of solution is 4.

So, concentration of

${ H }^{ + }$ =

${ 10 }^{ -4 }M$

(C).

$pH$ of pure water = 7

So, concetration of

${ H }^{ + }$ =

${ 10 }^{ -7 }M$

(D). Concentration of

${ H }^{ + }$ in 1.0 M

${ H }^{ + }$ is much less than the all the above options.

Since, concentration of

${ H }^{ + }$ is highest in option A. So, potential of option A is maximum .

So, correct answer is option A.

In the case of

$CH_3COOH$ , the oxidation number of carbon of the carboxylic group is:

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

Explanation

Acetic acid is represented as

$\displaystyle CH_3-\underset {\underset {\displaystyle O}{||}}{C}-OH$ . For the purpose of oxidation number calculation, the oxidation number of C atom of methyl group is considered 0, the oxidation number of oxygen atom of

$\displaystyle C=O$ double bond is considered

$\displaystyle -2$ . The oxidation number of oxygen atom of hydroxyl group is considered

$\displaystyle -1$ .

Let

$\displaystyle x$ be the oxidation number of C atom of carboxylic group.

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

$\displaystyle x-3=0$

$\displaystyle x=+3$

Hence, in the case of

$CH_3COOH$ , the oxidation number of carbon of the carboxylic group is

$+3$ .

The oxidation states of Cr in

$\left[ Cr\left( { H }_{ 2 }O \right) \right] { Cl }_{ 3 },\left[ { Cr\left( { C }_{ 6 }{ H }_{ 6 } \right) }_{ 2 } \right] ,$ and

${ K }_{ 2 }\left[ Cr{ \left( CN \right) }_{ 2 }{ \left( O \right) }_{ 2 }{ \left( O \right) }_{ 2 }\left( { NH }_{ 3 } \right) \right]$ respectively are :

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

Explanation

In the given molecules,

In $[Cr(H_2O)]Cl_3$ , let oxidation number of Cr be x, Here the molecule separates as its ions as, $[Cr(H_2O)]^{3+}$ and $3Cl^{-}$ . As $H_2O$ is a neutral molecule its oxidation number is zero. Hence, $x=+3$
In $[Cr(C_6H_6)_2]$ let oxidation number of Cr be y, Benzene is a neutral molecule and hence on observing the cross multiplied factor we get the oxidation number of Cr as 0. Hence, $y=0$ .
In $K_2[Cr(CN)_2(O)_2(NH_3)]$ ,let oxidation number of Cr be z, Here the ions are- $2K^{+}$ and $[Cr(CN)_2(O)_2(NH_3)]^{2-}$ . As $NH_3$ is neutral molecule and $O^{-}$ and $CN^{-}$ is $2 \times (-1)+4 \times (-1)=-6$ and hence oxidation number of Cr is, z-6=-0 $\implies z=+6$

Hence,option C is correct answer.

For the redox reaction,

$Zn+Cu^{2+}_{(0.1M)}\rightarrow Zn^{2+}_{(1M)}+Cu$ taking place in a cell

$E^{\ominus}_{cell}=1.10$ V.

$E^{\ominus}_{Cell}$ for the galvanic cell will be?

0%
$2.14$ V
0%
$1.80$ V
0%
$1.07$ V
0%
$0.82$ V

Explanation

BY Nernst's equation,

$_{c \in 11}=E_{\text {cell }}^{0}-\dfrac{0.0591}{n} \log \theta_{1}$ (at

$5^{\circ} \mathrm{C}$

Here,

$theta=\dfrac{\left[z n^{+2}\right]}{\left[\left(\mu^{+2}\right]\right.}=\dfrac{1}{0 \cdot 1}=10$

So,

$_{\text {cell }}=1.10-\dfrac{0.0591}{2} \log 10 .=1.10-0.0295$

$=1.0705 \mathrm{~V}$

Thus, option (C) is correct.

When zinc reacts with very dilute

$HNO_3$ , the oxidation state of nitrogen changes from:

0%
$+5$ to $+1$
0%
$+5$ to $-3$
0%
$+5$ to $+4$
0%
$+5$ to $+3$

Explanation

Generally, when zinc reacts with nitric acid, it forms zinc nitrate. But when we take very dilute solution of nitric acid, if forms ammonium nitrate as well.

So, here Nitrogen is reducing from

$+5$ to

$-3$ .

$4Zn(s)+\overset {+5}{10HNO_3}(aq)\longrightarrow 4Zn(NO_3)_2(aq)+\overset {-3}{N}H_4NO_3(aq)+3H_2O(l)$

Option B is correct.

	Compounds		O.N.
(A)	$KMn^*O_4$	(1)	+4
(B)	$Ni^*(CO)_4$	(2)	+7
(C)	$[Pt^*(NH_3)Cl_2]Cl_2$	(3)	0
(D)	$Na_2O^*_2$	(4)	-1

The correct code for the O.N. of asterisked atom would be:

0%
A -1, B - 2, C - 3, D - 4
0%
A -4, B - 3, C - 2, D - 1
0%
A -2, B - 3, C - 1, D - 4
0%
A -4, B - 1, C - 2, D - 3

Explanation

(c)

$\left[p t\left(N H_{3}\right) C l_{2}\right] Cl_{2} \rightarrow$ Let,

$0 . N$ of

$P_{t}$ is

$x$ .

$\begin{array}{l}x+0-2=2\\\therefore[x=4]\end{array}$

(D)

$\mathrm{Na}_{2} \mathrm{O}_{2} \rightarrow$ Sodium forms peroxide

$\left(\mathrm{O}_{2}^{2-}\right)$ .

$\operatorname{In} O_{2}^{2-},$ oxidation state of each

$O$ is (-1) .

So, correct option is C

1999841_1049603_ans_b4f4be488544444c851070fc4e617ab4.PNG

The oxidation number of

$Pt$ in

$[Pt({ C }_{ 2 }{ H }_{ 4 }{ ) }Cl_3]^{ - }$ is:

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

Explanation

$[Pt(C_{2}H_{4})cl_{3}]^{-}$

oxidation no. of

$C_{2}H_{4}=0$

oxidation no. of

$cl=-1$

$\therefore x+0+3(-1)=-1$

$\Rightarrow x=+2$

$\Rightarrow$ oxidation no. of

$pt=+2$

The oxidation number of phosphorus vary from:

0%
-3 to +5
0%
-1 to +1
0%
-3 to + 3
0%
-5 to +1

Explanation

Phosphorous belongs to group

$15.$ It has

$5$ electrons in its valence shell. So it can either lose

$5$

$e^-$ to attain

$+5$ oxidation state or gain

$3$

$e^-$ to attain

$-3$ oxidation state. Hence its oxidation state varies from

$-3$ in

$PH_3$ to

$+5$ in

$H_3PO_4.$

Calculate the equilibrium constant for the redox reaction at

$25^o$ C,

$Sr_{(s)}+Mg^{2+}_{(aq)}\rightarrow Sr^{2+}_{(aq)}+Mg_{(s)}$ , that occurs in a galvanic cell.

$E^o_{{Mg^{2+}}/{Mg}}=-2.37$ V and

$E^o_{{Sr}^2+/Sr}=-2.89$ V.

0%
$2.69\times 10^{15}$
0%
$2.69\times 10^{17}$
0%
$3.69\times 10^{17}$
0%
$3.69\times 10^{15}$

Explanation

(i)

${Mg^{2+}}_{(aq)}+2e^- \longrightarrow Mg_{(s)}; E^o_{Mg^{2+}/Mg}=-2.37V$

${Sr^{2+}}_{(aq)}+2e^- \longrightarrow Sr_{(s)}; E^o_{Sr^{2+}/Sr}=-2.89V$

(ii)

$Sr_{(s)} \longrightarrow S{r^{2+}}_{(aq)}+2e^-;E^o_{Sr/Sr^{2+}}=2.89V$

Adding (i) & (ii), we get,

${Sr}_{(s)}+Mg^{2+}_{(aq)} \longrightarrow Sr^{2+}_{(aq)}+Mg_{(s)}; E^o= -2.37+2.89=0.52V$

Now,

$\Delta G^o=-RTlnK_{eq}$

$\therefore -nFE^o= -RT lnK_{eq}$

$\therefore ln K_{eq}=\cfrac {nFE^o}{RT}=\cfrac {2 \times 96500 \times 0.52}{8.314 \times 298}$

$\therefore lnK_{eq}=40.507$

$\therefore K_{eq}= 3.69 \times 10^{17}$

In $Ni{(CO) }_{ 4 }$ the oxidation state of Ni is:

0%
4
0%
0
0%
2
0%
8

Explanation

Let the oxidation state of

$Ni$ be

$x$ .

The oxidation states of

$CO$ is

$0$ .

$\therefore$

$x+0=0$

$\therefore$

$x=0$

Hence, the oxidation state of

$Ni$ in

${ Ni\left( CO \right) }_{ 4 }$ is zero.

Two oxidation states for chlorine are found in the compound:

0%
${ CaOCl }_{ 2 }$
0%
$KCL$
0%
${ KClO }_{ 3 }$
0%
${ Cl }_{ 2 }{ O }_{ 7 }$

Explanation

Solution:- (A)

$CaO{Cl}_{2}$

$CaO{Cl}_{2}$ consist of hypochlorite

$\left( {OCl}^{-} \right)$ ion and chloride ion

$\left( {Cl}^{-} \right)$

The oxidation number of oxygen in hypochlorite ion is -2.

As we know that the sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.

Let the oxidation number of chlorine in hypochlorite ion be x.

$\therefore$ Oxidation no. of chlorine in

${OCl}^{-}$ ion-

$x + \left( -2 \right) = -1$

$\Rightarrow \; x = -1 + 2 = +1$

Therefore, oxidation number of chlorine in

${OCl}^{-}$ is +1.

Oxidation number of chlorine in

${Cl}^{-}$ is -1.

Hence in

$CaO{Cl}_{2}$ , the two

$Cl$ atoms are in different oxidation states i.e., one as

${Cl}^{–}$ having oxidation number of

$-1$ and the other as

${OCl}^{–}$ having oxidation number of

$+1$ .

Oxidation state of cobalt in

$[{ Co }(NH_{ 3 }{ ) }_{ 4 } ({ H }_{ 2 }O)Cl]{ SO }_{ 4 }$ ?

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

Explanation

$\left[ Co {\left( N{H}_{3} \right)}_{4} \left( {H}_{2}O \right) Cl \right] S{O}_{4}$

In the above compound,

Oxidation state of ammonia,

$\left( N{H}_{3} \right) = 0$

Oxidation state of water,

$\left( {H}_{2}O \right) = 0$

Oidation state of chlorine,

$\left( Cl \right) = -1$

Oxidation state of sulphate ion,

$\left( S{O}_{4} \right) = -2$

Let the oxidation state of cobalt,

$\left( Co \right)$ be x.

As we know that sum of oxidation states of all atoms is equal to the overall charge on the compound.

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

$\Rightarrow \; x + 0 + 0 -1 -2 = 0$

$\Rightarrow \; x - 3 = 0$

$\Rightarrow \; x = +3$

Hence, the oxidation state of cobalt,

$\left( Co \right)$ in the compound

$\left[ Co {\left( N{H}_{3} \right)}_{4} \left( {H}_{2}O \right) Cl \right] S{O}_{4}$ is +3.

Find the oxidation state of Cr in the given complex

$K_2[Cr(NO)(NH_3)(CN)_4]$ ,

$\mu$ = 1.73 BM.

0%
+ 1
0%
+ 2
0%
+3
0%
None of these

Explanation

Given,

$k_{2}\left[\operatorname{cr}(N O)\left(N H_{3}\right)(C N)_{4}\right]$ and magnetic moment

$(\mu)=1.73$ .

$\text { We know, }\mu=\sqrt{n(n+2)}=1.73=\sqrt{3}$

$\therefore n=1$ (Number of unpaired electrons)

since, chromium ion has one unpaired electron, so NO must

be unit positive ligand.

Let, oxidation number of

$\mathrm{Cr}$ is

$x$ .

$\begin{aligned}\Rightarrow \quad x+1 &+0-4=-2\\\therefore \quad & x=1\end{aligned}$

Thus oxidation state of

$Cr$ is +1 .

so, option (A) is correct.

Oxidation number of

$Fe$ in

${ Fe }_{ 3 }{ O }_{ 4 }$ are:

0%
$+2$ and $+3$
0%
$+1$ and $+2$
0%
$+1$ and $+3$
0%
None

Explanation

${Fe}_{3}{O}_{4}$ contains

$Fe$ atoms of both

$+2$ and

$+3$ oxidation number. It is a stoichiometric mixture of Ferrous

$\left( FeO \right)$ and Ferric

$\left( {Fe}_{2}{O}_{3} \right)$ oxides combined as

$FeO.{Fe}_{2}{O}_{3}$ .

Oxidation state of oxygen in

$FeO$ and

${Fe}_{2}{O}_{3}$ is -2.

Let the oxidation state of Fe in

$FeO$ and

${Fe}_{2}{O}_{3}$ be x and y respectively.

As we know that the sum of the oxidation states of all the atoms or ions in a neutral compound is zero.

Therefore,

Oxidation state of

$Fe$ in

$FeO$ -

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

$\Rightarrow \; x = 2$

Oxidation state of

$Fe$ in

${Fe}_{2}{O}_{3}$ -

$2 \left( y \right) + 3 \left( -2 \right) = 0$

$\Rightarrow \; 2y = 6$

$\Rightarrow \; y = 3$

Hence the oxidation number of

$Fe$ in

${Fe}_{3}{O}_{4}$ are +2 and +3 respectively.

In which of the following pair oxidation number of

$Fe$ is same:

0%
${ K }_{ 3 }{ Fe(CN) }_{ 6 },{ Fe }_{ 2 }{ O }_{ 3 }$
0%
${ Fe(CN) }_{ 5 },{ Fe }_{ 2 }{ O }_{ 3 }$
0%
${ Fe }_{ 2 }{ O }_{ 3, }FeO$
0%
${ Fe }_{ 2 }({ SO }_{ 4 })_{ 3 },{ K }_{ 4 }Fe({ C) }_{ 6 }$

Explanation

Let the oxidation state of Fe in

${K}_{3}Fe{\left( CN \right)}_{6}$ and

${Fe}_{2}{O}_{3}$ be x and y respectively.

As we know that the sum of the oxidation states of all the atoms or ions in a neutral compound is zero.

Therefore,

Oxidation state of

$Fe$ in

${K}_{3}Fe{\left( CN \right)}_{6}$ -

$\because \; CN$ is a negative ligand.

$\therefore$ Oxidation number of

$CN$ is -1.

Oxidation number of

$K$ is +1.

$\therefore \; 3 \left( +1 \right) + x + 6 \left( -1 \right) = 0$

$\Rightarrow \; x - 3 = 0$

$\Rightarrow \; x = 3$

Oxidation state of

$Fe$ in

${Fe}_{2}{O}_{3}$ -

Oxidation no. of

$O$ is -2.

$2 \left( y \right) + 3 \left( -2 \right) = 0$

$\Rightarrow \; 2y = 6$

$\Rightarrow \; y = 3$

Hence the oxidation number of

$Fe$ in

${K}_{3}Fe{\left( CN \right)}_{6}$ and

${Fe}_{2}{O}_{3}$ is same.

Oxidation number of carbon in sucrose is:

0%
$0$
0%
$+12$
0%
$-12$
0%
$+7$

Explanation

Molecular formula of sucrose

$= {C}_{12}{H}_{22}{O}_{11}$

Overall charge on sucrose

$= 0$

Oxidation no. of oxygen in sucrose

$= -2$

Oxidation no. of hydrogen in sucrose

$= +1$

Let the oxidation no. of carbon be x.

Sum of oxidation no. of atoms = overall charge on atom

$\Rightarrow \; 12 \times x + 22 \times \left( +1 \right) + 11 \times \left( -2 \right) = 0$

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

$\Rightarrow \; 12x = 0$

$\Rightarrow \; x = 0$

Hence, the oxidation number of carbon in sucrose is 0.

Oxidation number of P in

${ KH }_{ 2 }{ PO }_{ 3 }$ is:

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

Explanation

Molecular formula

$= K{H}_{2}P{O}_{3}$

Overall charge on compound

$= 0$

Oxidation no. of oxygen in compound

$= -2$

Oxidation no. of hydrogen in compound

$= +1$

Oxidation no. of potassium in compound

$= +1$

Let the oxidation no. of phosphorous be x.

Sum of oxidation no. of atoms = overall charge on atom

$\Rightarrow \; 1 \times \left( +1 \right) + 2 \times \left( +1 \right) + 1 \times x + 3 \times \left( -2 \right) = 0$

$\Rightarrow \; 1 + 2 + x - 6 = 0$

$\Rightarrow \; x - 3 = 0$

$\Rightarrow \; x = +3$

Hence, the oxidation number of P is +3.

Oxidation number of Cr atom in

$CrO_5$ and

$K_3CrO_8$ respectively:

0%
+6 , +6
0%
+5 , + 6
0%
+6 , + 5
0%
+ 5 , + 5

Explanation

Structure of

${ K }_{ 3 }{ Cr }{ O }_{ 8 }$

Refer to fig 1

Structure of

${ Cr }{ O }_{ 5 }$

Refer fig 2.

This oxidation state of

${ Cr }{ O }_{ 5 }$ and

${ K }_{ 3 }{ Cr }{ O }_{ 8 }$ is

$+6$ and

$+5$ respectively.

1093711_1055115_ans_f9d2abe943ee404c89607e2ca2c2ee43.JPG

Oxidation number of

$Cr$ in the following complex is:

0%
3
0%
6
0%
4
0%
5

Explanation

Oxidation state of

$Cr$ in

$Cr(H_2O)_6^{3+}$ is the charge on the ion=

$+3$

$H_2O$ units do not contribute to the charge on the ion, the only atom that contribute is the

$Cr$ atom.

In substance

$Mg({ HXO }_{ 3 })$ , the oxidation number of X is :-

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

Explanation

Molecular formula

$= Mg \left( HX{O}_{3} \right)$

Overall charge on compound

$= 0$

Oxidation no. of oxygen in compound

$= -2$

Oxidation no. of hydrogen in compound

$= +1$

Oxidation no. of magnesium im compound

$= +2$

Let the oxidation no. of X be x.

Sum of oxidation no. of atoms = overall charge on atom

$\Rightarrow \; 1 \times \left( +2 \right) + 1 \times \left( +1 \right) + 1 \times x + 3 \times \left( -2 \right) = 0$

$\Rightarrow \; 2 + 1 + x - 6 = 0$

$\Rightarrow \; x - 3 = 0$

$\Rightarrow \; x = 3$

Hence, the oxidation number of X is 3.

Which of the following, oxidation number of carbon is -1?

0%
${ CH }_{ 4 }$
0%
${ C }_{ 6 }{ H }_{ 6 }$
0%
${ C }Cl_{ 4 }$
0%
$HCHCl$

Explanation

For benzene let oxidation number of C be x,

$6x+6\times(+1) \implies x=-1$

The oxidation state of carbon in benzene is -1 and the oxidation state of hydrogen in benzene is +1. In a C-H bond(this bond is non-polar covalent), the H is treated as if it has an oxidation state of +1. This means that every C-H bond will decrease the oxidation state of carbon by 1.

In which reaction, there is change in oxidation number of N atom ?

0%
$2NO_2 \rightleftharpoons N_2O_4$
0%
$2NO_2 + H_2O \rightarrow HNO_3 + HNO_2$
0%
$NH_4OH \rightarrow NH^+_4 + OH^-$
0%
$N_2O_5 + H_2O \rightarrow 2HNO_3$

Reaction

$(A) S^{-2} + 4 H_2O_2 \rightarrow SO^{2-}_4 + 4H_2O$

$(B)Cl_2 + H_2O_2 \rightarrow 2HCl + O_2$
The true statement regarding the above reactions is :

0%
$H_2O_2$ act as reductant in both the reactions.
0%
$H_2O_2$ acts as oxidant in reaction (A) and reductant in reaction (B).
0%
$H_2O_2$ acts as an oxidant in reaction in both the reactions.
0%
$H_2O_2$ acts as reductant in reaction (A) and oxidant in reaction (B).

The brown ring complex compound is formulated as

$[Fe(H_2O)_5NO]SO_4$ . The oxidation state of iron is:

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

Explanation

Brown ring complex is an iron complex of great significance in inorganic chemistry and salt analysis.

Its formula:

$[Fe(H_2O)_5NO]SO_4$ (coordination compound)

In this complex,

$NO$ is in oxidation state of

$+1.$

So,

$[Fe(H_2O)_5NO]^{+2}$

$SO_4^{-2}$

$\overline {x}$

$\overline {0}$

$\overline {+1}$

$x+1=+2$

$x=+1$

Oxidation number of oxygen atom in

$O_3$ molecules is:

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

Explanation

Its common name is ozone and plays a very important role of protecting earth from harmful UV rays by forming its blanket around earth.

average oxidation state

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

1049514_1059540_ans_b2ef36d580ca40e8923ec1b115032f91.PNG

Which of the following is not a redox change ?

0%
$2H_2S + SO_2 \rightarrow 2H_2O + 3S$
0%
$2BaO + O_2 \rightarrow 2BaO_2$
0%
$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$
0%
$2KClO_3 \rightarrow 2KCl + 3O_2$

Explanation

$Ba{O}_{2} + {H}_{2}S{O}_{4} \longrightarrow BaS{O}_{4} + {H}_{2}{O}_{2}$

A redox reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.

$Ba{O}_{2} + {H}_{2}S{O}_{4} \longrightarrow BaS{O}_{4} + {H}_{2}{O}_{2}$

As we know that the oxidation no. of oxygen in peroxide is -1.

Since

$Ba{O}_{2}$ and

${H}_{2}{O}_{2}$ are peroxides, hence the oxidation number of oxygen in

$Ba{O}_{2}$ and

${H}_{2}{O}_{2}$ is -1 as the oxidation no. of oxygen in peroxide is -1.

$\therefore$ Oxidation number of barium and hydrogen in

$Ba{O}_{2}$ and

${H}_{2}{O}_{2}$ are +2 and +1 respectively.

The oxidation number of barium and hydrogen in

$BaS{O}_{4}$ and

${H}_{2}S{O}_{4}$ are +2 and +1 respectively.

The oxidation number of sulphur in

${H}_{2}S{O}_{4}$ and

$BaS{O}_{4}$ in same, i.e., +2.

The oxidation number of oxygen in

${H}_{2}S{O}_{4}$ and

$BaS{O}_{4}$ in same, i.e., -2.

Since the oxidation state of any atom in the above reaction is not changing, hence the above reaction is not a redox reaction.

Which of the following is not a redox reaction ?

0%
$H_2 + Cl_2 \rightarrow 2HCl$
0%
$NaOH + HCl \rightarrow NaCl + H_2O$
0%
Photosynthesis
0%
Cell respiration

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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