Redox Reactions Test

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Redox Reactions Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$HClO_4$$

B
$$HClO_3$$

C
$$HCl$$

D
$$HOCl$$

Solution

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 $$HCl$$.
Question 2
1 / -0

What are the oxidation states of phosphorus in the following compounds?
$$H_3PO_2,H_3PO_4,Mg_2P_2O_7,PH_3,HPO_3$$

A
$$+1, +3, +3, +3, +5$$

B
$$+3, +3, +5, +5, +5$$

C
$$+1, +2 ,+3, +5, +5$$

D
$$+1, +5 ,+5 ,-3 , +5$$

Solution

Let the oxidation number of phosphorous be $$x$$.
Oxidation number of $$H=+1$$, $$Mg=+2$$ and of $$O=-2$$.
1. $$H_3PO_2$$: $$3 \times 1+x +2 \times (-2)=0$$
$$3+x-4=0$$ or $$x=+1$$
2. $$H_3PO_4$$: $$3 \times 1+x +4 \times (-2)=0$$
$$3+x-8=0$$ or $$x=+5$$
3. $$Mg_2P_2O_7$$: $$2 \times 2+2x +7 \times (-2)=0$$
$$4+2x-14=0$$ or $$x=+5$$
4. $$PH_3$$: $$x+3 \times 1=0$$
$$x=−3$$
5. $$HPO_3$$: $$1+x+3 \times (-2)=0$$
$$1+x-6=0$$ 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.
Question 3
1 / -0

The correct sequence of the oxidation state of Fe, Ta, P and N elements in the given compounds are respectively :
$$ Na_2[Fe(CN)_5NO],\ K_2TaF_7,\ Mg_2P_2O_7,\ Na_2S_4O_6,\ N_3H$$

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

B
$$+5,+3,+5,+3,+\frac { 1 }{ 3 }$$

C
$$+3, +3 ,+5 ,+5 ,-\frac { 1 }{ 3 }$$

D
$$+5 , +5 ,+3 ,+2.5, +\frac { 1 }{ 3 }$$

Solution

Let the oxidation number of the desired atom be $$x$$.
Oxidation number of $$H=Na=+1,Mg=+2,CN=F=-1, $$ and of $$O=-2$$.

A. $$Na_2[Fe(CN)_5NO]$$: $$2 \times 1+x+5 \times (-1) + 0 = 0$$
$$x+2-5=0$$ "or" $$x=+3$$

B. $$K_2TaF_7$$: $$2 \times 1+x+7 \times (-1) = 0$$
$$x+2-7=0$$ "or" $$x=+5$$

C. $$Mg_2P_2O_7$$: $$2 \times 2+2x+7 \times (-2) = 0$$
$$2x+4-14=0$$ "or" $$x=+5$$

D. $$Na_2S_4O_6$$: $$2 \times 1+4x+6 \times (-2) = 0$$
$$4x+2-12=0$$ "or" $$x=+2.5$$

E. $$N_3H$$: $$3 \times x+1 = 0$$
"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.
Question 4
1 / -0

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

A
Phosphorus in +1 oxidation state - $$H_3PO_2$$

B
Chlorine in +7 oxidation state - $$HClO$$

C
Chromium in +6 oxidation state -$$ CrO_2Cl_2$$

D
Carbon in 0 oxidation state - $$C_{12}H_{22}O_{11}$$

Solution

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
(C) Chromium in $$CrO_2Cl_2$$: $$x+ 2 \times (−2)+2 \times (−1)=0$$
$$x-6=0$$ 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 - $$C_{12}H_{22}O_{11}$$
Question 5
1 / -0

Which of the following oxidation numbers is not correctly matched?

A
$$P$$ in $$ NaH_2PO_4 = +5$$

B
$$Ni$$ in $$ [ Ni(CN)_6]^{4-} = +2$$

C
$$P$$ in $$Mg_2P_2O_7 = +6$$

D
$$Cr$$ in $$(NH_4)_2Cr_2O_7 =+6$$

Solution

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 $$x=+2$$
(C) P in $$Mg_2P_2O_7$$: $$2 \times 2+2 \times x+7 \times (−2)=0$$
$$2x-10=0$$ 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 $$x=+6$$
Question 6
1 / -0

Oxidation number of sulphur in peroxomonosulphuric acid $$(H_2SO_5)$$ is:

A
+4

B
+2

C
+6

D
-2

Solution

$$H_2SO_5$$ 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 $$H=+1$$
In $$H_2SO_5$$: $$2 \times 1+x+3 \times (−2)+2 \times (−1)=0$$
$$x+2-6-2=0$$ or $$x=+6$$
thus Oxidation number of sulphur in peroxomonosulphuric acid ($$H_2SO_5$$) is $$+6$$.
Question 7
1 / -0

Mark the correct statement from the following.

A
Copper metal can be oxidised by $$Zn^{2+}$$ ions

B
Oxidation number of phosphorus in $$ P_4$$ is 4

C
An element in the highest oxidation state acts only as a reducing agent.

D
The element which shows highest oxidation number of +8 is $$OsO_4$$

Solution

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
$$x+4 \times (-2)=0$$
$$x=+8$$.
Question 8
1 / -0

The oxidation state of $$Fe$$ in $$K_4[Fe(CN)_6]$$ is:

A
+2

B
+3

C
+4

D
+6

Solution

Let the oxidation number of $$Fe$$ be $$x$$.
Oxidation number of $$CN=-1$$, $$K=+1$$.
In $$K_4[Fe(CN)_6]$$: $$4 \times 1+x+6 \times (-1)=0$$
$$4+x-6=0$$ or $$x=+2$$
The oxidation state of Fe in $$K_4[Fe(CN)_6]$$ is +2.
Question 9
1 / -0

What is the oxidation state of $$P$$ in $$ Ba(H_2PO_2)_2$$?

A
+3

B
+2

C
+1

D
-1

Solution

Let the oxidation number of $$P$$ be $$x$$.
Oxidation number of $$H=+1$$, $$Ba=+2$$ and of $$O=-2$$.
In $$Ba(H_2PO_2)_2$$: $$2 + 2\times (2\times 1+x+2 \times (-2))=0$$
$$2+2(2+x-4)=0$$ or $$2x=2$$
$$\therefore x=+1$$
The oxidation state of P in $$Ba(H_2PO_2)_2$$ is +1.
Question 10
1 / -0

In which of the following compounds, an element exhibits two different oxidation states ?

A
$$NH_2OH$$

B
$$NH_4NO_3$$

C
$$N_2H_4$$

D
$$N_3H$$

Solution

In $$NH_2OH$$ oxidation of N is -1
$$NH_4NO_3$$ exists as $$NH_4^+. NO_3^-$$,
thus the oxidation state of N in $$NH_{4}^{+}$$ is -3 while in $$NO_{3}^{-}$$ is +5.
In $$N_2H_4$$ oxidation of N is -2
In $$N_3H$$ oxidation of N is $$\frac {-1}{3}$$

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Redox Reactions Test - 43

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Redox Reactions Test - 43

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Question 1

$$HClO_4$$

$$HClO_3$$

$$HCl$$

$$HOCl$$

Solution

Question 2

$$+1, +3, +3, +3, +5$$

$$+3, +3, +5, +5, +5$$

$$+1, +2 ,+3, +5, +5$$

$$+1, +5 ,+5 ,-3 , +5$$

Solution

Question 3

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

$$+5,+3,+5,+3,+\frac { 1 }{ 3 }$$

$$+3, +3 ,+5 ,+5 ,-\frac { 1 }{ 3 }$$

$$+5 , +5 ,+3 ,+2.5, +\frac { 1 }{ 3 }$$

Solution

Question 4

Phosphorus in +1 oxidation state - $$H_3PO_2$$

Chlorine in +7 oxidation state - $$HClO$$

Chromium in +6 oxidation state -$$ CrO_2Cl_2$$

Carbon in 0 oxidation state - $$C_{12}H_{22}O_{11}$$

Solution

Question 5

$$P$$ in $$ NaH_2PO_4 = +5$$

$$Ni$$ in $$ [ Ni(CN)_6]^{4-} = +2$$

$$P$$ in $$Mg_2P_2O_7 = +6$$

$$Cr$$ in $$(NH_4)_2Cr_2O_7 =+6$$

Solution

Question 6

+4

+2

+6

-2

Solution

Question 7

Copper metal can be oxidised by $$Zn^{2+}$$ ions

Oxidation number of phosphorus in $$ P_4$$ is 4

An element in the highest oxidation state acts only as a reducing agent.

The element which shows highest oxidation number of +8 is $$OsO_4$$

Solution

Question 8

+2

+3

+4

+6

Solution

Question 9

+3

+2

+1

-1

Solution

Question 10

$$NH_2OH$$

$$NH_4NO_3$$

$$N_2H_4$$

$$N_3H$$

Solution

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