Redox Reactions Test

Result

Redox Reactions Test - 56

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The oxidation number of Cr is +6 in:

A
$$K_{2}CrO_{4}$$

B
$$K_{2}Cr_{2}O_{7}$$

C
$$KCrO_{3}CI$$

D
All of these

Solution

Chromium is hexavalent in this compound - each one has an oxidation state of $$+6 $$ to balance out. $$-2 × 3$$ oxygen ions and $$+1 $$potassium ions and $$Cl $$ has $$-1$$ (the net total has to be zero for a neutral compound).
Question 2
1 / -0

The oxidation number of Cl in $$Ca(ClO_{2})_{2}$$:

A
+2

B
+4

C
+6

D
+3

Solution

Its a $${(Cl{O}_{2})}^{-1}$$ ion. For finding the oxidation number of Cl , the sum of Oxidation Numbers of Cl and O should be equal to -1

x + (-4) = -1
x = +3

So Oxidation of Cl is +3
Question 3
1 / -0

Oxidation number of $$Fe$$ in $$Fe_{3}O_{4}$$ is:

A
$$\frac { 1 }{ 2 }$$

B
$$\frac { 2 }{ 6 }$$

C
$$\frac { 8 }{ 3 }$$

D
$$\frac { 3 }{ 2 }$$

Solution

Let the average oxidation number of $$Fe$$ in $$Fe_3O_4$$ be $$x$$

Since the net charge on the molecule is zero,

$$\Rightarrow 3\times x+ (-2)\times 4=0$$
$$\Rightarrow x=\dfrac{8}{3}$$
Question 4
1 / -0

The order of the oxidation state of the phosphorus atom in $$ H_{3}PO_{2} $$, $$ H_{3}PO_{4} $$, $$ H_{3}PO_{3} $$ and $$ H_{4}P_{2}O_{6} $$

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

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

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

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

Solution

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
Question 5
1 / -0

The oxidation states of $$Cr$$ in $$ \left [ Cr\left (H_{2}O_{6} \right ) \right ]Cl_{3} $$, $$ \left [ Cr\left (C_{6} H_{6} \right )_{2} \right ] $$ and $$ K_{2}\left [ Cr\left (CN_{2} \right ) \left ( O_{2} \right )\left ( O_{2} \right )\left ( NH_{3} \right ) \right] $$ respectively are

A
+3, 0 and +4

B
+3, +4 and +6

C
+3, +2 and +4

D
+3, 0 and +6

Solution

The oxidation states of $$Cr$$ in $$ \left [ Cr\left (H_{2}O_{6} \right ) \right ]Cl_{3} $$, $$ \left [ Cr\left (C_{6} H_{6} \right )_{2} \right ] $$ and $$ K_{2}\left [ Cr\left (CN_{2} \right ) \left ( O_{2} \right )\left ( O_{2} \right )\left ( NH_{3} \right ) \right] $$ respectively are $$\mathrm{+3, 0 ,+6}$$.

Hence, Option "D" is the correct answer.
Question 6
1 / -0

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

A
+3,0 and +4

B
+3,+4 and +6

C
+3,+2 and +4

D
+3,+0 and +6

Solution

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.
Question 7
1 / -0

Which of the following reaction is an example of redox reactions

A
$$ XeF_{4}+O_{2}F_{2}\rightarrow XF_{6} +O_{2} $$

B
$$ XeF_{2}+PF_{5}\rightarrow \left [ XeF \right ]^{+}PF_{6}^{-} $$

C
$$ XeF_{6}+H_{2}O\rightarrow XeoF_{4}+2HF $$

D
$$ XeF_{6}+2H_{2}O\rightarrow Xeo_{2}F_{2}+4HF $$

Solution
Question 8
1 / -0

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

A
$${ 4KClO }_{ 3 }\rightarrow { 3KClO }_{ 4 }+KCl$$

B
$${ SO }_{ 2 }+{ 2H }_{ 2 }S\rightarrow { 2H }_{ 2 }O+3S$$

C
$${ BaO }_{ 2 }+{ H }_{ 2 }{ SO }_{ 4 }\rightarrow { BaSO }_{ 4 }+H_{2}O_{2}$$

D
$$2BaO +{ O }_{ 2 }\rightarrow { 2BaO }_{ 2 }$$

Solution

Reaction shown in option (C) is not a redox reaction its is a double displacement reaction and hence no change in valency.
Question 9
1 / -0

Oxidation number of $$N$$ in $${({NH}_{4})}_{2}{SO}_{4}$$ is

A
$$-1/3$$

B
$$-1$$

C
$$+1$$

D
$$-3$$

Solution

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

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

Let, $$x$$ be the oxidation state of $$N.$$

$$x+4\times 1=+1;\ \Rightarrow x=-3$$
Question 10
1 / -0

The oxidation numbers of $$S$$ in $$S_{8},\ S_{2}F_{2}$$ and $$H_{2}S$$ respectively are

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

B
$$+2,\ +1$$ & $$-2$$

C
$$0,\ +1$$ & $$+2$$

D
$$-2,\ +1$$ & $$-2$$

Solution

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.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 56

Result

Redox Reactions Test - 56

Score

-

Rank

-

SHARING IS CARING

Question 1

$$K_{2}CrO_{4}$$

$$K_{2}Cr_{2}O_{7}$$

$$KCrO_{3}CI$$

All of these

Solution

Question 2

+2

+4

+6

+3

Solution

Question 3

$$\frac { 1 }{ 2 }$$

$$\frac { 2 }{ 6 }$$

$$\frac { 8 }{ 3 }$$

$$\frac { 3 }{ 2 }$$

Solution

Question 4

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

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

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

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

Solution

Question 5

+3, 0 and +4

+3, +4 and +6

+3, +2 and +4

+3, 0 and +6

Solution

Question 6

+3,0 and +4

+3,+4 and +6

+3,+2 and +4

+3,+0 and +6

Solution

Question 7

$$ XeF_{4}+O_{2}F_{2}\rightarrow XF_{6} +O_{2} $$

$$ XeF_{2}+PF_{5}\rightarrow \left [ XeF \right ]^{+}PF_{6}^{-} $$

$$ XeF_{6}+H_{2}O\rightarrow XeoF_{4}+2HF $$

$$ XeF_{6}+2H_{2}O\rightarrow Xeo_{2}F_{2}+4HF $$

Solution

Question 8

$${ 4KClO }_{ 3 }\rightarrow { 3KClO }_{ 4 }+KCl$$

$${ SO }_{ 2 }+{ 2H }_{ 2 }S\rightarrow { 2H }_{ 2 }O+3S$$

$${ BaO }_{ 2 }+{ H }_{ 2 }{ SO }_{ 4 }\rightarrow { BaSO }_{ 4 }+H_{2}O_{2}$$

$$2BaO +{ O }_{ 2 }\rightarrow { 2BaO }_{ 2 }$$

Solution

Question 9

$$-1/3$$

$$-1$$

$$+1$$

$$-3$$

Solution

Question 10

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

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

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

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

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.