Redox Reactions Test

Result

Redox Reactions Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The oxidation state of $$I$$ in $${H_4IO}_6^{-}$$ is :

A
+7

B
-1

C
+5

D
+1

Solution

Let oxidation state of I in $${H_4IO}_6^{-}$$ is $$x$$.
Since the overall charge on the complex is $$-1$$, the sum of oxidation states of all elements in it should be equal to $$-1$$.
Therefore, $$4+x-12 = -1$$ (Here, oxidation state of H is +1 and oxygen is -2)
or, $$ x= 7$$
Question 2
1 / -0

The oxidation number of $$Ni$$ in $$K_4[Ni(CN)_4]$$ is :

A
$$+1$$

B
$$+2$$

C
$$-1$$

D
$$0$$

Solution

Let the oxidation state of $$Ni$$ be $$x$$ in $$K_4[Ni(CN)_4]$$.
Since the overall charge on the complex is $$0$$, the sum of oxidation states of all elements in it should be equal to $$0$$.
Therefore, $$4+x-4=0$$
or, $$x=0$$
Question 3
1 / -0

Maximum oxidation state is present in :

A
$$CrO_2Cl_2$$ and $${MnO}^{-}_4$$

B
$$MnO_2$$

C
$$[Fe(CN)_6]^{3-}$$ and $$[Co(CN)_6]^{3-}$$

D
$$MnO$$

Solution

COMPOUND OXIDATION STATE
$$CrO_2Cl_2$$ +6
$${MnO}^{-}_4$$ +7 (Maximum)
$$MnO_2$$ +4
$$[Fe(CN)_6]^{3-}$$ +3
$$[Co(CN)_6]^{3-}$$ +3
$$MnO$$ +2
Question 4
1 / -0

When $$SO_2$$ is passed in a solution of potassium iodate, the oxidation state of iodine changes from :

A
$$+5$$ to $$0$$

B
$$+5$$ to $$-1$$

C
$$-5$$ to $$0$$

D
$$-7$$ to $$-1$$

Solution

The reaction between $$SO_2$$ and $$KIO_3$$ is given below:
$$5SO_2 + 4H_2O + 2KIO_3 \rightarrow 4H_2SO_4 + I_2 + K_2SO_4$$
+4 +5 +6 0
Thus, we can see that oxidation state of iodine changes fron $$+5$$ in $$KIO_3$$ to $$0$$ in $$I_2$$.
Question 5
1 / -0

The oxidation number of phosphorus in Ca(H2PO2)2 is:

A
$$+3$$

B
$$+2$$

C
$$+1$$

D
$$-1$$
Question 6
1 / -0

Which process is not involved in the reaction?

$$KO_2+H_2O+CO_2\rightarrow KHCO_3+O_2$$

A
Hydrolysis

B
Auto-redox

C
Acid-base reaction

D
Non redox change

Solution

$$KO_2+H_2O+CO_2\rightarrow KHCO_3+O_2$$
$$-\frac{1}{2}$$ $$-2$$ $$0$$
So, its a auto-redox reaction.
Also, its acid-base and hydrolysis reaction.
However, non redox change doesn't happen here.
Question 7
1 / -0

Oxidation number of P in $$P_2O^{4-}_7 $$ is :

A
+3

B
+4

C
+5

D
+6

Solution

Let the oxidation number of P in $$P_2O^{4-}_7 $$ be $$x$$.
Since, the overall charge on complex is $$-4$$, sum of oxidation states of all elements in it should be equal to $$-4$$.
Hence, $$2x+7(-2) = -4$$
or, $$x = +5$$
Question 8
1 / -0

On combustion of $$CH_4$$ to $$CO_2$$ and $$H_2$$, the oxidation number of carbon changes by :

A
8

B
3

C
0

D
4

Solution

Reaction:
$$CH_4+O_2\rightarrow CO_2+2H_2$$
-4 +4
Oxidation number of C changes by $$4-(-4) =8$$.
Question 9
1 / -0

The oxidation number of Pt in $$[Pt(C_2H_4)Cl_3]^-$$ is :

A
+1

B
+2

C
+3

D
+4

Solution

Let oxidation state of Pt be $$x$$ in $$[PtC_2H_4)Cl_3]^-$$.
Since the overall charge on the complex is $$-1$$, the sum of oxidation states of all elements in it should be equal to $$-1$$.
Therefore, $$ x + 0 - 3 = -1$$
or, $$x = +2$$
Question 10
1 / -0

Among the following, identify the species with an atom in $$+4$$ oxidation state.

A
$$MnO^-_4$$

B
$$Cr(CN)_6^{3-}$$

C
$$NiF^{2-}_6$$

D
$$CrO_2Cl_2$$

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 22

Result

Redox Reactions Test - 22

Score

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Rank

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SHARING IS CARING

Question 1

+7

-1

+5

+1

Solution

Question 2

$$+1$$

$$+2$$

$$-1$$

$$0$$

Solution

Question 3

$$CrO_2Cl_2$$ and $${MnO}^{-}_4$$

$$MnO_2$$

$$[Fe(CN)_6]^{3-}$$ and $$[Co(CN)_6]^{3-}$$

$$MnO$$

Solution

Question 4

$$+5$$ to $$0$$

$$+5$$ to $$-1$$

$$-5$$ to $$0$$

$$-7$$ to $$-1$$

Solution

Question 5

$$+3$$

$$+2$$

$$+1$$

$$-1$$

Question 6

Hydrolysis

Auto-redox

Acid-base reaction

Non redox change

Solution

Question 7

+3

+4

+5

+6

Solution

Question 8

8

3

0

4

Solution

Question 9

+1

+2

+3

+4

Solution

Question 10

$$MnO^-_4$$

$$Cr(CN)_6^{3-}$$

$$NiF^{2-}_6$$

$$CrO_2Cl_2$$

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