Redox Reactions Test

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Redox Reactions Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

The oxidation number of manganese in potassium manganate is:

A
$$+7$$

B
$$+6$$

C
$$+4$$

D
$$+2$$

Solution

$$K_2MnO_4$$: Potassium manganate.

For 2$$K^{+}$$, the oxidation state is $$+2$$.

For $$MnO_{4}^{ 2-}$$; $$Mn + 4\times -2 = -2$$ or, $$x = +6$$.

therefore, $$Mn$$ has the oxidation state of $$+6$$.

Option B is correct.
Question 2
1 / -0

In $$ICl_{3}$$, oxidation number of iodine and chlorine are, respectively

A
$$0$$ and $$0$$

B
$$+3$$ and $$-1$$

C
$$-1$$ and $$+3$$

D
$$-3$$ and $$+1$$

Solution

Chlorine is more electronegative than iodine. Hence, its oxidation number will be $$-1$$.
Since $$3$$, chlorine atoms are present and the molecule is neutral, the oxidation number of iodine will be $$+3$$.

Hence, the correct option is $$B$$
Question 3
1 / -0

Oxidation state of phosphorus in pyrophospate ion $$(P_{2}O_{7}^{-4})$$ is:

A
$$+7$$

B
$$+3$$

C
$$+8$$

D
$$+5$$

Solution

Let $$x$$ be the oxidation state of phosphorus in pyrophosphate ion $$ (P_2O^{4-}_7)$$.

The oxidation state of oxygen is $$-2$$. The charge on the ion is $$-4$$.

Hence, the sum of the oxidation states of phosphorus and oxygen must be $$-4$$.

$$2(x)+7(-2)=-4$$

or, $$x=+5$$
Question 4
1 / -0

Characteristic oxidation number of atoms in free metal is

A
$$-1$$

B
$$1$$

C
$$0$$

D
any number

Solution

The characteristic oxidation number of atoms in free elements is zero.
Question 5
1 / -0

In which of the following substances, sulphur has the lowest oxidation number?

A
$$H_{2}SO_{4}$$

B
$$SO_{2}$$

C
$$H_{2}SO_{3}$$

D
$$H_{2}S$$

Solution

In $$H_2S$$, the oxidation state of sulfur is $$-2$$.

In other given compounds, sulfur has positive oxidation states.

The oxidation states of sulfur in $$H_2SO_4$$, $$SO_2$$ and $$H_2SO_3$$ are $$+6$$, $$+4$$ and $$+4$$ respectively.

Option D is correct.
Question 6
1 / -0

In calcium hypophosphite, oxidation number of phosphorus is

A
$$0$$

B
$$+1$$

C
$$+3$$

D
$$+5$$

Solution

The charge on the hypophosphite, $$H_2PO_2^-$$ ion is $$-1$$ .
The oxidation numbers of H and O are $$+1$$ and $$-2$$ respectively.
Let $$x$$ be the oxidation number of phosphorus, then the sum of the oxidation numbers will be equal to $$-1$$, which is the charge on the ion. Hence, $$2(+1)+x+2(-2)= - 1$$ or, $$x= +1$$. So, the oxidation state of phosphorus is +1.
Question 7
1 / -0

Oxidation number of carbon in $$SCN^{-}$$ ion is

A
$$+2$$

B
$$-2$$

C
$$+4$$

D
$$-4$$

Solution

In thiocyanate ion ($$SCN^-$$), carbon atom is attached to $$N$$ via single bond and $$S$$ via triple bond. The oxidation numbers of S and N are $$-2$$ and $$-3$$ respectively.
Let, $$x$$ be the oxidation number of C. The total charge on ion is $$-1$$. Hence, $$-2+x+(-3)=-1$$ or, $$x=+4$$.
The structure of $$SCN^-$$ is given below:
Question 8
1 / -0

The oxidation state of phosphorus is minimum in:

A
$$P_{4}O_{6}$$

B
$$NaH_{2}PO_{2}$$

C
$$PH_{3}$$

D
$$Na_{3}PO_{4}$$

Solution

Hint: The sum of oxidation states of all the atoms in a neutral compound is equal to zero.

Step 1: Calculate the oxidation state of phosphorous in all compounds.

Let the oxidation states of $$P$$ be $$x$$ in all the compounds.

A) $$P_4O_{6} \rightarrow (4\times x) + (6\times (-2)) = 0 \implies x= +3$$

B) $$NaH_2PO_2 \rightarrow (1\times (+1)) + (2\times (+1)) + (1\times x) + (2 \times (-2)) =0 \implies x= +1$$

C) $$PH_3 \rightarrow (1\times x) + (3 \times (+1)) = 0 \implies x= -3$$

D) $$Na_3PO_4 \rightarrow (3\times (+1)) + (1\times x) + (4 \times (-2)) = 0 \implies x= +5$$

Step 2: Hence, the order of oxidation states is $$C<B<A<D$$.

Final answer: $$PH_3$$

So, the correct option is $$C$$
Question 9
1 / -0

The sum of the oxidation number of the carbon atoms in $$CH_{3} CHO$$ is:

A
$$-2$$

B
$$+2$$

C
$$-4$$

D
$$-1$$

Solution

Let $$x$$ be the sum of the oxidation states of carbon atoms in acetaldehyde $$(CH_3CHO)$$.

The oxidation numbers of hydrogen and oxygen are $$+1$$ and $$-2$$ respectively.

Let, $$x$ be the sum of the oxidation state of C.

For a neutral molecule, the sum of the oxidation numbers of all the molecules is zero.

Hence, $$x+4\times (+1)+(-2)=0$$ or, $$x=-2$$.

Hence, the correct option is $$A$$
Question 10
1 / -0

Which of the following change is neither oxidation nor reduction?

A
$$K_{2}Cr_{2}O_{7}$$ to $$K_{2}CrO_{4}$$

B
$$Na_{2}SO_{4}$$ to $$NaHSO_{4}$$

C
$$CuCl_{2}$$ to $$CuSO_{4}$$

D
All the above

Solution

$$K_{2}Cr_{2}^{+6}O_{7}\rightarrow K_{2}Cr^{+6}O_{4}$$
$$Na^{1+}$$ $$\rightarrow$$ $$Na^{1+}$$
$$Cu^{2+}$$ $$\rightarrow$$ $$Cu^{2+}$$
Therefore, all the above are not redox reactions.

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 16

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Redox Reactions Test - 16

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Question 1

$$+7$$

$$+6$$

$$+4$$

$$+2$$

Solution

Question 2

$$0$$ and $$0$$

$$+3$$ and $$-1$$

$$-1$$ and $$+3$$

$$-3$$ and $$+1$$

Solution

Question 3

$$+7$$

$$+3$$

$$+8$$

$$+5$$

Solution

Question 4

$$-1$$

$$1$$

$$0$$

any number

Solution

Question 5

$$H_{2}SO_{4}$$

$$SO_{2}$$

$$H_{2}SO_{3}$$

$$H_{2}S$$

Solution

Question 6

$$0$$

$$+1$$

$$+3$$

$$+5$$

Solution

Question 7

$$+2$$

$$-2$$

$$+4$$

$$-4$$

Solution

Question 8

$$P_{4}O_{6}$$

$$NaH_{2}PO_{2}$$

$$PH_{3}$$

$$Na_{3}PO_{4}$$

Solution

Question 9

$$-2$$

$$+2$$

$$-4$$

$$-1$$

Solution

Question 10

$$K_{2}Cr_{2}O_{7}$$ to $$K_{2}CrO_{4}$$

$$Na_{2}SO_{4}$$ to $$NaHSO_{4}$$

$$CuCl_{2}$$ to $$CuSO_{4}$$

All the above

Solution

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