Redox Reactions Test

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Redox Reactions Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Oxidation state of central atom in $$Ni(CN)_{4}^{2-}$$ is :

A
$$- 2$$

B
$$0$$

C
$$+ 2$$

D
$$+ 4$$

Solution

Let $$x$$ be the oxidation state of nickel in $$Ni(CN)^{-2}_4$$.
The oxidation state of cyanide group is $$-1$$.
The sum of the oxidation states is equal to $$-2$$ which is the charge on the ion.
Therefore $$x+4(-1)=-2$$ or, $$x=+2$$
Hence, the oxidation state is $$+2$$.
Question 2
1 / -0

What is the oxidation number of chlorine in $$ClO_{3}^{-}$$?

A
$$+ 5$$

B
$$+3$$

C
$$+4$$

D
$$+2$$

Solution

Let $$x$$ be the oxidation number of chlorine in $$ClO_3^-$$.
The oxidation number of oxygen is $$-2$$.
The sum of the oxidation numbers of chlorine and oxygen is $$-1$$, which is equal to the charge on ion.
Hence, $$x+3(-2)=-1$$ or, $$x=+5$$.
Question 3
1 / -0

The number of moles of $$K_{2}Cr_{2}O_{7}$$ that will be needed to react with one mole of $$H_{2}S$$ in acidic medium is :

A
$$\dfrac{1}{6}$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{2}{5}$$

D
$$\dfrac{2}{3}$$

Solution

The balanced chemical equation for the reaction between potassium dichromate and hydrogen sulphide is as given below:

$$K_2Cr_2O_7 +4H_2SO_4+3H_2S \rightarrow K_2SO_4+Cr_2(SO_4)_3+7H_2O+3S$$

$$1$$ mol of potassium dichromate reacts with $$3$$ moles of hydrogen sulphide.

Hence, the number of moles of $$K_2Cr_2O_7$$ that will be needed to react one mole of $$H_2S$$ in acidic medium is $$\dfrac{1}{3}$$.

Option B is correct.
Question 4
1 / -0

Oxidation number of S in $$H_{2}SO_{5}$$ is 6. This is observed because:

A
there are five oxygen atom in molecule

B
the hydrogen atom is directly linked with non-metal

C
there is peroxide linkage in the molecule

D
the sulphur atom shows co-ordinate linkage

Solution

H$$_2SO_5$$
The O.S of S sulphur is +6. This is because there is a peroxide link in the molecule.
O
||
HO-S-O-O-H
|| [Peroxide link]
O
Question 5
1 / -0

In the reaction of $$HI$$ with $$H_2SO_4$$ the oxidation state of sulphur is changed to -2. Identify the correct balanced equation.

A
$$H_2SO_4+ 2HI \rightarrow SO_2 + 2H_2O + I_2$$

B
$$H_2SO_4 + 6HI \rightarrow S + 4H_2O + 3I_2$$

C
$$H_2SO_4 + 8HI \rightarrow H_2S + 4H_2O + 4I_2$$

D
All of the above

Solution

$$H_2SO_4 + 8HI \rightarrow H_2S + 4H_2O + 4I_2$$
The ON of S in $$H_2SO_4$$ is +6 and in $$H_2S$$ it is -2.
Question 6
1 / -0

Oxidation number of $$P$$ in $$PO^{3-}_4$$, of $$S$$ in $$SO^{2-}_4$$ and that of $$Cr$$ in $$Cr_2O^{2-}_7$$ are respectively:

A
+3, +6 and +6

B
+5, +6 and +6

C
+3, +6 and +5

D
+5, +3 and +6

Solution

$$PO_4^{3-}$$, Oxidation No. of $$P= +5$$
$$SO_4^{2-}$$, Oxidation No. of $$S=+6$$
$$Cr_2O_7^{2-}$$ , Oxidation No. of $$Cr =+6$$
Question 7
1 / -0

Oxidation states of P in $$H_{4}P_{2}O_{5},H_{4}P_{2}O_{6},H_{4}P_{2}O_{7},$$ are respectively:

A
+3, +5, + 4

B
+5, +3, + 4

C
+5, +4, + 3

D
+3, +4, + 5

Solution

$$H_4P_2O_5$$ $$H_4P_2O_6$$ $$H_4P_2O_7$$

$$4+2x -10=0$$ $$4+2x -12=0$$ $$4+2x -14=0$$
$$2x=6$$ $$2x=8$$ $$2x=10$$
$$x=+3$$ $$x=+4$$ $$x=+5$$

Hence, the correct option is $$C$$
Question 8
1 / -0

Which of the following can act as both oxidizing and reducing agent?

A
$$H_2S$$

B
$$H_2SO_4$$

C
$$SO_2$$

D
$$SO_3$$

Solution

$$SO_2$$ oxidizes $$H_2S$$ to S and reduces oxygen to $$SO_3$$.
Question 9
1 / -0

Oxidation state of Nitrogen in Nitrogen dioxide is:

A
$$+2$$

B
$$-2$$

C
$$+4$$

D
$$-4$$

Solution

NO$$_2$$ (Nitrogen Dioxide)
x-4=0
x=+4
The oxidation state of Nitrogen in NO$$_2$$ is (+4)
Question 10
1 / -0

The stability of $$+1$$ oxidation state increases in the sequence:

A
$$Ga < In < Al < Tl$$

B
$$Al < Ga < In < Tl$$

C
$$Tl < In < Ga < Al$$

D
$$In < Tl < Ga < Al$$

Solution

Hint: As we go down the group the common oxidation state decrease by $$2$$ due to the inert pair effect.

Explanation:

As we moving down the group, due to the poor shielding of $$d,f$$ orbitals leads to the increase in effective nuclear charge on the $$s$$ electrons in the outermost shell.
Therefore these two $$s$$ electrons don't participate as often, and hence the stability of the $$n-2$$ oxidation state decreases on moving down the group.
Since the order of moving down the group is $$Al \rightarrow Ga \rightarrow In \rightarrow Tl$$ .
The increasing order of atomic size is as; $$Tl > In > Ga > Al $$

The stability of $$+1$$ oxidation state will be the reverse of this order,
i.e. $$Al<Ga<In<Tl$$.

Correct Option: $$B$$

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Redox Reactions Test - 19

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Redox Reactions Test - 19

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Question 1

$$- 2$$

$$0$$

$$+ 2$$

$$+ 4$$

Solution

Question 2

$$+ 5$$

$$+3$$

$$+4$$

$$+2$$

Solution

Question 3

$$\dfrac{1}{6}$$

$$\dfrac{1}{3}$$

$$\dfrac{2}{5}$$

$$\dfrac{2}{3}$$

Solution

Question 4

there are five oxygen atom in molecule

the hydrogen atom is directly linked with non-metal

there is peroxide linkage in the molecule

the sulphur atom shows co-ordinate linkage

Solution

Question 5

$$H_2SO_4+ 2HI \rightarrow SO_2 + 2H_2O + I_2$$

$$H_2SO_4 + 6HI \rightarrow S + 4H_2O + 3I_2$$

$$H_2SO_4 + 8HI \rightarrow H_2S + 4H_2O + 4I_2$$

All of the above

Solution

Question 6

+3, +6 and +6

+5, +6 and +6

+3, +6 and +5

+5, +3 and +6

Solution

Question 7

+3, +5, + 4

+5, +3, + 4

+5, +4, + 3

+3, +4, + 5

Solution

Question 8

$$H_2S$$

$$H_2SO_4$$

$$SO_2$$

$$SO_3$$

Solution

Question 9

$$+2$$

$$-2$$

$$+4$$

$$-4$$

Solution

Question 10

$$Ga < In < Al < Tl$$

$$Al < Ga < In < Tl$$

$$Tl < In < Ga < Al$$

$$In < Tl < Ga < Al$$

Solution

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