Redox Reactions Test

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Redox Reactions Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

Oxygen has an oxidation state of $$+2$$ in.

A
$$H_2O_2$$

B
$$OF_2$$

C
$$SO_2$$

D
$$H_2O$$

Solution

Oxidation state of oxygen is always -2 except in peroxides,superoxides and when it reacts with fluorine.
In $$H_2O_2$$,oxidation state of $$H$$ is +1, so oxidation state of oxygen is -1.
In $$OF_2$$,oxidation state of $$F$$ is -1, so oxidation state of oxygen is +2.
In $$SO_2$$ and $$H_2O$$,oxidation state of oxygen is -2.
So B) is correct answer.
Question 2
1 / -0

Bromine reacts with the hot aqueous alkali to give bromide and bromate. What is the change that is brought about in the oxidation state of bromine to bromate?

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

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

C
$$0$$ to $$+7$$

D
None of these

Solution

$$3{ Br }_{ 2 }+6NaOH\rightarrow 5NaBr+NaBr{ O }_{ 3 }+3{ H }_{ 2 }O$$

Oxidation number of bromine changes from 0 in $$Br_2$$ to +5 in $$NaBr{ O }_{ 3 }$$.

Oxidation number of bromine in bromate can be calculated with the help of known oxidation state of oxygen (-2) and $$Na$$(+1).

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

Oxidation number of C in HNC is_________.

A
$$+2$$

B
$$-3$$

C
$$+3$$

D
Zero

Solution

Oxidation number of hydrogen is +1. As nitrogen is more electronegative than carbon so its oxidation number is -3. Net charge on compound is zero.

Now we can find oxidation number of carbon:

Let oxidation number of carbon is $$x$$,
$$1+(-3)+x=0$$
$$x=+2$$

Hence, the correct option is $$A$$
Question 4
1 / -0

Oxidation number of iodine in $$IO^-_3, IO^-_4, KI$$ and $$I_2$$ respectively are____________.

A
$$-1, -1, 0, +1$$

B
$$+3, +5, +7, 0$$

C
$$+5, +7, -1, 0$$

D
$$-1, -5, -1, 0$$

E
$$-2, -5, -1, 0$$

Solution

Oxidation number of oxygen is $$-2$$ and $$K$$ is $$+1$$.
In $$IO_3^{-}$$,oxidation number of iodine is $$(x+3(-2)=-1,x=+5) +5$$.
In $$IO_4^{-}$$,oxidation number of iodine is $$(x+4(-2)=-1,x=+7) +7$$.
In $$KI$$,oxidation number of iodine is $$(1+x=0,x=-1) -1$$.
In $$I_{2}$$,oxidation number of iodine is $$0$$.
Question 5
1 / -0

Nitrogen forms a variety of compounds in all oxidation states ranging from.

A
$$-3$$ to $$+5$$

B
$$-3$$ to $$+3$$

C
$$-3$$ to $$+4$$

D
$$-3$$ to $$+6$$
Solution
Hint:

The nitrogen atom has five electrons in its valence shell.

Explanation:
- A nitrogen atom can donate all five electrons to form bonds with other elements.
- As it requires three electrons to satisfy the octet rule, it can receive three electrons from other elements to form a molecule.
- It loses five electrons to show $$+5$$ oxidation state. And it accepts three electrons to show $$-3$$ oxidation state.
- Some compounds that exhibits these oxidation state are
$$NH_3$$ have $$-3$$ state, $$N_2H_4$$ have $$-2$$ state, $$NH_2OH$$ have $$-1$$ state, $$N_2O_5$$ have $$+5$$ state, $$NO_2$$ have $$+4$$ state, $$N_2O_3$$ have $$+3$$ state, $$NO$$ have $$+2$$ state and $$N_2O$$ have $$+1$$ state.
Hence, nitrogen exhibits oxidation states that ranges from $$-3$$ to $$+5$$.

Final Answer:
Therefore, the correct answer is option $$A.$$
Question 6
1 / -0

On reduction of $$KMnO_4$$ by oxalic acid in acidic medium, the oxidation number of Mn chnages. What is the magnitude of this change?

A
$$7$$ to $$2$$

B
$$6$$ to $$2$$

C
$$5$$ to $$2$$

D
$$7$$ to $$4$$

Solution

In acidic medium reduction of $$KMnO_4$$ takes place as follows:
$$Mn{ O }_{ 4 }^{ - }+5{ e^{ - }+8{ H }^{ + } }\rightarrow { Mn }^{ +2 }+{ 4H }_{ 2 }O+5{ e^{ - } }$$
So oxidation state of $$Mn$$ changes from +7 to +2.
Question 7
1 / -0

The oxidation number of sulphur in $$S_8, S_2F_2$$ and $$H_2S$$ respectively are__________.

A
$$0, +1$$ and $$-2$$

B
$$+2, +1$$ and $$-2$$

C
$$0, +1$$ and $$+2$$

D
$$-2, +1$$ and $$-2$$

Solution

Oxidation number of sulphur in $$S_8$$ is 0.
Oxidation number of $$F$$ is -1 so oxidation number of sulphur in $$S_2F_2$$ is +1.
Oxidation number of $$H$$ is +1 so oxidation number of sulphur in $$H_2S$$ is -2.
Question 8
1 / -0

In alkaline medium, $$H_2O_2$$ reacts with $$Fe^{3+}$$ and $$Mn^{2+}$$ separately to give:

A
$$Fe^{4+}$$ and $$Mn^{4+}$$

B
$$Fe^{2+}$$ and $$Mn^{2+}$$

C
$$Fe^{2+}$$ and $$Mn^{4+}$$

D
$$Fe^{4+}$$ and $$Mn^{2+}$$

Solution

$$2K_3[\overset{3+}{Fe}(CN)_6]+2KOH+2H_2O_2\rightarrow 2K_4[\overset{2+}{Fe}(CN)_6]+2H_2O+O_2$$
$$\overset{2+}{Mn}SO_4+H_2O_2\rightarrow \overset{4+}{Mn}O_2+H_2SO_4$$.
Question 9
1 / -0

Which of the following shows nitrogen with its increasing order of oxidation number?

A
$$NO < N_2O < NO_2 < NO^-_3, NH^+_4$$

B
$$NH^+_4 < N_2O < NO_2 < NO^-_3 < NO$$

C
$$NH^+_4 < N_2O < NO < NO_2 < NO^-_3$$

D
$$NH^+_4 < NO < N_2O < NO_2 < NO^-_3$$

Solution

Oxidation state of $$N$$ in $$NO$$ is +2,in $$N_2O$$ is +1,in $$NO_2$$ is +4,in $$NO_3^{-}$$ is +5 and in $$NH_4^{+}$$ is -3.
Nitrogen with increasing number of oxidation number is:
$$NH_4^{+}$$<$$N_2O$$<$$NO$$<$$NO_2$$<$$NO_3^{-}$$
Question 10
1 / -0

The oxidation numbers of the sulphur atoms in peroxy monosulphuric acid $$(H_2SO_5)$$ and peroxydisulphuric and $$(H_2S_2O_8)$$ are respectively.

A
$$+8$$ and $$+7$$

B
$$+3$$ and $$+3$$

C
$$+6$$ and $$+6$$

D
$$+4$$ and $$+6$$

Solution

By looking the structure of $$H_2SO_5$$, we can observe that there are two oxygen atoms which are linked by peroxide linkage so their oxidation numbers are -1.Rest oxygen atoms attached normally so their oxidation state is -2. The oxidation number of hydrogen is +1.So oxidation number of sulphur is
$$2(+1)+x+2(-1)+3(-2)=0,x=+6$$
By looking the structure of $$H_2S_2O_8$$, we can observe that there are two oxygen atoms which are linked by peroxide linkage so their oxidation numbers are -1.Rest oxygen atoms attached normally so their oxidation state is -2. Oxidation number of hydrogen is +1.So oxidation number of sulphur is
$$2(+1)+2(x)+2(-1)+6(-2)=0,x=+6$$

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 35

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Redox Reactions Test - 35

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Question 1

$$H_2O_2$$

$$OF_2$$

$$SO_2$$

$$H_2O$$

Solution

Question 2

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

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

$$0$$ to $$+7$$

None of these

Solution

Question 3

$$+2$$

$$-3$$

$$+3$$

Zero

Solution

Question 4

$$-1, -1, 0, +1$$

$$+3, +5, +7, 0$$

$$+5, +7, -1, 0$$

$$-1, -5, -1, 0$$

$$-2, -5, -1, 0$$

Solution

Question 5

$$-3$$ to $$+5$$

$$-3$$ to $$+3$$

$$-3$$ to $$+4$$

$$-3$$ to $$+6$$

Solution

Question 6

$$7$$ to $$2$$

$$6$$ to $$2$$

$$5$$ to $$2$$

$$7$$ to $$4$$

Solution

Question 7

$$0, +1$$ and $$-2$$

$$+2, +1$$ and $$-2$$

$$0, +1$$ and $$+2$$

$$-2, +1$$ and $$-2$$

Solution

Question 8

$$Fe^{4+}$$ and $$Mn^{4+}$$

$$Fe^{2+}$$ and $$Mn^{2+}$$

$$Fe^{2+}$$ and $$Mn^{4+}$$

$$Fe^{4+}$$ and $$Mn^{2+}$$

Solution

Question 9

$$NO < N_2O < NO_2 < NO^-_3, NH^+_4$$

$$NH^+_4 < N_2O < NO_2 < NO^-_3 < NO$$

$$NH^+_4 < N_2O < NO < NO_2 < NO^-_3$$

$$NH^+_4 < NO < N_2O < NO_2 < NO^-_3$$

Solution

Question 10

$$+8$$ and $$+7$$

$$+3$$ and $$+3$$

$$+6$$ and $$+6$$

$$+4$$ and $$+6$$

Solution

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