Redox Reactions Test

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Redox Reactions Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

The oxidation state of the underlined element in the given compound is:
$$\underline{Ru}O_4$$

A
+8

B
+6

C
+4

D
none of these

Solution

$$RuO_4$$ oxidation state be $$x$$
Then, $$ x + 4 (-2) = 0$$
$$ x - 8 = 0$$
$$ x = +8$$
Question 2
1 / -0

The oxidation state of the underlined element in the given compound state is:
$$\underline{U_2}{O_7}^{4-}$$

A
+5

B
+6

C
+7

D
none of these

Solution

$$U_2o_7^4-$$
Let oxidation state be $$x$$
$$2(x) + 7 (-2) = -4$$
$$2x = 10$$
$$x = +5$$
Question 3
1 / -0

Which of the following has been arranged in order of increasing oxidation number of nitrogen?

A
$$NH_3 < N_2O_5 < NO < N_2$$

B
$$NO^+_2 < NO^-_3 < NO^-_2 < N^-_3$$

C
$$NH^+_4 < N_2H_4 < NH_2OH < N_2O$$

D
$$NO_2 < NaN_3 < NH^+_4 < N_2O$$

Solution

Oxidation state of $$H$$ $$=+1$$
Oxidation state of $$O$$ $$=-2$$
For option A oxidation state of $$N$$ in $$NH_3$$ is -3,in $$N_2O_5$$ is +5,in $$NO$$ is +2. So this is not increasing order of oxidation state of nitrogen.
For option B oxidation state of $$N$$ in $$NO_2^{+}$$ is +5,in $$NO_3^{-}$$ is +5. So this is not increasing order of oxidation state of nitrogen.
For option C oxidation state of $$N$$ in $$NH_4^{+}$$ is -3,in $$N_2H_4$$ is -2,in $$NH_2OH$$ is -1 and in $$N_2O$$ is +1. So this is correct answer as compounds are arranged in increasing order of oxidation number of nitrogen.
For option A oxidation state of $$N$$ in $$NO_2$$ is -4,in $$NaN_3$$ is -1/3,in $$NH_4^{+}$$ is -3. So this is not increasing order of oxidation state of nitrogen.
Question 4
1 / -0

The oxidation number of nitrogen atoms in $$NH_4NO_3$$ are:

A
$$+3, +3$$

B
$$+3, -3$$

C
$$-3, +5$$

D
$$-5, +3$$

Solution

$$NH_4NO_3\rightleftharpoons NH^+_4+NO^-_3$$
$$NH^+_4x+4=+1$$
$$x=-3$$
$$NO^-_3x-6=-1$$
$$x=+5$$.
Question 5
1 / -0

The oxidation states of sulphur in Caro's and Marshall's acid are:

A
$$+6, +6$$

B
$$+4, +6$$

C
$$+6, -6$$

D
$$+6, +4$$

Solution

$$H_2O_5$$ Caro's Acid $$H-O-\overset{\displaystyle \overset{\displaystyle O}{||}}{\underset{\displaystyle \underset{\displaystyle O}{||}}{\displaystyle S}}-O-O-H$$. $$\underset{State}{(+6)}$$
$$H_2S_2O_8$$ Marshall's Acid
$$H-O-\overset{\displaystyle \overset{\displaystyle O}{||}}{\underset{\displaystyle \underset{\displaystyle O}{||}}{S}}-O-O-\overset{\displaystyle\overset{\displaystyle O}{||}}{\underset{\displaystyle \underset{\displaystyle O}{||}}{S}}-O-H$$ $$\underset{State}{(+6)}$$
Both these acids have peroxy link.
Question 6
1 / -0

Oxidation state(s) of chlorine in $$CaOCl_2$$(bleaching powder) is/are?

A
$$+1$$ and $$-1$$

B
$$+1$$ only

C
$$-1$$ only

D
None of these

Solution

In $$CaOCl_2$$, two $$Cl$$ are in different oxidation states because one $$Cl$$ is attached to oxygen atom exists as $$OCl^{-}$$ and other is attached directly to $$Ca$$ exists as $$Cl^{-}$$.One attached directly to $$Ca$$ has -1 oxidation state and one attached to oxygen has +1 oxidation state. So $$Cl$$ exists in two different oxidation states -1 and +1.
Question 7
1 / -0

Oxidation states of carbon atoms in diamond and graphite are.

A
$$+2, +4$$

B
$$+4, +2$$

C
$$-4, 4$$

D
Zero, zero

Solution

Graphite and diamond by itself is elemental carbon and oxidation state of an element in free state is 0. So in graphite and diamond, carbon is in 0 oxidation state.
Question 8
1 / -0

Oxidation number of $$C$$ in $$HCCOH$$ is________.

A
$$+2$$

B
$$+4$$

C
$$+3$$

D
$$0$$
Question 9
1 / -0

Oxidation number of sulphur in $$S_8, S_2F_2$$ and $$H_2S$$ are:

A
$$+2, 0, +2$$

B
$$0, +1, -2$$

C
$$-2, 0, +2$$

D
$$0, +1, +2$$

Solution

Oxidation number of sulphur in $$S_8$$ is 0 as in $$S_8$$, sulphur exists in elemental form so it has 0 oxidation state.
Oxidation number of sulphur in $$S_2F_2$$ is +1 as oxidation number of $$F$$ is -1.
Oxidation number of sulphur in $$H_2S$$ is -2 as oxidation number of hydrogen is +1.
Question 10
1 / -0

In the ethylene molecule the two carbon atoms have the oxidation numbers.

A
$$-1, -1$$

B
$$-2, -2$$

C
$$-1, -2$$

D
$$+2, -2$$

Solution

Formula of ethylene $$=$$ $$C_2H_4$$
Let oxidation number of $$C$$ is $$x$$ and we know oxidation number of $$H$$ is $$+1$$.
So, $$2(x)+4(+1)=0$$
$$x=-2$$
So oxidation number of each carbon is $$-2$$.

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 37

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Redox Reactions Test - 37

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Question 1

+8

+6

+4

none of these

Solution

Question 2

+5

+6

+7

none of these

Solution

Question 3

$$NH_3 < N_2O_5 < NO < N_2$$

$$NO^+_2 < NO^-_3 < NO^-_2 < N^-_3$$

$$NH^+_4 < N_2H_4 < NH_2OH < N_2O$$

$$NO_2 < NaN_3 < NH^+_4 < N_2O$$

Solution

Question 4

$$+3, +3$$

$$+3, -3$$

$$-3, +5$$

$$-5, +3$$

Solution

Question 5

$$+6, +6$$

$$+4, +6$$

$$+6, -6$$

$$+6, +4$$

Solution

Question 6

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

$$+1$$ only

$$-1$$ only

None of these

Solution

Question 7

$$+2, +4$$

$$+4, +2$$

$$-4, 4$$

Zero, zero

Solution

Question 8

$$+2$$

$$+4$$

$$+3$$

$$0$$

Question 9

$$+2, 0, +2$$

$$0, +1, -2$$

$$-2, 0, +2$$

$$0, +1, +2$$

Solution

Question 10

$$-1, -1$$

$$-2, -2$$

$$-1, -2$$

$$+2, -2$$

Solution

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