Redox Reactions Test

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Redox Reactions Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

The sum of the oxidation states of all the carbon atoms present in the compound $$C_{6}H_{5}CHO$$ is:

A
$$-4$$

B
$$3$$

C
$$+5$$

D
$$-4/7$$

Solution

As we know that sum of oxidation state of all atoms in a compound is equal to the total charge on that compound.
$$\therefore \text{ In } {C}_{6}{H}_{5}CHO$$-
Let the oxidation state of carbon in $${C}_{6}{H}_{5}CHO$$ be x.
$$\therefore \; 6 \times x + 5 \times 1 + x + 1 + \left( -2 \right) = 0$$
$$\Rightarrow \; 7x = -5 - 1 + 2 = -4$$
As there are seven carbon atoms are present in the compound, the sum of oxidation state of all the carbon atoms will be $$7x$$, i.e. $$-4$$.
Question 2
1 / -0

The sum of the oxidation states of all the carbon atoms present in the compound $$C_6H_5CHO$$ is:

A
-4

B
3

C
+5

D
-4/7

Solution

Here the 5 carbon atoms have -1 oxidation number . CHO, C has +1.
Thus sum of total oxidation states is [(-1) x 5] + 1= -4
Answer is option A.
Question 3
1 / -0

Oxidation number of carbon in $$ CH_2Cl_2 $$ is :

A
0

B
+1

C
+2

D
+4

Solution

Consider the oxidation state of carbon in dichloromethane as $$x$$.
We know that the charges on $$H$$ and $$Cl$$ are $$+1$$ and $$-1$$ respectively.
Therefore, $$CH_2Cl_2 \rightarrow x+2(+1) +2(-1) = 0 \Rightarrow x = 0$$
Hence option A is the correct answer of this question.
Question 4
1 / -0

Oxidation state of $$S$$ in $$ H_2S_2O_8$$ is:

A
+6

B
+7

C
+8

D
0

Solution

The structure of $$H_2S_2O_8$$ is given below.
$$H-O-\underset{O}{\underset{||}{\overset{O}{\overset{||}{S}}}}$$-O-O-$$\underset{O}{\underset{||}{\overset{O}{\overset{||}{S}}}}$$-O-H
In the structure of $$H_2S_2O_8$$, two oxygen atoms are involved in peroxide linkage, hence oxidation number will be $$-1$$ (each). Remaining oxygen atoms will have $$-2$$ charge. Thus, we can calculate the oxidation state of sulphur ($$x$$) as follows:
$$H_2S_2O_8 =2\underset { H }{ (+1 )} + 2\underset { S }{ (x)} +2\underset { O-O}{ (-1) } +6\underset { O}{(-2)}= 0$$
$$x = +6$$.
Question 5
1 / -0

Which of the following is not a rule for calculating oxidation number ?

A
For ions, oxidation number is equal to the charge on the ion.

B
The oxidation number of oxygen is -2 in all of its compound.

C
The oxidation number of fluorine is -1 in all of its compounds.

D
Oxidation number of hydrogen is +1 except in binary hydrides of alkali metals and alkaline earth metals where it is -1.

Solution

In many cases oxygen has the charge $$-2$$. But in some compounds like peroxides and superoxides which are the exceptional cases, the oxidation state is not $$-2$$.
In peroxides like hydrogen peroxide, benzoyl peroxide, oxidation state of oxygen is $$-1$$, whereas in superoxide, it is $$-1/2$$.
Question 6
1 / -0

Which of the following is not a redox reaction ?

A
$$ CuO + H_2 \rightarrow Cu + H_2O$$

B
$$ Na + H_2O \rightarrow NaOH + \frac { 1 }{ 2 } H_2$$

C
$$ CaCO_3 \rightarrow CaO + CO_2 $$

D
$$ 2K + F_2 \rightarrow 2KF $$

Solution

Redox reaction involves both oxidation and reduction reactions which are complimentary to each other.
In the reaction given in option C, there is no change in the oxidation states of any atom. Hence it is not redox reaction. This reaction does not involve either oxidation or reduction. The type of this reaction is called as decomposition.
$$\overset { +2 }{ Ca }\overset { +4 }{ C }\overset { -2 }{ O}_3$$ $$\rightarrow$$ $$\overset { +2 }{ Ca }\overset { -2 }{O }+\overset { +4}{ C }\overset { -2 }{ O}_2$$
Question 7
1 / -0

Which one of the following statements is not correct?

A
Oxidation state of $$S$$ in $$(NH_{4})_{2}S_{2}O_{8}$$ is $$+6$$

B
Oxidation number of $$Os$$ is $$OsO_{4}$$ is $$+8$$

C
Oxidation state of $$S$$ in $$H_{2}SO_{5}$$ is $$+8$$

D
None

Solution

$$ \begin{array}{l} In\left(N H_{4}\right)_{2} S_{2} O_{8} \\ \text { Oxidation state of } S \\ \qquad \begin{array}{l} +1(2)+x(2)+(-2)(8)=0 \\ \therefore \quad x=+7 \end{array} \end{array} $$
$$ \begin{array}{c} I n \text { Os} O_{4} \\ \text { Oxidation state of Os, } \\ \qquad \begin{array}{c} A+(-2)(4)=0 \\ x=+8 \end{array} \end{array} $$
$$I n \quad H_{2} S O_{5}$$
oxidation state of $$S$$
$$ \begin{array}{c} +1(2)+x+(-2) 5=0 \\ x=+8 \end{array} $$

Hence option $$(A)$$ is incorrect.
Question 8
1 / -0

Oxidation state of iron in $$ Fe(CO)_4 $$ is :

A
$$+1$$

B
$$-1$$

C
$$+2$$

D
$$0$$

Solution

In the formation of metal carbonyls, two types of bond are formed. Initially, carbonyl ligand $$(CO)$$ donates its electron pair to the central metal atom due to which metal gets a negative charge which makes it unstable. Hence the metal again forms back bonding with the ligand and due to which the net result is the zero charge on metal atom.
Question 9
1 / -0

Match the compounds given in column I with oxidation states of carbon given in column II and mark the apropriate choice.
Column I Column II
(A) $$ C_6H_{12}O_6$$ (i) +3
(B) $$CHCl_3$$ (ii) -3
(C) $$CH_3CH_3$$ (iii) +2
(D) $$(COOH)_2$$ (iv) 0

A
( A )$$\rightarrow $$( iv ), ( B )$$\rightarrow$$ ( iii ), ( C ) $$\rightarrow$$ ( ii ) , ( D ) $$\rightarrow$$ (i)

B
( A )$$ \rightarrow$$ ( i ) , ( B )$$ \rightarrow $$ ( ii ) , ( C ) $$ \rightarrow $$ ( iii ) , ( D ) $$ \rightarrow $$ ( iv )

C
( A ) $$ \rightarrow$$ ( ii ), ( B ) $$ \rightarrow $$ ( iii ), ( C ) $$ \rightarrow $$ ( iv ), ( D ) $$ \rightarrow $$ ( i )

D
( A ) $$ \rightarrow $$ ( iii ), ( B ) $$ \rightarrow $$ ( ii ) , ( C ) $$ \rightarrow $$ ( i ) , ( D ) $$ \rightarrow $$ ( iv )

Solution

Let the oxidation number of carbon be $$x$$.
Oxidation state of $$H=+1$$, $$Cl=-1$$ and of $$O=-2$$.
(A) $$C_6H_{12}O_6$$:
$$6x+12 \times 1+6 \times (−2)=0$$
$$6x+12-12=0$$
or $$x=0$$ = (iv)
(B) $$CHCl_3$$:
$$x+1+3 \times (−1)=0$$
$$x+1-3=0$$
or $$x=+2$$=(iii)
(C) $$CH_3CH_3$$:
$$x+3 \times 1+x+3 \times 1=0$$
$$2x+6=0$$
or $$x=−3$$=(ii)
(D) $$(COOH)_2$$:
$$2 \times [x+(-2)+(-2)+1]=0$$
$$x-3=0$$
or $$x=+3$$=(i)
A-(iv), B-(iii), C-(ii), D(i)
Question 10
1 / -0

The oxidation number of nitrogen in $$ ( N_2H_5)^+$$ is:

A
-2

B
+2

C
+3

D
-3

Solution

Let the oxidation number of nitrogen be $$x$$.
Oxidation number of $$H=+1$$
$$(N_2H_5)^+$$: $$2 \times x+ 5 \times 1=+1$$
$$2x+5=1$$ or $$x=-2$$
Thus Oxidation number of $$N$$ in $$(N_2H_5)^+$$ is +2

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Re-Attempt Weekly Quiz Competition

	Column I		Column II
(A)	$$ C_6H_{12}O_6$$	(i)	+3
(B)	$$CHCl_3$$	(ii)	-3
(C)	$$CH_3CH_3$$	(iii)	+2
(D)	$$(COOH)_2$$	(iv)	0

Redox Reactions Test - 41

Result

Redox Reactions Test - 41

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SHARING IS CARING

Question 1

$$-4$$

$$3$$

$$+5$$

$$-4/7$$

Solution

Question 2

-4

3

+5

-4/7

Solution

Question 3

0

+1

+2

+4

Solution

Question 4

+6

+7

+8

0

Solution

Question 5

For ions, oxidation number is equal to the charge on the ion.

The oxidation number of oxygen is -2 in all of its compound.

The oxidation number of fluorine is -1 in all of its compounds.

Oxidation number of hydrogen is +1 except in binary hydrides of alkali metals and alkaline earth metals where it is -1.

Solution

Question 6

$$ CuO + H_2 \rightarrow Cu + H_2O$$

$$ Na + H_2O \rightarrow NaOH + \frac { 1 }{ 2 } H_2$$

$$ CaCO_3 \rightarrow CaO + CO_2 $$

$$ 2K + F_2 \rightarrow 2KF $$

Solution

Question 7

Oxidation state of $$S$$ in $$(NH_{4})_{2}S_{2}O_{8}$$ is $$+6$$

Oxidation number of $$Os$$ is $$OsO_{4}$$ is $$+8$$

Oxidation state of $$S$$ in $$H_{2}SO_{5}$$ is $$+8$$

None

Solution

Question 8

$$+1$$

$$-1$$

$$+2$$

$$0$$

Solution

Question 9

( A )$$\rightarrow $$( iv ), ( B )$$\rightarrow$$ ( iii ), ( C ) $$\rightarrow$$ ( ii ) , ( D ) $$\rightarrow$$ (i)

( A )$$ \rightarrow$$ ( i ) , ( B )$$ \rightarrow $$ ( ii ) , ( C ) $$ \rightarrow $$ ( iii ) , ( D ) $$ \rightarrow $$ ( iv )

( A ) $$ \rightarrow$$ ( ii ), ( B ) $$ \rightarrow $$ ( iii ), ( C ) $$ \rightarrow $$ ( iv ), ( D ) $$ \rightarrow $$ ( i )

( A ) $$ \rightarrow $$ ( iii ), ( B ) $$ \rightarrow $$ ( ii ) , ( C ) $$ \rightarrow $$ ( i ) , ( D ) $$ \rightarrow $$ ( iv )

Solution

Question 10

-2

+2

+3

-3

Solution

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