Redox Reactions Test

Result

Redox Reactions Test - 39

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which of the statement is correct?

A
The oxidation state of iron in $$Na_2[Fe(CN)_5NO]$$ is +2

B
$$[Ag(NH_3)_2]^+$$ linear in shape

C
In $$[Fe(H_2O)_6]^{3+}$$, Fe is $$sp^2d^2$$ hybridised

D
In $$[Co(H_2O)_6]^{3+}$$, Co is $$d^2sp^3$$ hybridised.
Question 2
1 / -0

Which one of the following leads to redox reaction?

A
$$AgN{ O }_{ 3 }+HCl$$

B
$$KOH+HCl$$

C
$$KI+{ Cl }_{ 2 }$$

D
$${ NH }_{ 3 }+HCl\quad $$

Solution

Redox reactions are characterized by the transfer of electrons between chemical species, most often with one species undergoing oxidation while another species undergoes reduction.

$$2\overset { -1 }{ KI } +\overset { 0 }{ { Cl }_{ 2 } } \longrightarrow 2\overset { 1 }{ KCl } +\overset { 0 }{ { I }_{ 2 } } $$

It is an example of a redox reaction.

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

The oxidation state of $$Fe$$ in $${ Fe }_{ 3 }{ O }_{ 4 }$$ is

A
$$+3$$

B
$$+\dfrac{8}{3}$$

C
$$+6$$

D
$$+2$$

Solution

Let the oxidation state of $$Fe$$ in $${ Fe }_{ 3 }{ O }_{ 4 }=x$$

$$\therefore 3x+4\times \left( -2 \right) =0$$

or $$3x-8=0$$

$$\therefore x=+\dfrac{ 8 }{ 3 }$$
Question 4
1 / -0

The oxidation state of central atom in the anion of compound $$NaH_2PO_2$$ will be:

A
$$-3$$

B
$$+1$$

C
$$+3$$

D
$$+5$$

Solution

$$Na(H_2)PO_2$$
$$=+1+(2x+1)+x+2(-2)=0 $$
$$x -1= 0$$
or $$ x= 1$$
Question 5
1 / -0

Oxidation state of $$Cl$$ in $$CaO{ Cl }_{ 2 }$$ is

A
$$0$$

B
$$+1$$

C
$$-1$$

D
$$+1,-1$$

Solution

The oxidation state of $$Cl$$ in $$CaOCl_2$$ is +1 and -1 respectively.
Question 6
1 / -0

Oxidation number of $$'N'$$ in $$N_3H$$ (hydrazoic acid) is:

A
$$-\cfrac{1}{3}$$

B
$$+3$$

C
$$0$$

D
$$-3$$

Solution

The oxidation state of $$N$$ in $$N_3H$$ (hydrazoic acid) is:

$$\Rightarrow 3x+1=0$$

$$\Rightarrow 3x=-1$$

$$\Rightarrow x=- \dfrac 13$$

Hence, option $$A$$ is correct.
Question 7
1 / -0

Oxidation number of carbon in graphite is:

A
zero

B
$$+1$$

C
$$+4$$

D
$$+2$$

Solution

Oxidation number of an atom in its elemental state is zero .Graphite is already in its elemental state and therefore graphite shows oxidation state of zero.
Question 8
1 / -0

One mole of $$N_{2}H_{4}$$ loses $$10$$ mole of electrons to form a new compound $$Y$$. Assuming that all nitrogen appears in the new compound, what is the oxidation state of $$N$$ and $$Y$$ (there is not change in the oxidation state of hydrogen).

A
$$-3$$

B
$$+3$$

C
$$+5$$

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

The oxidation states of P atom in $$POCl_3, H_2PO_3$$ and $$H_4P_2O_6$$, respectively are:

A
$$+5, +4, +4$$

B
$$+5, +5, +4$$

C
$$+4, +4, +5$$

D
$$+3, +4, +5$$

Solution

Oxygen has an oxidation state of $$-2$$ and hydrogen has $$+1$$.

$$POCl_3=x+(-2)+3(-1)=0,x=+5$$

$$H_2PO_3=2(+1)+x+3(-2)=0,x=+4$$

$$H_4P_2O_6=4(+1)+2(x)+6(-2)=0,x=+4$$

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

In which of the following, the oxidation number of oxygen has been arranged in increasing order?

A
$$BaO_{2} < KO_{2} < O_{3} < OF_{2}$$

B
$$OF_{2} < KO_{2} < BaO_{2} < O_{3}$$

C
$$BaO_{2} < O_{3} < OF_{2} < KO_{2}$$

D
$$KO_{2} < OF_{2} < O_{3} < BaO_{2}$$

Solution

Hint: The oxidation state of oxygen differs from compound to compound.

Step 1: Oxidation state calculation.
A. In $$BaO_2$$, oxygen is in a peroxide state. therefore has a charge of $$-1$$.

B. In $$KO_2$$, oxygen is in a superoxide state. therefore has charge of $$\frac {-1}{2}$$.

C. In $$O_3$$, as there is $$3$$ oxygen. First one has charged $$-1$$, second one has a charge of $$0$$ and third one has therefore had a charge of $$+1$$.

Hence, overall charge is $$0$$.

In $$OF_2$$, oxygen is less electronegative than fluorine. Therefore has a charge of $$-2$$.

Step 2: Oxidation state of given compounds.
As shown above, the calculated oxidation state of compounds is as follows.

A. $$BaO_2$$ $$\rightarrow$$ $$-1$$

B. $$KO_2$$ $$\rightarrow$$ $$\frac {-1}{2}$$

C. $$O_3$$ $$\rightarrow$$ $$0$$

D. $$OF_2$$ $$\rightarrow$$ $$+2$$

Final step: From the above-calculated value, the increasing order of oxidation state is in option A.
Hence A is the correct option.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 39

Result

Redox Reactions Test - 39

Score

-

Rank

-

SHARING IS CARING

Question 1

The oxidation state of iron in $$Na_2[Fe(CN)_5NO]$$ is +2

$$[Ag(NH_3)_2]^+$$ linear in shape

In $$[Fe(H_2O)_6]^{3+}$$, Fe is $$sp^2d^2$$ hybridised

In $$[Co(H_2O)_6]^{3+}$$, Co is $$d^2sp^3$$ hybridised.

Question 2

$$AgN{ O }_{ 3 }+HCl$$

$$KOH+HCl$$

$$KI+{ Cl }_{ 2 }$$

$${ NH }_{ 3 }+HCl\quad $$

Solution

Question 3

$$+3$$

$$+\dfrac{8}{3}$$

$$+6$$

$$+2$$

Solution

Question 4

$$-3$$

$$+1$$

$$+3$$

$$+5$$

Solution

Question 5

$$0$$

$$+1$$

$$-1$$

$$+1,-1$$

Solution

Question 6

$$-\cfrac{1}{3}$$

$$+3$$

$$0$$

$$-3$$

Solution

Question 7

zero

$$+1$$

$$+4$$

$$+2$$

Solution

Question 8

$$-3$$

$$+3$$

$$+5$$

$$+1$$

Question 9

$$+5, +4, +4$$

$$+5, +5, +4$$

$$+4, +4, +5$$

$$+3, +4, +5$$

Solution

Question 10

$$BaO_{2} < KO_{2} < O_{3} < OF_{2}$$

$$OF_{2} < KO_{2} < BaO_{2} < O_{3}$$

$$BaO_{2} < O_{3} < OF_{2} < KO_{2}$$

$$KO_{2} < OF_{2} < O_{3} < BaO_{2}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.