Redox Reactions Test

Result

Redox Reactions Test - 46

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

Oxidation number of three elements $$A,\ B$$ and $$C$$ are $$+2, +3$$ and $$-2$$ respectively. What is the formula of the compound formed from these three elements?

A
$$A_2BC_3$$

B
$$AB_2C_4$$

C
$$A_3BC_3$$

D
$$A_2B_2C_4$$

Solution

Given, $$A \rightarrow +2, B \rightarrow +3$$ & $$C\rightarrow -2$$
The formula which satisfies total sum of oxidation states of elements is equal to zero is,
$$AB_2C_4$$ i.e, $$(+2 \times 1)+(+3\times 2)+(-2 \times 4)$$
$$=2+6-8$$
$$=0$$ .
Question 2
1 / -0

Calculate the Oxidation number of $$Co$$ in the compound:
$$\left[ Co{ \left( { C }_{ 2 }{ O }_{ 4 } \right) }_{ 2 }{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] $$

A
+2

B
+3

C
+4

D
+1

Solution

The given molecule has a negative charge on it. So, $$[Co(C_2O_4)_2(NH_3)_2]^{-}$$ and let the oxidation munber of Co be x.
We know, $$C_2O_4^{2-}$$ is the ion and hence,
$$\implies x+2\times(-2)=-1 \implies x=+3$$
Question 3
1 / -0

Oxidation number of $$Cr$$ in $$CrO_5$$ is:

A
$$+10$$

B
$$+6$$

C
$$+4$$

D
$$+5$$

Solution

$$\displaystyle CrO_5$$ has one $$\displaystyle Cr=O$$ double bond. The oxygen atom of this double bond has oxidation number of $$\displaystyle -2$$. The remaining 4 oxygen atoms are part of peroxide linkages. The oxygen atom of peroxide linkage has oxidation number of $$\displaystyle -1$$. Let, X be the oxidation number of $$\displaystyle Cr$$
$$\displaystyle X+(-2)+4(-1)=0$$
$$\displaystyle X-2-4=0$$
$$\displaystyle X=+6$$
Hence, the oxidation number of $$\displaystyle Cr=+6$$
$$\displaystyle $$
Question 4
1 / -0

The sum of oxidation states of sulphur in $$H_2S_2O_8$$ is?

A
$$+2$$

B
$$+6$$

C
$$+7$$

D
$$+12$$

Solution

Structure of $$H_2S_2O_8$$ is:
$$HO-\overset { \overset { O }{ || } }{ \underset { \overset { || }{ O } }{ S } } -O-O-\overset { \overset { O }{ || } }{ \underset { \overset { || }{ O } }{ S } } -OH$$
there is one peroxy linkage $$S-O-O-S$$. Hence two of the eight oxygen atoms have -1 oxidation state and rest 6 have -2 oxidation state. Let the total oxidation state of S be x
then $$2\times 1+ x+6 \times (-2)+2 \times (-1)=0$$
or $$2+x-12-2=0$$
$$x=+12$$
Sum of oxidation state of both S is x i.e. 12 in $$H_2S_2O_8$$.
Question 5
1 / -0

Among the following, identify the species with an atom in $$+6$$ oxidation state.

A
$$[MnO_4]^-$$

B
$$[Cr(CN)_6]^{3-}$$

C
$$Cr_2O_3$$

D
$$CrO_2Cl_2$$

Solution

The correct answer is $$(D)\;CrO_2Cl_2$$
$$Cr+2(O)+2(Cl)=0$$
$$Cr+2(-2)+2(1)=0$$
$$Cr=+6$$
Question 6
1 / -0

Oxygen exhibit $$-1$$ oxidation state in:

A
$$OF_{2}$$

B
$$H_{2}O_{2}$$

C
$$HC1O$$

D
$$H_{2}O$$

Solution

$${ O }^{ -1 }$$ exist where there is peroxide linkage. Hence $${ H }_{ 2 }O_{ 2 }$$(Hydrogen peroxide).
$$H-O-O-H$$
Question 7
1 / -0

In the case of $$CH_3COOH$$, the oxidation number of carbon of the carboxylic group is:

A
$$-3$$

B
zero

C
$$+1$$

D
$$+3$$

Solution

Acetic acid is represented as $$\displaystyle CH_3-\underset {\underset {\displaystyle O}{||}}{C}-OH$$. For the purpose of oxidation number calculation, the oxidation number of C atom of methyl group is considered 0, the oxidation number of oxygen atom of $$\displaystyle C=O$$ double bond is considered $$\displaystyle -2$$. The oxidation number of oxygen atom of hydroxyl group is considered $$\displaystyle -1$$.
Let $$\displaystyle x$$ be the oxidation number of C atom of carboxylic group.
$$\displaystyle x+0+(-2)+(-1)=0$$
$$\displaystyle x-3=0$$
$$\displaystyle x=+3$$
Hence, in the case of $$CH_3COOH$$, the oxidation number of carbon of the carboxylic group is $$+3$$.
Question 8
1 / -0

Which one of the following statements is not correct?

A
Oxidation state of $$S$$ in $$(NH_4)_2S_2O_8$$ is $$+6$$

B
Oxidation number of $$Os$$ in $$OsO_4$$ is $$+8$$

C
Oxidation state of $$S$$ in $$H_2SO_5$$ is $$+8$$

D
Oxidation number of $$O$$ in $$KO_2$$ is $$-\dfrac{1}{2}$$

Solution

In $$(NH_4)_2S_2O_8$$
Oxidation state of $$S$$ $$+1(2)+x(2)+(-2\times 8)=0$$
$$2x=14 \Rightarrow x=+7$$

In $$OsO_4$$
Oxidation state of $$Os$$ $$x+(-2 \times 4)=0$$
$$x=+8$$

In $$H_2SO_5$$
Oxidation state of $$S$$ $$+1(2)+x+(-2)5=0$$
$$x=+8$$
In $$KO_2$$
Oxidation state of $$O$$ $$1+(x)2=0$$
$$x=-1/2$$
Question 9
1 / -0

The oxidation states of Cr in $$\left[ Cr\left( { H }_{ 2 }O \right) \right] { Cl }_{ 3 },\left[ { Cr\left( { C }_{ 6 }{ H }_{ 6 } \right) }_{ 2 } \right] ,$$ and $${ K }_{ 2 }\left[ Cr{ \left( CN \right) }_{ 2 }{ \left( O \right) }_{ 2 }{ \left( O \right) }_{ 2 }\left( { NH }_{ 3 } \right) \right] $$ respectively are :

A
+3,+4, and +6

B
+3,+2, and +4

C
+3, 0, and +6

D
+3, 0, and +4
Solution
In the given molecules,
In $$[Cr(H_2O)]Cl_3$$, let oxidation number of Cr be x, Here the molecule separates as its ions as, $$[Cr(H_2O)]^{3+}$$ and $$3Cl^{-}$$. As $$H_2O$$ is a neutral molecule its oxidation number is zero. Hence, $$x=+3$$
In $$[Cr(C_6H_6)_2]$$ let oxidation number of Cr be y, Benzene is a neutral molecule and hence on observing the cross multiplied factor we get the oxidation number of Cr as 0. Hence,$$y=0$$.
In $$K_2[Cr(CN)_2(O)_2(NH_3)]$$,let oxidation number of Cr be z, Here the ions are- $$2K^{+}$$ and $$[Cr(CN)_2(O)_2(NH_3)]^{2-}$$. As $$NH_3$$ is neutral molecule and $$O^{-}$$and $$CN^{-}$$ is $$2 \times (-1)+4 \times (-1)=-6$$ and hence oxidation number of Cr is, z-6=-0$$\implies z=+6$$
Hence,option C is correct answer.
Question 10
1 / -0

Calculate the Oxidation number of $$Fe$$ in the compound.
$$\left[ Fe\left( { No } \right) { ({ H }_{ 2 }O )}_{ 5 } \right] { SO }_{ 4 }$$

A
+3

B
+2

C
+1

D
0

Solution

Let the oxidation number of Fe in the given compound $$[Fe(No)(H_2O)_5]SO_4$$ be x
As it dissociated into ions then we get, $$[Fe(No)(H_2O)_5]^{2+}$$ and $$SO_4^{2-}$$
We know No exists as $$No^{-}$$ and hence oxidation state of Fe is,
$$x-1=+2 \implies x=+3$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 46

Result

Redox Reactions Test - 46

Score

-

Rank

-

SHARING IS CARING

Question 1

$$A_2BC_3$$

$$AB_2C_4$$

$$A_3BC_3$$

$$A_2B_2C_4$$

Solution

Question 2

+2

+3

+4

+1

Solution

Question 3

$$+10$$

$$+6$$

$$+4$$

$$+5$$

Solution

Question 4

$$+2$$

$$+6$$

$$+7$$

$$+12$$

Solution

Question 5

$$[MnO_4]^-$$

$$[Cr(CN)_6]^{3-}$$

$$Cr_2O_3$$

$$CrO_2Cl_2$$

Solution

Question 6

$$OF_{2}$$

$$H_{2}O_{2}$$

$$HC1O$$

$$H_{2}O$$

Solution

Question 7

$$-3$$

zero

$$+1$$

$$+3$$

Solution

Question 8

Oxidation state of $$S$$ in $$(NH_4)_2S_2O_8$$ is $$+6$$

Oxidation number of $$Os$$ in $$OsO_4$$ is $$+8$$

Oxidation state of $$S$$ in $$H_2SO_5$$ is $$+8$$

Oxidation number of $$O$$ in $$KO_2$$ is $$-\dfrac{1}{2}$$

Solution

Question 9

+3,+4, and +6

+3,+2, and +4

+3, 0, and +6

+3, 0, and +4

Solution

Question 10

+3

+2

+1

0

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.