Redox Reactions Test

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Redox Reactions Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

Compounds O.N.
(A) $$KMn^*O_4$$ (1) +4
(B) $$Ni^*(CO)_4$$ (2) +7
(C) $$[Pt^*(NH_3)Cl_2]Cl_2$$ (3) 0
(D) $$Na_2O^*_2$$ (4) -1
The correct code for the O.N. of asterisked atom would be:

A
A -1, B - 2, C - 3, D - 4

B
A -4, B - 3, C - 2, D - 1

C
A -2, B - 3, C - 1, D - 4

D
A -4, B - 1, C - 2, D - 3

Solution

(c) $$\left[p t\left(N H_{3}\right) C l_{2}\right] Cl_{2} \rightarrow$$ Let, $$0 . N$$ of $$P_{t}$$ is $$x$$.
$$\begin{array}{l}x+0-2=2\\\therefore[x=4]\end{array}$$
(D) $$\mathrm{Na}_{2} \mathrm{O}_{2} \rightarrow$$ Sodium forms peroxide $$\left(\mathrm{O}_{2}^{2-}\right)$$.
$$\operatorname{In} O_{2}^{2-},$$ oxidation state of each $$O$$ is (-1) .
So, correct option is C
Question 2
1 / -0

The oxidation number of $$Pt$$ in $$[Pt({ C }_{ 2 }{ H }_{ 4 }{ ) }Cl_3]^{ - }$$ is:

A
$$+1$$

B
$$+2$$

C
$$+3$$

D
$$+4$$

Solution

$$[Pt(C_{2}H_{4})cl_{3}]^{-}$$

oxidation no. of $$C_{2}H_{4}=0$$
oxidation no. of $$cl=-1$$
$$\therefore x+0+3(-1)=-1$$
$$\Rightarrow x=+2$$
$$\Rightarrow $$ oxidation no. of $$pt=+2$$
Question 3
1 / -0

The oxidation number of phosphorus vary from:

A
-3 to +5

B
-1 to +1

C
-3 to + 3

D
-5 to +1

Solution

Phosphorous belongs to group $$15.$$ It has $$5$$ electrons in its valence shell. So it can either lose $$5$$ $$e^-$$ to attain $$+5$$ oxidation state or gain $$3$$ $$e^-$$ to attain $$-3$$ oxidation state. Hence its oxidation state varies from $$-3$$ in $$PH_3$$ to $$+5$$ in $$H_3PO_4.$$
Question 4
1 / -0

Find the oxidation state of Cr in the given complex $$K_2[Cr(NO)(NH_3)(CN)_4]$$, $$\mu$$ = 1.73 BM.

A
+ 1

B
+ 2

C
+3

D
None of these

Solution

Given, $$k_{2}\left[\operatorname{cr}(N O)\left(N H_{3}\right)(C N)_{4}\right]$$ and magnetic moment $$(\mu)=1.73$$.

$$\text { We know, }\mu=\sqrt{n(n+2)}=1.73=\sqrt{3}$$
$$\therefore n=1$$ (Number of unpaired electrons)
since, chromium ion has one unpaired electron, so NO must
be unit positive ligand.
Let, oxidation number of $$\mathrm{Cr}$$ is $$x$$.
$$\begin{aligned}\Rightarrow \quad x+1 &+0-4=-2\\\therefore \quad & x=1\end{aligned}$$
Thus oxidation state of $$ Cr$$ is +1 .
so, option (A) is correct.
Question 5
1 / -0

Compound $$YBa_2Cu_3O_7$$ is a superconductor. The oxidation number of copper in the compound will be :

[O.No. of Y = + 3]

A
+ 7/ 3

B
-7/3

C
+ 2/3

D
+ 1/5

Solution

$$YBa_2Cu_3O_7$$ that is Yttrium Barium Copper Oxide belongs to family of crystalline chemical compounds. It is known for displaying high temperature super conductivity.

$$YBa_2Cu_3O_7$$
$$\overline {3}$$ $$\overline {+4}$$ $$\overline{3x}$$$$=14$$

$$3+4+3x-14=0$$
$$3x=7$$
$$x=\cfrac {7} {3}$$
Question 6
1 / -0

Which of the following is not a redox reaction?

A
$$2NO + O_2 \rightarrow 2NO_2$$

B
$$Cl_2 + H_2O \rightarrow HCl + HClO$$

C
$$2CrO^{2-}_{4} + 2H^+ \rightarrow Cr_2O^{2-}_{7} + H_2O$$

D
$$MnO^{-}_{4} + 8H^+ \rightarrow Mn^{2+} + 4H_2O$$

Solution

$$Ans: A.$$
A redox reaction is said to be occur if the oxidation number of atoms change between reactants and products.
Reactio $$2NO+O_2\longrightarrow 2NO_2$$ is just an oxidation reaction.
Question 7
1 / -0

Oxidation number of carbon in sucrose is:

A
$$0$$

B
$$+12$$

C
$$-12$$

D
$$+7$$

Solution

Molecular formula of sucrose $$= {C}_{12}{H}_{22}{O}_{11}$$
Overall charge on sucrose $$= 0$$
Oxidation no. of oxygen in sucrose $$= -2$$
Oxidation no. of hydrogen in sucrose $$= +1$$
Let the oxidation no. of carbon be x.
Sum of oxidation no. of atoms = overall charge on atom
$$\Rightarrow \; 12 \times x + 22 \times \left( +1 \right) + 11 \times \left( -2 \right) = 0$$
$$\Rightarrow \; 12x + 22 - 22 = 0$$
$$\Rightarrow \; 12x = 0$$
$$\Rightarrow \; x = 0$$
Hence, the oxidation number of carbon in sucrose is 0.
Question 8
1 / -0

In substance $$Mg({ HXO }_{ 3 })$$, the oxidation number of X is :-

A
$$0$$

B
$$+2$$

C
$$+3$$

D
$$+4$$

Solution

Molecular formula $$= Mg \left( HX{O}_{3} \right)$$
Overall charge on compound $$= 0$$
Oxidation no. of oxygen in compound $$= -2$$
Oxidation no. of hydrogen in compound $$= +1$$
Oxidation no. of magnesium im compound $$= +2$$
Let the oxidation no. of X be x.
Sum of oxidation no. of atoms = overall charge on atom
$$\Rightarrow \; 1 \times \left( +2 \right) + 1 \times \left( +1 \right) + 1 \times x + 3 \times \left( -2 \right) = 0$$
$$\Rightarrow \; 2 + 1 + x - 6 = 0$$
$$\Rightarrow \; x - 3 = 0$$
$$\Rightarrow \; x = 3$$
Hence, the oxidation number of X is 3.
Question 9
1 / -0

Oxidation number of P in $${ KH }_{ 2 }{ PO }_{ 3 }$$ is:

A
$$-1$$

B
$$-3$$

C
$$+5$$

D
$$+3$$

Solution

Molecular formula $$= K{H}_{2}P{O}_{3}$$
Overall charge on compound $$= 0$$
Oxidation no. of oxygen in compound $$= -2$$
Oxidation no. of hydrogen in compound $$= +1$$
Oxidation no. of potassium in compound $$= +1$$
Let the oxidation no. of phosphorous be x.
Sum of oxidation no. of atoms = overall charge on atom
$$\Rightarrow \; 1 \times \left( +1 \right) + 2 \times \left( +1 \right) + 1 \times x + 3 \times \left( -2 \right) = 0$$
$$\Rightarrow \; 1 + 2 + x - 6 = 0$$
$$\Rightarrow \; x - 3 = 0$$
$$\Rightarrow \; x = +3$$
Hence, the oxidation number of P is +3.
Question 10
1 / -0

Oxidation number of Cr atom in $$CrO_5$$ and $$K_3CrO_8$$ respectively:

A
+6 , +6

B
+5 , + 6

C
+6 , + 5

D
+ 5 , + 5

Solution

Structure of $${ K }_{ 3 }{ Cr }{ O }_{ 8 }$$
Refer to fig 1
Structure of $${ Cr }{ O }_{ 5 }$$
Refer fig 2.
This oxidation state of $${ Cr }{ O }_{ 5 }$$ and $${ K }_{ 3 }{ Cr }{ O }_{ 8 }$$ is $$+6$$ and $$+5$$ respectively.

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Re-Attempt Weekly Quiz Competition

	Compounds		O.N.
(A)	$$KMn^*O_4$$	(1)	+4
(B)	$$Ni^*(CO)_4$$	(2)	+7
(C)	$$[Pt^*(NH_3)Cl_2]Cl_2$$	(3)	0
(D)	$$Na_2O^*_2$$	(4)	-1

Redox Reactions Test - 48

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Redox Reactions Test - 48

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Question 1

A -1, B - 2, C - 3, D - 4

A -4, B - 3, C - 2, D - 1

A -2, B - 3, C - 1, D - 4

A -4, B - 1, C - 2, D - 3

Solution

Question 2

$$+1$$

$$+2$$

$$+3$$

$$+4$$

Solution

Question 3

-3 to +5

-1 to +1

-3 to + 3

-5 to +1

Solution

Question 4

+ 1

+ 2

+3

None of these

Solution

Question 5

+ 7/ 3

-7/3

+ 2/3

+ 1/5

Solution

Question 6

$$2NO + O_2 \rightarrow 2NO_2$$

$$Cl_2 + H_2O \rightarrow HCl + HClO$$

$$2CrO^{2-}_{4} + 2H^+ \rightarrow Cr_2O^{2-}_{7} + H_2O$$

$$MnO^{-}_{4} + 8H^+ \rightarrow Mn^{2+} + 4H_2O$$

Solution

Question 7

$$0$$

$$+12$$

$$-12$$

$$+7$$

Solution

Question 8

$$0$$

$$+2$$

$$+3$$

$$+4$$

Solution

Question 9

$$-1$$

$$-3$$

$$+5$$

$$+3$$

Solution

Question 10

+6 , +6

+5 , + 6

+6 , + 5

+ 5 , + 5

Solution

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