Redox Reactions Test

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Redox Reactions Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

$${ KMnO }_{ 4 }$$ can be prepared from $${ K }_{ 2 }MnO_{ 4 }$$ as per the reaction,

$${ 3MnO }_{ 2 }^{ 2- }+{ 2H }_{ 2 }O\rightleftharpoons { 2MnO }_{ 4 }^{ - }+{ MnO }_{ 2 }{ +4OH }^{ - }$$

The reaction can go to completion by removing $${ OH }^{ - }$$ ions by adding:

A
$$KOH$$

B
$${ CO }_{ 2 }$$

C
$$SO$$

D
$$HCl$$
Solution
Since $${OH}^{-}$$ are generated from water $$\left( {H}_{2}O \right)$$ which is a weak acid, therefore a weak acid like $$C{O}_{2}$$ should be used to remove it because-
A strong acid reverse the reaction $$\left( HCl \right)$$,
$$KOH$$ increases the concentration of $${OH}^{-}$$, thus again shifts the reaction in the backward direction.
$$C{O}_{2}$$ combines with $${OH}^{-}$$ to give carbonate which is easily removed.
$$S{O}_{2}$$ reacts with water to give strong acid, so it cannot be used.

$$3 {K}_{2}Mn{O}_{4} + 2 C{O}_{2} \longrightarrow 2 KMn{O}_{4} + Mn{O}_{2} + 2 {K}_{2}C{O}_{3}$$
Hence the reaction can go to completion by removing $${OH}^{−}$$ ions by adding $$C{O}_{2}$$.
Question 2
1 / -0

The sum of the oxidation numbers of all the atoms in a neutral molecule is equal to:

A
zero

B
one

C
three

D
four

Solution

The sum of oxidation number of a neutral compound is zero and the sum of oxidation number in a polyatomic atom is equal to the charge of the atom.
Example:
In HCl the sum of oxidation number of H and Cl is zero.
Question 3
1 / -0

In which of the following compounds, the oxidation state of $$I-$$atom is highest?

A
$${KI}_{3}$$

B
$$KI{O}_{4}$$

C
$$KI{O}_{3}$$

D
$$All\ of\ the \ above$$

Solution

$$KI_3 \Rightarrow I_3^- + K^+$$

$$3X= -1$$ $$\Rightarrow X= -1/3$$

$$KIO_4 \Rightarrow IO_4^- + K^+$$

$$X + 4 \times {(-2)}= -1$$ $$\Rightarrow X= +7$$

$$KIO_3 \Rightarrow {IO_3}^- + K^+$$

$$X + 3 \times {(-2)}= -1$$ $$\Rightarrow X= +5$$

Option B is correct.
Question 4
1 / -0

Oxidation state of oxygen in $$Cr{ O }_{ 5 }$$ is :

A
-1

B
-2

C
Both $$A$$ and $$B$$

D
$$-\frac { 1 }{ 2 } $$

Solution

$$CrO_5$$ is chromium with four peroxy oxygen and one coordinated oxygen.

It is a butterfly type structure in which central chromium surrounds with 4 peroxy cyclic oxygen and 1 coordinate oxygen. In this, the oxidation number of chromium is +6.

$$CrO_5$$ has 2 pairs in peroxide. It implies that 4 oxygen is in $$-1$$ oxidation state and one in $$-2$$ oxidation state.

Hence, option $$(C)$$ is correct.
Question 5
1 / -0

Average oxidation number of iodine in $$KI_3$$?

A
$$+1/3$$

B
$$-1/3$$

C
$$+3$$

D
$$-1$$

Solution

Criven : $$-K I_{3}$$
Here I $$_{3}$$ is present as $$I_{3}$$
So, the negative charge is on theree Iodine
atoms . Average onidation number of Iodine
$$=-\dfrac{1}{3}$$
B) is covrect answer.
Question 6
1 / -0

What is the oxidation number of chromium in the dimeric hydroxo bridged species :-

A
$$+6$$

B
$$+4$$

C
$$+3$$

D
$$+2$$

Solution

let $$'Cr'$$ oxidation state $$\rightarrow x$$
$$2x-2=+4$$
$$2x=6$$
$$x=3$$
Each chromium shows $$+3$$ oxidation state.
Question 7
1 / -0

What is the oxidation state of $$Co$$ in $$[Co(H_2O)_5Cl]^{2+}$$?

A
$$+2$$

B
$$+3$$

C
$$+1$$

D
$$+4$$

Solution

$$\underset { x }{ [Co } \underset { 0 }{ { \left( { H }_{ 2 }O \right) }_{ 5 } } \underset { -1 }{ { Cl] }^{ 2+ } } $$

Let, $$x$$ be the oxidation state of Co.

Now,

$$x+5\times 0-1=2;\quad \Rightarrow x=+3$$
Question 8
1 / -0

The oxidation state of Cr is $$[Cr(NH_3)_4Cl_2]$$ is?

A
$$+3$$

B
$$+2$$

C
$$+1$$

D
$$0$$

Solution

$$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\right]$$
We know that $$C 1=-1$$ $$\mathrm{NH}_{3}=0$$
So onidation state of $$\underline{Cr}$$
$$\begin{array}{l}x+4 \times 0+2(-1)=0\\x=+2\end{array}$$
Corret Answer $$-\underline{B}$$
Question 9
1 / -0

The oxidation number of nitrogen in $$NCI_3$$ is ?

A
$$+3$$

B
$$-3$$

C
$$Zero$$

D
-$$1/3$$

Solution

Oxidation number is the apparent charge on an atom of an element present in a molecule.In $$NCl_3$$, Electronegativity of Nitrogen is 3 while that of Chlorine is 3.2. Thus Chlorine being more electronegative has oxidation state=-1.
Sum of oxidation numbers of all elements present in a neutral compound is zero.
So, O.N. of Nitrogen+ 3(O.N. of $$Cl$$)=0
O.N. of $$N^{-3}=0$$
O.N. of $$N=+3$$
Question 10
1 / -0

Of the following elements, which one has the same oxidation state in all of its compounds?

A
Hydrogen

B
Fluorine

C
Carbon

D
Oxygen

Solution

Possible oxidation states of the elements are :-
Hydrogen $$=$$ $$-1,+1$$
Fluorine $$=$$ $$-1$$
Carbon $$=$$ $$-4,-3,-2,-1,0,+1,+2,+3,+4$$
Oxygen $$=$$ $$-2,-1,0,+1,+2$$
Here, only fluorine shows a single oxidation state in all of its compounds.

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 51

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Redox Reactions Test - 51

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Question 1

$$KOH$$

$${ CO }_{ 2 }$$

$$SO$$

$$HCl$$

Solution

Question 2

zero

one

three

four

Solution

Question 3

$${KI}_{3}$$

$$KI{O}_{4}$$

$$KI{O}_{3}$$

$$All\ of\ the \ above$$

Solution

Question 4

-1

-2

Both $$A$$ and $$B$$

$$-\frac { 1 }{ 2 } $$

Solution

Question 5

$$+1/3$$

$$-1/3$$

$$+3$$

$$-1$$

Solution

Question 6

$$+6$$

$$+4$$

$$+3$$

$$+2$$

Solution

Question 7

$$+2$$

$$+3$$

$$+1$$

$$+4$$

Solution

Question 8

$$+3$$

$$+2$$

$$+1$$

$$0$$

Solution

Question 9

$$+3$$

$$-3$$

$$Zero$$

-$$1/3$$

Solution

Question 10

Hydrogen

Fluorine

Carbon

Oxygen

Solution

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