1. What is the change in oxidation number of carbon in the following reaction? ⇒ (2020 Phase I)

CH_4(g) + 4Cl_2(g) → CCl_4(l) + 4HCl_(g)

A. +4 to +4

B. -4 to +4

C. 0 to -4

D. 0 to +4

Correct Answer is Option (A)

In CH₄, oxidation number of carbon is -4 while in CCl₄, oxidation number of carbon is +4. Thus, the change in oxidation number of carbon in the given reaction is from -4 to +4.

2. The oxidation state of Cr in CrO₅ is : ⇒ (Odisha NEET 2019 & AIPMT 2014)

A. +10

B. +6

C. +3

D. +3.5

Correct Answer is Option (B)

CrO₅ has geometry :

structure

3. The correct order of N-compounds in its decreasing order of oxidation states is: ⇒ (2018)

A. $H N O_{3}$ , $N H_{4} C l$ , $N O$ , $N_{2}$

B. $H N O_{3}$ , $N O$ , $N H_{4} C l$ , $N_{2}$

C. $H N O_{3}$ , $N O$ , $N_{2}$ , $N H_{4} C l$

D. $N H_{4} C l$ , $N_{2}$ , $N O$ , $H N O_{3}$

Correct option is (C)

$H N O_{3}$ , $N O$ , $N_{2}$ , $N H_{4} C l$
$H N O_{3}$ , $N O$ , $N_{2}$ , $N H_{4} C l$
Oxidation +5 +2 0 -3
state of N

4. A mixture of potassium chlorate, oxalic acid, and sulphuric acid is heated. During the reaction which element undergoes maximum change in oxidation number ⇒ (AIPMT 2012 Prelims)

A. H

B. S

C. Cl

D. C

The Correct Answer is Option (C)

Cl
$K C l O_{3} + H_{2} C_{2} O_{4} + H_{2} S O_{4} \to K_{2} S O_{4} + K C l + C O_{2} + H_{2} O$
Maximum change in oxidation number is observed in Cl (+5 to –1)

5. When $C l_{2}$ gas reacts with hot and concentrated sodium hydroxide, the oxidation number of chlorine changes from:
⇒ (AIPMT 2012 Prelims)

A. Zero to -1 and zero to +3

B. Zero to +1 and zero to -3

C. Zero to -1 and zero to +5

D. Zero to +1 and zero to -5

The Correct Option is (C)

Zero to -1 and zero to +5
The oxidation state of chlorine atoms in chlorine molecules is zero.
When chlorine gas is reacted with hot and concentrated sodium hydroxide solution to give the product of sodium chloride and sodium chlorate.
$6 N a O H (a q) + 3 C l_{2} 0 (g) \to 5 N a C l - 1 (a q) + N a C + 5 l O_{3} (a q) + 3 H_{2} O (l)$

Hence, option (b) is correct.

6. Oxidation numbers of P in $P O_{4}^{3 -}$ , of S in $S O_{4}^{2 -}$ and that of Cr in $C r_{2} O_{7}^{2 -}$ are respectively: ⇒ (AIPMT 2009)

A. +3,+6 and +5

B. +5,+3and +6

C. -3,+6 and +6

D. +5,+6 and +6

Correct option is (D)

(I) $^{x} P O_{4}^{3 -} \Rightarrow x + 4 \times (- 2) = - 3$ $\Rightarrow x = - 3 + 8 = + 5$ $\Rightarrow x = + 5$ Oxidation number of $P = + 5$ (II) $^{x} S O_{4}^{2 -} \Rightarrow x + 4 \times (- 2) = - 2$ $\Rightarrow x = - 2 + 8$ $\Rightarrow x = + 6$ Oxidation number of $S = + 6$ (III) $^{x} C r_{2} O_{7}^{2 -} \Rightarrow 2 x + 7 \times (- 2) = - 2$ $\Rightarrow 2 x = - 2 + 14$ $\Rightarrow 2 x = 12$ $\Rightarrow x = \frac{1 2}{2} = + 6$

7. The oxidation states of sulphur in the anions $S O_{3}^{2 -}, S_{2} O_{4}^{2 -}$ and $S_{2} O_{6}^{2 -}$ follow the order
⇒ (AIPMT 2003)

A. $S_{2} O_{6}^{2 -} < S_{2} O_{4}^{2 -} < S O_{3}^{2 -}$

B. $S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}$

C. $S O_{3}^{2 -} < S_{2} O_{4}^{2 -} < S_{2} O_{6}^{2 -}$

D. $S_{2} O_{4}^{2} < S_{2} O_{6}^{2 -} < S O_{3}^{2 -}$

The correct option is (B)

$S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}$
Solution: Correct option is b).
$S_{2} O_{4}^{2 -} < S O_{4}^{2 -} < S_{2} O_{6}^{2 -}$

Oxi. state of sulphur in $S_{2} O_{4}^{2 -} = + 3$
Oxi. state of sulphur in $S O_{3}^{2 -} = + 4$
Oxi state of sulphur in $S_{2} O_{6}^{2 -} . = + 5$

8. Oxidation number of Fe in Fe₃O₄ is: ⇒ (1999)

A. 3/2

B. 4/5

C. 5/4

D. 8/3

The Correct Answer is Option (D)

Concept:

The oxidation number/ oxidation state of an element in a compound is the number of charges (positive or negative) assigned to an atom in a molecule or ion according to a set of some arbitrary rules.

Rules for assigning Oxidation number:

The oxidation number of all elements in the free state is zero.
The most electronegative element, Fluorine has a -1 oxidation number in all compounds.
The oxidation number of monoatomic ions is the same as its charge. For example:
Oxidation state of Li+ , Ca2+ and Al3+ are +1, +2 and +3 respectively.
The oxidation state of Cl- and SO42- are -1 and -2 respectively.
Hydrogen has a +1 oxidation state in all compounds except ionic hydrides (-1 oxidation state).
The oxidation state of oxygen is -2 in a compound except in peroxides which are -1.
Alkali metals (Li, Na, K, etc.) always have +1, and alkaline earth metals (Be, Mg, etc.) have +2 oxidation states respectively.

Calculation:

Let the oxidation number of Iron in Fe3O4 is 'x'.
The oxidation number of oxide is -2.
The oxide is neutral as a whole, so,

3x + 4 × (-2) = 0

or, 3x - 8 = 0

or, 3x = 8

or, x = 8 / 3

Hence, the oxidation number of Fe in Fe3O4 is 8 / 3.

9. The oxidation state of iodine in ${H_4}I{O_6}^ -$ is ?⇒ ()

A. +7

B. -1

C. +5

D. +1

The Correct Answer is Option (A)

The oxidation state of iodine is equal to the number of its valence electrons. It can be calculated by adding the known values and putting it equal to (-1).

Complete step by step answer :
We will find oxidation state step by step as –
Initially, we will suppose the oxidation state of I in ${H_4}I{O_6}^ -$ be x.
Now, we know that the overall molecule has a charge of -1. So, the sum of oxidation states of all elements will be equal to -1.
Hydrogen will be +1 *4=4
Oxygen will be (-2)*6=(-12)
Thus, the sum is 4+x-12=(-1)
x-8=(-1)
x=+7

Therefore, the oxidation state of Iodine is +7 and option a.) is the correct answer.

Note : If the molecule would have been neutral; we would have put it equal to zero.
The ${H_4}I{O_6}^ -$ is called Orthoperiodate. It is a monovalent anion obtained after deprotonation. It is a conjugate base of orthoperiodic acid. The ${H_4}I{O_6}^ -$ has a pKa value of 8.31. The orthoperiodic acid has monoclinic structure. Like all other periodic acids, it is used in determining the structure of carbohydrates.

⇒Redox

Topic 03 : Oxidation Number/Oxidation State