Chapter 8.3 - Stock Notation

In the previous section we saw the oxidation number and oxidation state. In this section, we will see Stock notation.

• The German scientist Alfred Stock developed a convenient method to write the oxidation number of metals in compounds. The method is known as Stock notation. It can be written in 5 steps:
1. The oxidation number of the metal is written in Roman numerals.
2. The number so written, is placed inside parenthesis.
3. The number, together with the parenthesis, is placed just after the 'symbol of the metal' in the molecular formula of the compound.
4. Let us see an example. It can be written in 3 steps:
(i) In FeCl₃, the oxidation number of Fe is 3 and that of Cl is -1
(ii) So in our usual method, we write it as: $\mathbf\small{\rm{\overset{+3}{Fe}\,\overset{-1}{Cl}_3}}$
(iii) Using Stock notation, we write it as: Fe(III)Cl₃
5. Let us see another example. It can be written in 3 steps:
(i) In FeCl₂, the oxidation number of Fe is 2 and that of Cl is -1
(ii) So in our usual method, we write it as: $\mathbf\small{\rm{\overset{+2}{Fe}\,\overset{-1}{Cl}_2}}$
(iii) Using Stock notation, we write it as: Fe(II)Cl₂

Solved example 8.3
Using Stock notation, represent the following compounds: HAuCl₄, Tl₂O, FeO,
Fe₂O₃, CuI, CuO, MnO and MnO₂.
Solution:
Part (a): HAuCl₄
1. We can write this as: $\mathbf\small{\rm{\overset{+1}{H}\,\overset{x}{Au}\,\overset{-1}{Cl}_4}}$
    ♦ So we get: [+1 + x + (4 × -1)] = 0  ⇒ x = 3
    ♦ Thus, oxidation number of Au in HAuCl₄ is +3.
2. In Stock notation, we write: HAu(III)Cl₄
Part (b): Tl₂O
1. We can write this as: $\mathbf\small{\rm{\overset{x}{Tl}_2\,\overset{-2}{O}}}$
    ♦ So we get: [2x + -2] = 0  ⇒ x = 1
    ♦ Thus, oxidation number of Tl in Tl₂O is +1.
2. In Stock notation, we write: Tl₂(I)O
Part (c): FeO
1. We can write this as: $\mathbf\small{\rm{\overset{x}{Fe}\,\overset{-2}{O}}}$
    ♦ So we get: [x + -2] = 0  ⇒ x = 2
    ♦ Thus, oxidation number of Fe in FeO is +2.
2. In Stock notation, we write: Fe(II)O
Part (d): Fe₂O₃
1. We can write this as: $\mathbf\small{\rm{\overset{x}{Fe}_2\,\overset{-2}{O}_3}}$
    ♦ So we get: [2x + (3 × -2)] = 0  ⇒ x = 3
    ♦ Thus, oxidation number of Fe in Fe₂O₃ is +3.
2. In Stock notation, we write: Fe₂(III)O₃
Part (e): CuI
1. We can write this as: $\mathbf\small{\rm{\overset{x}{Cu}\,\overset{-1}{I}}}$
    ♦ So we get: [x + -1] = 0  ⇒ x = 1
    ♦ Thus, oxidation number of Cu in CuI is +1.
2. In Stock notation, we write: Cu(I)I   
Part (f): CuO
1. We can write this as: $\mathbf\small{\rm{\overset{x}{Cu}\,\overset{-2}{O}}}$
    ♦ So we get: [x + -2] = 0  ⇒ x = 2
    ♦ Thus, oxidation number of Cu in CuO is +2.
2. In Stock notation, we write: Cu(II)O   
Part (g): MnO
1. We can write this as: $\mathbf\small{\rm{\overset{x}{Mn}\,\overset{-2}{O}}}$
    ♦ So we get: [x + -2] = 0  ⇒ x = 2
    ♦ Thus, oxidation number of Cu in MnO is +2.
2. In Stock notation, we write: Mn(II)O   
Part (h): MnO₂
1. We can write this as: $\mathbf\small{\rm{\overset{x}{Mn}\,\overset{-2}{O}_2}}$
    ♦ So we get: [x + (2 × -2)] = 0  ⇒ x = 4
    ♦ Thus, oxidation number of Cu in MnO is +4.
2. In Stock notation, we write: Mn(IV)O₂

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• The idea of oxidation number can be used to define oxidation, reduction, oxidizing agent, reducing agent and redox reactions. This can be explained in 5 steps:
1. Consider the balanced equation of a reaction.
2. Pick any one element. Examine it’s oxidation number on the reactant side and product side.
• Two conditions can arise:
(i) The oxidation number on the product side is greater.
(ii) The oxidation number on the product side is lesser.
3. Interpreting the results:
    ♦ If the result is 2(i), that element has undergone oxidation.
    ♦ If the result is 2(ii), that element has undergone reduction.
• Each element in the balanced equation must be examined in this way.
4. Oxidizing and reducing agents:
• The element which has undergone reduction, is the oxidizing agent.  
    ♦ They are also known as oxidants.
• The element which has undergone oxidation, is the reducing agent.
    ♦ They are also known as reductants.
5. A chemical reaction which involve change in oxidation number of the reacting elements is known as redox reaction.

Solved example 8.4
Justify that the reaction:
2Cu₂O (s) + Cu₂S (s) → 6Cu (s) + SO₂ (g)
is a redox reaction. Identify the species oxidized/reduced, which acts as an
oxidant and which acts as a reductant.
Solution:
1. Let us write the oxidation numbers of all elements:
• Cu₂O
We can write this as: $\mathbf\small{\rm{\overset{x}{Cu}_2\,\overset{-2}{O}}}$
    ♦ So we get: [2x + -2] = 0  ⇒ x = 1
    ♦ Thus, oxidation number of Cu in Cu₂O is +1.
    ♦ We can write: $\mathbf\small{\rm{\overset{+1}{Cu}_2\,\overset{-2}{O}}}$
• Cu₂S
We can write this as: $\mathbf\small{\rm{\overset{x}{Cu}_2\,\overset{-2}{S}}}$
(S is given the oxidation number of -2 because, it comes in the same group as O)
    ♦ So we get: [2x + -2] = 0  ⇒ x = 1
    ♦ Thus, oxidation number of Cu in Cu₂S is +1.
    ♦ We can write: $\mathbf\small{\rm{\overset{+1}{Cu}_2\,\overset{-2}{S}}}$
• SO₂
We can write this as: $\mathbf\small{\rm{\overset{x}{S}\,\overset{-2}{O}}}$
    ♦ So we get: [x + (2 × -2)] = 0  ⇒ x = 4
    ♦ Thus, oxidation number of Cu in SO₂ is +4.
    ♦ We can write: $\mathbf\small{\rm{\overset{+4}{S}\,\overset{-2}{O}}}$
2. So the balanced equation can be written as:
$\mathbf\small{\rm{2\,\overset{+1}{Cu}_2\,\overset{-2}{O} +\overset{+1}{Cu}_2\,\overset{-2}{S}\rightarrow 6\,\overset{0}{Cu}+\overset{+4}{S}\,\overset{-2}{O}}}$
3. Now we examine each element:
(i) Consider Cu:
   ♦ It's oxidation number on the reactant side is +1.
   ♦ It's oxidation number on the product side is 0.
• There is a decrease in oxidation number.
   ♦ So we can write: Cu is reduced.
(ii) Consider O:
   ♦ It's oxidation number on the reactant side is -2.
   ♦ It's oxidation number on the product side is -2.
• There is no change in oxidation number.
   ♦ So we can write: O is neither oxidized nor reduced.
(iii) Consider S:
   ♦ It's oxidation number on the reactant side is -2.
   ♦ It's oxidation number on the product side is +4.
• There is an increase in oxidation number.
   ♦ So we can write: S is oxidized.
4. We can write the conclusion:
   ♦ Copper is reduced from +1 oxidation state to zero oxidation state.
   ♦ Sulfur is oxidized from –2 oxidation state to +4 oxidation state.
   ♦ The above reaction is thus a redox reaction.
   ♦ The element which gets oxidized is the reducing agent (reductant).
         ✰ So S is the reductant.
   ♦ The element which gets reduced is the oxidizing agent (oxidant).
         ✰ So Cu is the oxidant.

Solved example 8.5
Justify that the following reactions are redox reactions:
(a) CuO(s) + H₂ (g) → Cu(s) + H₂O(g)
(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂ (g)
(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆ (g) + 3LiCl(s) + 3 AlCl₃ (s)
(d) 2K(s) + F₂ (g) → 2K⁺F^- (s)
(e) 4 NH₃ (g) + 5O₂ (g) → 4NO(g) + 6H₂O (g)
Solution:
Part (a): CuO(s) + H₂ (g) → Cu(s) + H₂O(g)
1. Let us write the oxidation numbers of all elements:
• CuO
We can write this as: $\mathbf\small{\rm{\overset{x}{Cu}\,\overset{-2}{O}}}$
    ♦ So we get: [x + -2] = 0  ⇒ x = 2
    ♦ Thus, oxidation number of Cu in CuO is +2.
    ♦ We can write: $\mathbf\small{\rm{\overset{+2}{Cu}\,\overset{-2}{O}}}$
• H₂O
We can write this as: $\mathbf\small{\rm{\overset{+1}{H}_2\,\overset{x}{O}}}$
    ♦ So we get: [(2 × +1) + x] = 0  ⇒ x = -2
    ♦ Thus, oxidation number of O in H₂O is -2.
    ♦ We can write: $\mathbf\small{\rm{\overset{+1}{H}_2\,\overset{-2}{O}}}$
2. So the balanced equation can be written as:
$\mathbf\small{\rm{\overset{+2}{Cu}\,\overset{-2}{O}+\overset{0}{H}_2 \rightarrow \overset{0}{Cu}  +\overset{+1}{H}_2\,\overset{-2}{O}}}$
3. Now we examine each element:
(i) Consider Cu:
   ♦ It's oxidation number on the reactant side is +2.
   ♦ It's oxidation number on the product side is 0.
• There is a decrease in oxidation number.
   ♦ So we can write: Cu is reduced.
(ii) Consider O:
   ♦ It's oxidation number on the reactant side is -2.
   ♦ It's oxidation number on the product side is -2.
• There is no change in oxidation number.
   ♦ So we can write: O is neither oxidized nor reduced.
(iii) Consider H:
   ♦ It's oxidation number on the reactant side is 0.
   ♦ It's oxidation number on the product side is +1.
• There is an increase in oxidation number.
   ♦ So we can write: H is oxidized.
4. We can write the conclusion:
   ♦ Copper is reduced from +2 oxidation state to zero oxidation state.
   ♦ Hydrogen is oxidized from 0 oxidation state to +1 oxidation state.
   ♦ The above reaction is thus a redox reaction.
   ♦ The element which gets oxidized is the reducing agent (reductant).
         ✰ So H is the reductant.
   ♦ The element which gets reduced is the oxidizing agent (oxidant).
         ✰ So Cu is the oxidant.

Part (b): Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂ (g)
1. Let us write the oxidation numbers of all elements:
• Fe₂O₃
We can write this as: $\mathbf\small{\rm{\overset{x}{Fe}_2\,\overset{-2}{O}_3}}$
    ♦ So we get: [2x + (3 × -2)] = 0  ⇒ x = 3
    ♦ Thus, oxidation number of Fe in Fe₂O₃ is +3.
    ♦ We can write: $\mathbf\small{\rm{\overset{+3}{Fe}_2\,\overset{-2}{O}_3}}$
• CO
We can write this as: $\mathbf\small{\rm{\overset{x}{C}\,\overset{-2}{O}}}$
    ♦ So we get: [x + -2] = 0  ⇒ x = 2
    ♦ Thus, oxidation number of C in CO is +2.
    ♦ We can write: $\mathbf\small{\rm{\overset{+2}{C}\,\overset{-2}{O}}}$
• CO₂
We can write this as: $\mathbf\small{\rm{\overset{x}{C}\,\overset{-2}{O}_2}}$
    ♦ So we get: [x + (2 × -2)] = 0  ⇒ x = 4
    ♦ Thus, oxidation number of C in CO₂ is +4.
    ♦ We can write: $\mathbf\small{\rm{\overset{+4}{C}\,\overset{-2}{O}_2}}$
2. So the balanced equation can be written as:
$\mathbf\small{\rm{\overset{+3}{Fe}_2\,\overset{-2}{O}_3+3\;\overset{+2}{C}\,\overset{-2}{O}\rightarrow \overset{0}{Fe}+\overset{+4}{C}\,\overset{-2}{O}_2}}$
3. Now we examine each element:
(i) Consider Fe:
   ♦ It's oxidation number on the reactant side is +3.
   ♦ It's oxidation number on the product side is 0.
• There is a decrease in oxidation number.
   ♦ So we can write: Fe is reduced.
(ii) Consider O:
   ♦ It's oxidation number on the reactant side is -2.
   ♦ It's oxidation number on the product side is -2.
• There is no change in oxidation number.
   ♦ So we can write: O is neither oxidized nor reduced.
(iii) Consider C:
   ♦ It's oxidation number on the reactant side is +2.
   ♦ It's oxidation number on the product side is +4.
• There is an increase in oxidation number.
   ♦ So we can write: C is oxidized.
4. We can write the conclusion:
   ♦ Iron is reduced from +3 oxidation state to zero oxidation state.
   ♦ Carbon is oxidized from +2 oxidation state to +4 oxidation state.
   ♦ The above reaction is thus a redox reaction.
   ♦ The element which gets oxidized is the reducing agent (reductant).
         ✰ So C is the reductant.
   ♦ The element which gets reduced is the oxidizing agent (oxidant).
         ✰ So Fe is the oxidant.

Part (c): 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆ (g) + 3LiCl(s) + 3AlCl₃ (s)
1. Let us write the oxidation numbers of all elements:
• BCl₃
We can write this as: $\mathbf\small{\rm{\overset{x}{B}\,\overset{-1}{Cl}_3}}$
    ♦ So we get: [x + (3 × -1)] = 0  ⇒ x = 3
    ♦ Thus, oxidation number of B in BCl₃ is +3.
    ♦ We can write: $\mathbf\small{\rm{\overset{+3}{B}\,\overset{-1}{Cl}_3}}$
• LiAlH₄
We can write this as: $\mathbf\small{\rm{\overset{+1}{Li}\,\overset{+3}{Al}\,\overset{x}{H}_4}}$
    ♦ So we get: [+1 + +3 + 4x] = 0  ⇒ x = -1
    ♦ Thus, oxidation number of H in LiAlH₄ is -1.
    ♦ We can write: $\mathbf\small{\rm{\overset{+1}{Li}\,\overset{+3}{Al}\,\overset{-1}{H}_4}}$
• B₂H₆
We can write this as: $\mathbf\small{\rm{\overset{x}{B}_2\,\overset{+1}{H}_6}}$
    ♦ So we get: [2x + (6 × +1)] = 0  ⇒ x = -3
    ♦ Thus, oxidation number of B in B₂H₆ is -3.
    ♦ We can write: $\mathbf\small{\rm{\overset{-3}{B}_2\,\overset{+1}{H}_6}}$
• LiCl
We can write this as: $\mathbf\small{\rm{\overset{+1}{Li}\,\overset{x}{Cl}}}$
    ♦ So we get: [1 + x] = 0  ⇒ x = -1
    ♦ Thus, oxidation number of Cl in LiCl is -1.
    ♦ We can write: $\mathbf\small{\rm{\overset{+1}{Li}\,\overset{-1}{Cl}}}$
• AlCl₃
We can write this as: $\mathbf\small{\rm{\overset{+3}{Al}\,\overset{x}{Cl}_3}}$
    ♦ So we get: [+3 + 3x] = 0  ⇒ x = -1
    ♦ Thus, oxidation number of Cl in AlCl₃ is -1.
    ♦ We can write: $\mathbf\small{\rm{\overset{+3}{Al}\,\overset{-1}{Cl}_3}}$
2. So the balanced equation can be written as:
$\mathbf\small{\rm{4\,\overset{+3}{B}\,\overset{-1}{Cl}_3     +3\,\overset{+1}{Li}\,\overset{+3}{Al}\,\overset{-1}{H}_4\rightarrow 2\,\overset{-3}{B}_2\,\overset{+1}{H}_6+3\,\overset{+1}{Li}\,\overset{-1}{Cl}+3\,\overset{+3}{Al}\,\overset{-1}{Cl}_3}}$
3. Now we examine each element:
(i) Consider B:
   ♦ It's oxidation number on the reactant side is +3.
   ♦ It's oxidation number on the product side is -3.
• There is a decrease in oxidation number.
   ♦ So we can write: B is reduced.
(ii) Consider Cl:
   ♦ It's oxidation number on the reactant side is -1.
   ♦ It's oxidation number on the product side is -1.
• There is no change in oxidation number.
   ♦ So we can write: Cl is neither oxidized nor reduced.
(iii) Consider Li:
   ♦ It's oxidation number on the reactant side is +1.
   ♦ It's oxidation number on the product side is +1.
• There is no change in oxidation number.
   ♦ So we can write: Li is neither oxidized nor reduced.
(iv) Consider Al:
   ♦ It's oxidation number on the reactant side is +3.
   ♦ It's oxidation number on the product side is +3.
• There is no change in oxidation number.
   ♦ So we can write: Al is neither oxidized nor reduced.
(v) Consider H:
   ♦ It's oxidation number on the reactant side is -1.
   ♦ It's oxidation number on the product side is +1.
• There is an increase in oxidation number.
   ♦ So we can write: H is oxidized.
4. We can write the conclusion:
   ♦ Boron is reduced from +3 oxidation state to -3 oxidation state.
   ♦ Hydrogen is oxidized from -1 oxidation state to +1 oxidation state.
   ♦ The above reaction is thus a redox reaction.
   ♦ The element which gets oxidized is the reducing agent (reductant).
         ✰ So H is the reductant.
   ♦ The element which gets reduced is the oxidizing agent (oxidant).
         ✰ So B is the oxidant.

Part (d): 2K(s) + F₂ (g) → 2K⁺F^- (s)
1. In this problem, the reactants on the left side are in elemental form. So they have zero oxidation states.
2. Also it is easy to find the oxidation states on the right side.
K has an oxidation state of +1
F has an oxidation state of -1
3. We can write the conclusion:
   ♦ F is reduced from 0 oxidation state to -1 oxidation state.
   ♦ K is oxidized from 0 oxidation state to +1 oxidation state.
   ♦ The above reaction is thus a redox reaction.
   ♦ The element which gets oxidized is the reducing agent (reductant).
         ✰ So K is the reductant.
   ♦ The element which gets reduced is the oxidizing agent (oxidant).
         ✰ So F is the oxidant.

Part (e): 4NH₃ (g) + 5O₂ (g) → 4NO(g) + 6H₂O (g)
1. Let us write the oxidation numbers of all elements:
• NH₃
We can write this as: $\mathbf\small{\rm{\overset{x}{N}\,\overset{+1}{H}_3}}$
    ♦ So we get: [x + (3 × +1)] = 0  ⇒ x = -3
    ♦ Thus, oxidation number of N in NH₃ is -3.
    ♦ We can write: $\mathbf\small{\rm{\overset{-3}{N}\,\overset{+1}{H}_3}}$
• NO
We can write this as: $\mathbf\small{\rm{\overset{x}{N}\,\overset{-2}{O}}}$
    ♦ So we get: [x + -2] = 0  ⇒ x = 2
    ♦ Thus, oxidation number of N in NO is +2.
    ♦ We can write: $\mathbf\small{\rm{\overset{+2}{N}\,\overset{-2}{O}}}$
• H₂O
We can write this as: $\mathbf\small{\rm{\overset{x}{H}_2\,\overset{-2}{O}}}$
    ♦ So we get: [2x +  -2)] = 0  ⇒ x = 1
    ♦ Thus, oxidation number of H in H₂O is +1.
    ♦ We can write: $\mathbf\small{\rm{\overset{+1}{H}\,\overset{-2}{O}}}$
2. So the balanced equation can be written as:
$\mathbf\small{\rm{4\,\overset{-3}{N}\,\overset{+1}{H}_3+5\,\overset{0}{O}_2\rightarrow 4\,\overset{+2}{N}\,\overset{-2}{O}+\overset{+1}{H}_2\,\overset{-2}{O}}}$
3. Now we examine each element:
(i) Consider N:
   ♦ It's oxidation number on the reactant side is -3.
   ♦ It's oxidation number on the product side is +2.
• There is an increase in oxidation number.
   ♦ So we can write: N is oxidized.
(ii) Consider O:
   ♦ It's oxidation number on the reactant side is 0.
   ♦ It's oxidation number on the product side is -2.
• There is a decrease in oxidation number.
   ♦ So we can write: O is reduced.
(iii) Consider H:
   ♦ It's oxidation number on the reactant side is +1.
   ♦ It's oxidation number on the product side is +1.
• There is no change in oxidation number.
   ♦ So we can write: H is neither oxidized nor reduced.
4. We can write the conclusion:
   ♦ O is reduced from 0 oxidation state to -2 oxidation state.
   ♦ N is oxidized from -3 oxidation state to +2 oxidation state.
   ♦ The above reaction is thus a redox reaction.
   ♦ The element which gets oxidized is the reducing agent (reductant).
         ✰ So N is the reductant.
   ♦ The element which gets reduced is the oxidizing agent (oxidant).
         ✰ So O is the oxidant.

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In the next section, we will see types of redox reactions.


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