Chapter 7.1 - Sulphuric acid

In the previous section, we discussed about ammonia. In this section we will a few more non-metallic compounds.

Sulphuric acid

Sulphuric acid is a chemical of utmost importance in industry. It's various industrial uses are listed below:

•Manufacture of fertilisers •Manufacture of explosives •Manufacture of paints •Manufacture of fibres •Manufacture of other chemicals •Used for dehydration

■ Sulphuric acid is called the King of chemicals.

Industrial preparation of Sulphuric acid

Sulphuric acid is industrially prepared by Contact process. Let us see the various stages in the process:

Stage 1: Sulphur is burnt in air to produce sulphurdioxide (SO₂). The equation is:
S + O₂ → SO₂. This is a balanced equation.

Stage 2: This SO₂ is allowed to combine with more oxygen. High temperature is required for this process. Also vanadium pentoxide (V₂O₅) is used as a catalyst. The product is sulphur trioxide (SO₃).

Let us write the equation:

Reactants:

    ♦ Sulphur dioxide. One molecule is SO₂

    ♦ Oxygen. One molecule is O₂

Products:

    ♦ Sulphur trioxide. One molecule is SO₃

• So skeletal equation is:

SO₂ + O₂ → SO₃. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: SO₂ + O₂ → SO₃
Step 2: 2SO₂ + O₂ → 2SO₃

	Reactants			Products
	S	O		S	O
Step 1	1	4		1	3
Step 2	2	6		2	6

So the balanced equation is: 2SO₂ + O₂ → 2SO₃

Stage 3: SO₃ is dissolved in concentrated sulphuric acid. The product is H₂S₂O₇. It is called oleum. The equation is: SO₃ + H₂SO₄ → H₂S₂O₇

This is a balanced equation.
Stage 4: This Oleum is mixed with water to produce sulphuric acid of the required concentration. Let us write the equation:

Reactants:

    ♦ Oleum. One molecule is H₂S₂O₇. 

    ♦ Water. One molecule is H₂O

Products:

    ♦ Sulphuric acid. One molecule is H₂SO₄.

• So skeletal equation is:

H₂S₂O₇ + H₂O → H₂SO₄. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: H₂S₂O₇ + H₂O → H₂SO₄
Step 2: H₂S₂O₇ + H₂O → 2H₂SO₄

	Reactants				Products
	H	S	O		H	S	O
Step 1	4	2	8		2	1	4
Step 2	4	2	8		4	2	8

So the balanced equation is: H₂S₂O₇ + H₂O → 2H₂SO₄.

• Theoretically, there is no need for stages 3 and 4. Let us see why this is so:

(i) At the end of stage 2, we get sulphur trioxide (SO₃). When this is dissolved in water, we should get sulphuric acid. The equation is: SO₃ + H₂O → H₂SO₄. This is a balanced equation.

(ii) But there are practical difficulties in carrying out this reaction. The dissolution of SO₃ in water is an exothermic process. 
(iii) The H₂SO₄ which is initially formed, becomes fine smog like particles. These particles hinder further dissolution. So the complete dissolution according to the equation does not take place. 
(iv) So, in the industrial production, we go through all the four stages. 

The stages can be presented in the form of a flow chart given below:

Physical properties of sulphuric acid

The important physical properties of H₂SO₄ are: •Colourless •viscous in nature •Highly corrosive •Denser than water •Water soluble

Chemical properties of sulphuric acid

Affinity towards water: 
• Take 5 ml of water in a test tube and slowly add concentrated sulphuric acid to it. 
• If we touch the bottom of the test tube, we can feel the heat. 
• This is because, the reaction between water and H₂SO₄ is highly exothermic.

■ While diluting sulphuric acid, the acid should be added to water in very small quantities, while stirring it. We should not do the other way round. That is., we should not add water to the acid. If we do it, there will be spurting and may cause burns to our body.

Concentrated H₂SO₄ has great affinity towards water. It has the ability to absorb moisture from the substances that come in contact with it. It will absorb water even if the water is chemically combined with the substance. 
Let us see an example:

1. Take a little sugar in a watch glass. Add a few drops of concentrated sulphuric acid to it. A reaction takes place and a black residue is left in the watch glass. Let us write the equation: 
C₁₂H₂₂O₁₁  → 12C + 11H₂O. This is a balanced equation. 
2. The 12 carbon atoms present in the left side as 'C₁₂' is present as '12C' on the right side. 
3. Now look at the other elements: Oxygen and hydrogen: 
    ♦ In the left side, there are 22 Hydrogen atoms and 11 Oxygen atoms. 
    ♦ So we can say: In a sugar molecule, the number of H atoms and number of O atoms are in the ratio 22:11 which is same as 2:1. 
    ♦ This is the same 'ratio of hydrogen and oxygen' in water. 
   ♦ On the right side, we see that 11 molecules of water are formed. Those 11 water molecules contain the same 22 H and 11 O on the left  side 
4. So, when sulphuric acid reacts with a substance, the H and O atoms get separated from that substance. Those separated atoms combine together to form water.
5. The residue that remains, after the removal of water is carbon. 

■ After the separation, we say that the substance is dehydrated. Thus, H₂SO₄ has the ability to dehydrate substances.

• Similar to sugar, glucose (C₆H₁₂O₆) and fructose (C₆H₁₂O₆) also reacts with H₂SO₄ and get dehydrated. [Note that, glucose and fructose have the same chemical formula. But the arrangement of atoms is different]
• Number of H atoms : Number of O atoms ratio is the same 2:1 in glucose and fructose also.

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The examples of sugar, glucose and fructose that we saw above are those in which, water is present in a chemically combined form. 
Another case is there, in which water is not chemically combined. But it is present in crystalline form. That is., molecules of water are attached to the molecules of the substance by physical bonding. 
• Copper sulphate is an example. 
    ♦ It contains water in crystalline form. The chemical formula is: CuSO₄.5H₂O.
    ♦ It is called hydrous copper sulphate. It is blue in colour 
    ♦ The H₂SO₄ reacts with CuSO₄.5H₂O and causes dehydration.
    ♦ After dehydration, it is called 'anhydrous copper sulphate' (CuSO₄). It is white in colour.

In the previous section, we saw the application of quick lime (CaO) as a drying agent in the preparation of ammonia. Now we see that H₂SO₄ also can absorb water. So it can also be used as a drying agent. In fact, it is used as a drying agent in the manufacture of chlorine (Cl₂), Sulphur dioxide (SO₂) and Hydrogen chloride (HCl). We saw the case of Cl₂ in a previous section (see fig.5.4). 

But it cannot be used as a drying agent in the manufacture of ammonia because of the following reason: H₂SO₄ is an acid, while ammonia is alkaline. The two will react together.

Reaction of sulphuric acid with salts

• We have seen that salts are formed when acids and alkalies react together. 
• Now, if we take such a salt, and allow it to react with H₂SO₄, we will get back the acid. Let us see an example:

• We know that when the acid HCl react with the alkali NaOH, we get the salt NaCl. 
• Now, this NaCl is allowed to react with H₂SO₄. We will get back the acid HCl. The reaction is as shown below:
NaCl + H₂SO₄ → NaHSO₄ + HCl. This is a balanced equation.

• Another example:
Sulphuric acid reacts with potassium nitrate (KNO₃). Nitric acid is obtained as a product. The equation is: 
KNO₃ + H₂SO₄ → KHSO₄ + HNO₃. This is a balanced equation.

■So we can say: Concentrated sulphuric acid can displace acids from their salts. This method is employed in the manufacture of hydrochloric acid, nitric acid etc., 

Oxidising nature of sulphuric acid

We have seen oxidation and reduction in a previous chapter. Details here. Based on that discussion, we can write the 'oxidation number' of any element in a compound.

Consider the following reaction:

C + H₂SO₄ → CO₂ + 2H₂O + 2SO₂.

This is the reaction between carbon and sulphuric acid. The reaction is carried out as follows:

• Concentrated sulphuric acid is added to a small quantity of carbon in a test tube and heated. 

From the equation, we can see that, the products are CO₂, H₂O and SO₂. Let us write the oxidation number of each element on both sides of the equation:

C⁰ + H₂⁺¹S⁺⁶O₄^-2 → C⁺⁴O₂^-2 + 2H₂⁺¹O^-2 + 2S⁺⁴O₂^-2.

From this we can see that, the oxidation state of carbon increased from 0 to +4. That means, the carbon is oxidised. The sulphuric acid acts as the oxidising agent.

Another example:

Cu + 2H₂SO₄ → CuSO₄ + 2H₂O + SO₂.

Let us write the oxidation number of each element on both sides of the equation:

Cu⁰ + 2H₂⁺¹S⁺⁶O₄^-2 → Cu⁺²S⁺⁶O₄^-2 + 2H₂⁺¹O^-2 + S⁺⁴O₂^-2.

From this we can see that, the oxidation state of copper increased from 0 to +2. That means, the copper is oxidised. The sulphuric acid acts as the oxidising agent.

In the above examples, sulphuric acid oxidised carbon (a non-metal) and copper (a metal). So we get a common property of sulphuric acid:

■ Sulphuric acid oxidises some metals and non-metals

Identification of sulphate salts

• We know that, when hydrochloric acid reacts with alkalies, we get chlorides.Example:

HCl + NaOH → NaCl + H₂O.

• HCl also reacts with metals to form chlorides. Example:

Zn + 2HCl  → ZnCl₂ + H₂.

■ We can write many examples where hydrochloric acid takes part in reactions to give chlorides. 

■ In the same way, sulphuric acid also takes part in chemical reactions to give sulphates. 

Let us see some examples:

• Sulphuric acid reacts with the alkali sodium hydroxide (NaOH) to give sodium sulphate. The equation is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

• Sulphuric acid reacts with the metal zinc to give zinc sulphate. The equation is:

Zn + H₂SO₄ → ZnSO₄ + H₂O.

• Sulphuric acid reacts with calcium carbonate to give calcium sulphate. The equation is:

CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂.

So we saw some methods by which sulphates are formed. Let us now see how these sulphate salts can be identified:

1. Prepare an aqueous solution of the given salt. We want to check whether this salt is sulphate salt or not.

2. Add some barium chloride solution to it.

3. If it is a sulphate salt, white precipitate of barium sulphate will be formed. The equation is:

BaCl₂ + Na₂SO₄ → BaSO₄↓ +  2NaCl.

4. Can we confirm that it is a sulphate salt? That is., will any other salt give a white precipitate when barium chloride is added?

In fact, like sulphate salts, carbonate salts also give white precipitate. The equation is:
BaCl₂ + Na₂CO₃ → BaCO₃↓ +  2NaCl.

5. So, when we test a given salt by adding barium chloride, and if we get a white precipitate, the given salt can either be sulphate or carbonate. We want to know which one.
6. For that, add concentrated HCl to the precipitate.
7. If the precipitate dissolves in HCl and bubbles of CO₂ are formed, it is a carbonate.
8. If the precipitate does not dissolve, it is a sulphate.

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In the next section, we will see Hydrogen chloride. 

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Chapter 3.6 - Oxidation and Reduction

In the previous section, we saw how to determine the chemical formula of any compound from the valency of it's component elements. In this section, we will see Oxidation and Reduction.

Oxidation and Reduction

We have seen the reaction between magnesium and chlorine. (Details here). The electron dot diagram is shown again below:

• After the reaction, Mg has become a positive ion. This is due to the loss of electrons. Two electrons are lost. So the charge is 2+. 
    ♦ The process of losing the two electrons can be written as: Mg → Mg²⁺ + 2e^-
• Each Cl has become a negative ion. This is due to the gain of electron. One electrons is gained by each Cl. So the charge is 1- for each.
    ♦ The process of gaining electrons can be written as: 2Cl + 2e^- → 2Cl^-
• In the reaction, Mg donates electrons, and, Cl accepts electrons.
■ Oxidation is the process of donation of electrons
■ Reduction is the process of accepting electrons 

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Is the word 'oxidation' related to 'oxygen' in any way?
To find the answer, we must look into some history. 
• The term 'oxidation' was first used by Lavoisier to mean 'reaction of a substance with oxygen'. 
• Much later, it was realized that, when any element 'X' reacts with oxygen (that is., when 'X' is oxidised), electrons are donated by 'X'
• So later, the term 'oxidation' began to be applied to all elements which donates electrons in a reaction. Regardless of whether it's reaction is with oxygen, or not.

---

An analogy to easily remember the 'difference between oxidation and reduction' is as follows:

When electrons are accepted, negative charge increases. 'Increasing negative' can be considered as 'decreasing (reducing) value'. So the element which accepts electrons can be considered to be 'reduced'.
And the opposite is applicable to oxidation:
The element which donates electrons can be considered to be 'oxidised'

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• So in the above reaction, chlorine is reduced. Reduced by whom?

    ♦ By magnesium. Magnesium reduces chlorine by 'forcing chlorine to accept electrons'. So magnesium is the reducing agent

The reverse can also be written:

• Magnesium is oxidised. Oxidised by whom?

    ♦ By chlorine. Chlorine oxidises magnesium by 'forcing magnesium to donate electrons'. So chlorine is the oxidising agent.

---

Now let us see one more example. That of Sodium fluoride (NaF):
• NaF is produced by the reaction between sodium (Na) and fluorine (F)
• Na, with atomic number 11, has the electronic configuration 2,8,1  (See here)
• F, with atomic number 9, has the electronic configuration 2,7
• So during the reaction, Na donates one electron, and, F accepts one electron
• Thus we say: Na is oxidised, and, F is reduced
• Na is oxidised by F. Because, F forces Na, to donate one electron. So F is oxidising agent
• F is reduced by Na. Because, Na forces F, to accept one electron. So Na is the reducing agent

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Oxidation number

Every element in a compound will get a particular number. What is this number? How is it determined? What is it's significance? Does this number help us in any way? Let us see the answers:

• This number is called the oxidation number

• Upon seeing the oxidation number of an element in a compound, we will immediately be able to say how many electrons that element has gained (or lost)

• For example, in NaCl, the Na atom has lost one electron. So it is given an oxidation number: +1

(Why 'positive' one? The answer is simple: losing electrons means gaining positive charge)

• Similarly, the Cl atom has gained one electron. So it is given an oxidation number: -1
• Together, they are represented as: Na⁺¹Cl^-1

---

So when we see Na⁺¹Cl^-1 , we can write the following details:
• Na is in an oxidation state of +1. That is., it has lost one electron
• Cl is in an oxidation state of -1. That is., it has gained one electron

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Let us consider some more examples:
Consider MgO. We have seen the details here. The diagram is shown again below:

We can write the oxidation state as: Mg⁺²O^-2
• Mg is in an oxidation state of +2. That is., it has lost two electrons
• O is in an oxidation state of -2. That is., it has gained two electrons

Consider Aluminium chloride (AlCl₃)
• Al, with atomic number 13, has the electronic configuration 2,8,3  (See here)
• Cl, with atomic number 17, has the electronic configuration 2,8,7
• So Al needs to lose 3 electrons and Cl needs to gain one electron
• Thus 3 chlorine atoms combine with one aluminium atom. Each chlorine atom accept one of the three electrons donated by the aluminium atom
• One aluminium atom loses 3 electrons. So it's oxidation state is +3
• Each chlorine atom gains one electron so oxidation state of each is -1
• So we can write: Al⁺³Cl₃^-1 

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When we see Al⁺³Cl₃^-1, we can write the following:
• Aluminium is in an oxidation state of +3. That is., it has lost 3 electrons
• Chlorine is in an oxidation state of -1. That is., each chlorine atom has gained one electron

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For covalent compounds, there is no transfer of electrons. Only sharing of electrons takes place. That is., there is no donating or accepting of electrons. In such a case, how do we write the oxidation number?
Consider HCl. We have seen the details here. The diagram is shown again below:

In such cases, it is assumed that, the shared pairs of electrons are completely displaced towards the more electronegative atom. Let us see the application of this assumption in our present case.
• Electronegativity of H = 2.20 (see the chart here)
• Electronegativity of Cl = 3.16 
• Chlorine is more electronegative. So the shared pair of electrons move towards the chlorine atom. 
• That is., two electrons move towards the chlorine atom
• Out of those two, one was already possessed by Cl. So the number of extra electrons is only 1
• Thus the chlorine is given an oxidation number of -1
• What about hydrogen?
• The hydrogen is now left with zero electrons. Because the pair as a whole, has moved away from the hydrogen atom
• Out of the two electrons that has moved away, one belonged to the hydrogen atom. Now it is gone.
• So in effect, hydrogen has lost one electron, and so it is given an oxidation number of +1
• We can write the equation as:  H₂⁰ + Cl₂⁰ → 2H⁺¹Cl^-1 

Let us see another example. Consider carbon dioxide CO₂
Two oxygen atoms forms a covalent bond with carbon. We saw the details here. It is shown again below:

• Electronegativity of C = 2.55 
• Electronegativity of O = 3.44 
• Oxygen is more electronegative. So the shared pair of electrons move towards the oxygen atom. 
• That is., in each bond, four electrons move towards the oxygen atom
• Out of those four, two were already possessed by O. So the number of extra electrons is only 2
• Thus each oxygen is given an oxidation number of -2
• What about carbon?
• The carbon is now left with zero electrons. Because both the pairs have moved away from the carbon atom
• Out of the eight electrons that has moved away, four belonged to the carbon atom. Now they are gone.
• So in effect, carbon has lost four electrons, and so, it is given an oxidation number of +4
• We can write the equation as:  C⁰ + 2O⁰ → C⁺⁴O₂^-2 

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So we see that for both ionic and covalent compounds, the component elements will have oxidation numbers. But it is very important to remember an exception to this rule:

• Consider the molecules of elements. In such molecules, the component atoms are all of the same element.

• All the component atoms will be having the same power to attract the shared electrons. We have seen examples of such bonding here. Some are shown again below:

• So we see that, when the component atoms are the same, the shared electron pairs will be attracted equally, and so they will remain exactly midway between the atoms

• No atom will get any extra electrons, and so the oxidation numbers of all component atoms will be zero

• We write them as: F₂⁰, Cl₂⁰ etc.,

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Telling 'whether it is an oxidation or a reduction' by using oxidation number

So we have seen the oxidation number of elements in both ionic and covalent compounds. We will now see a practical application of this number:
Consider the formation of HCl that we saw above. If we examine the equation H₂⁰ + Cl₂⁰ → 2H⁺¹Cl^-1 , we will see that:
• The oxidation number of H has increased from 0 to +1
• The oxidation number of Cl has decreased from 0 to -1
■ The element, whose oxidation number increases, is oxidised. In other words, the element, whose oxidation number increases, has undergone oxidation.
    ♦ In our example, hydrogen is oxidised. In other words, hydrogen has undergone oxidation
■ The element, whose oxidation number decreases, is reduced. In other words, the element, whose oxidation number decreases, has undergone reduction.
    ♦ In our example, chlorine is reduced. In other words, chlorine has undergone reduction.

Another example: Consider the formation of carbon dioxide. If we examine the equation C⁰ + 2O⁰ → C⁺⁴O₂^-2 , we will see that:
• The oxidation number of C has increased from 0 to +4. So C has undergone oxidation
• The oxidation number of O has decreased from 0 to -2. So O has undergone reduction

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We see that, in a reaction, one element is oxidised and the other element is reduced. That is., oxidation and reduction takes place simultaneously. So the overall process is known as a redox process.

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We see that, in a reaction, the element which increases in the oxidation number is the one which gets oxidised.
• That element  gets oxidised by whom?
• It is oxidised by the other element. We know that, this 'other element' which causes the oxidation is called the 'oxidising agent'.
• Now, what is the state of the 'oxidation number' of this 'other element', which is the 'oxidising agent'?
• The 'oxidation number' of this 'other element' decreases.
■ So we can write: The element whose oxidation number decreases, is the oxidising agent.
■ Similarly, The element whose oxidation number increases is the reducing agent

This can be explained using a general example. Consider the reaction between two elements 'P' and 'Q' (symbols are not real):
• P^x1 + Q^y1 → P^x2Q^y2   [(x2 > x1) and (y2 <y1)]
• We see that, after the reaction, P has an increased oxidation number.
• So P has undergone oxidation
• Obviuosly, this oxidation was caused by Q, and so, we call Q, the oxidising agent
• Oxidation number of Q has decreased
■ So we can write: The element whose oxidation number decreases, is the oxidising agent.
And the reverse can also be worked out:
• We see that, after the reaction, Q has a decreased oxidation number.
• So Q has undergone reduction
• Obviuosly, this reduction was caused by P, and so, we call P, the reducing agent
• Oxidation number of P has increased
• So we can write: The element whose oxidation number increases, is the reducing agent. 

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We will now see an example:

• The equation for the reaction between hydrochloric acid (HCl) and zinc (Zn), to form zinc chloride and hydrogen is given below:

• Zn + 2HCl → ZnCl₂ + H₂

• Note that, it is a balanced equation. The modified form of this equation, which shows the oxidation number of each element is given below:

• Zn⁰ + 2H⁺¹Cl^-1→ Zn⁺²Cl₂^-1 + H₂⁰

• From this modified equation, we can write the following details:

■ Oxidation number of Zn increases from 0 to +2. So, Zn undergoes oxidation. And also, it is the reducing agent

■ Oxidation number of H decreases from +1 to 0. So H undergoes reduction. And also, it is the oxidising agent

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• So now we are able to find which element gets oxidised, and which element gets reduced

• Also, we can find which is the oxidising agent, and which is the reducing agent
• We are able to find the above because, the 'oxidation number' of one element increases, and that of the other element decreases.
• But there is some thing more than just 'increase and decrease':
■ The sum of oxidation numbers in a compound will be always zero. 
Let us see an example. Consider Zn⁺²Cl₂^-1:
• Total oxidation number of Zn:
    ♦ There is only one Zn atom, and it has an oxidation number of +2. Thus:
    ♦ Total oxidation number of Zn = number of Zn atoms × oxidation number of each = 1 × +2 = +2 
• Total oxidation number of Cl:
    ♦ There are two Cl atoms, each with an oxidation number -1. Thus:
    ♦ Total oxidation number of Cl = number of Cl atoms × oxidation number of each = 2 × -1 = -2
■ Total sum of the oxidation numbers of all the atoms in the molecule = 2 + -2 = 0 

Another example: Consider Al⁺³Cl₃^-1:
• Total oxidation number of Al:
    ♦ There is only one Al atom, and it has an oxidation number of +3. Thus:
    ♦ Total oxidation number of Al = number of Al atoms × oxidation number of each = 1 × +3 = +3 
• Total oxidation number of Cl:
    ♦ There are three Cl atoms, each with an oxidation number -1. Thus:
    ♦ Total oxidation number of Cl = number of Cl atoms × oxidation number of each = 3 × -1 = -3
■ Total sum of the oxidation numbers of all the atoms in the molecule = 3 + -3 = 0

Another example: Consider Mg⁺²O^-2:
• Total oxidation number of Mg:
    ♦ There is only one Mg atom, and it has an oxidation number of +2. Thus:
    ♦ Total oxidation number of Mg = number of Mg atoms × oxidation number of each = 1 × +2 = +2 
• Total oxidation number of O:
    ♦ There is only one O atom, and it has an oxidation number of -2. Thus:
    ♦ Total oxidation number of O = number of O atoms × oxidation number of each = 1 × -2 = -2 
■ Total sum of the oxidation numbers of all the atoms in the molecule = 2 + -2 = 0

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Thus we find that the sum of oxidation numbers in any molecule is equal to zero. In the next section, we will see a practical application of this property 

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