Chapter 9.6 - Properties of d-block and f-block elements

In the previous section, we saw the properties of s-block and p-block elements.  In this section we will see the properties of d-block elements.

Properties of d-block elements 

■ Position: The d-block consists of the groups from 3 to 12 of the periodic table
■ Common names:
The elements of the groups from 3 to 12 are known as transition elements
■ Oxidation state
• Let us find the oxidation state of Fe in FeCl₂. 
We have seen the method of calculation here. From the table we have: oxidation state of Cl = -1
Let the oxidation state of Fe be 'x'. Then we get: x + 2 × -1 = 0 ⇒ x -2 = 0 ⇒ x = +2   
So the oxidation state of Fe in FeCl₂ is +2
• Let us find the oxidation state of Fe in FeCl₃. 
From the table we have: oxidation state of Cl = -1
Let the oxidation state of Fe be 'x'. Then we get: x + 3 × -1 = 0 ⇒ x -3 = 0 ⇒ x = +3   
So the oxidation state of Fe in FeCl₃ is +3
• So we see that Fe shows variable oxidation states. Let us analyse and find the reason for this:
1. The S.E.C of  Fe is: 1s²2s²2p⁶3s²3p⁶4s²3d⁶ . 
Fe loses two electrons and becomes Fe²⁺. The two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
So the S.E.C of  Fe²⁺ is: 1s²2s²2p⁶3s²3p⁶3d⁶ .
2. The S.E.C of  Fe is: 1s²2s²2p⁶3s²3p⁶4s²3d⁶ . 
Fe loses three electrons and becomes Fe³⁺. 
First, two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
Then one electron is lost from the d subshell in the penultimate main shell. That is., the 3d subshell. 
So the S.E.C of  Fe³⁺ is: 1s²2s²2p⁶3s²3p⁶3d⁵.
• When we consider the main shells alone, this phenomenon cannot be explained. 
• But the explanation becomes clear when we consider subshells also:
• The difference in energy between 3d subshell and 4s subshell is very small. So, under suitable conditions, the electron in 3d subshell also take part in chemical reactions. 
• This accounts for the variable oxidation states shown by transition elements  

Another example:
• Let us find the oxidation state of Mn in MnCl₂. 
We have seen the method of calculation here. From the table we have: oxidation state of Cl = -1
Let the oxidation state of Mn be 'x'. Then we get: x + 2 × -1 = 0 ⇒ x -2 = 0 ⇒ x = +2   
So the oxidation state of Mn in MnCl₂ is +2
• Let us find the oxidation state of Mn in MnO₂. 
From the table we have: oxidation state of O = -2
Let the oxidation state of Mn be 'x'. Then we get: x + 2 × -2 = 0 ⇒ x -4 = 0 ⇒ x = +4   
So the oxidation state of Mn in MnO₂ is +4
• Let us find the oxidation state of Mn in Mn₂O₃. 
From the table we have: oxidation state of O = -2
Let the oxidation state of Mn be 'x'. Then we get: 2 × x + 3 × -2 = 0 ⇒ 2x -6 = 0 ⇒ x = +3   
So the oxidation state of Mn in Mn₂O₃ is +3
• Let us find the oxidation state of Mn in Mn₂O₇. 
From the table we have: oxidation state of O = -2
Let the oxidation state of Mn be 'x'. Then we get: 2 × x + 7 × -2 = 0 ⇒ 2x -14 = 0 ⇒ x = +7   
So the oxidation state of Mn in Mn₂O₇ is +7

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1. The S.E.C of Mn is: 1s²2s²2p⁶3s²3p⁶4s²3d⁵.

Mn loses two electrons and becomes Mn²⁺. The two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
So the S.E.C of  Mn²⁺ is: 1s²2s²2p⁶3s²3p⁶3d⁵ . 
2. The S.E.C of  Mn is: 1s²2s²2p⁶3s²3p⁶4s²3d⁵ . 
Mn loses four electrons and becomes Mn⁴⁺. 
First, two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
Then two electrons are lost from the d subshell in the penultimate main shell. That is., the 3d subshell. 
So the S.E.C of  Mn⁴⁺ is: 1s²2s²2p⁶3s²3p⁶3d³.
3. The S.E.C of  Mn is: 1s²2s²2p⁶3s²3p⁶4s²3d⁶ . 
Mn loses three electrons and becomes Mn³⁺. 
First, two three electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
Then one electron is lost from the d subshell in the penultimate main shell. That is., the 3d subshell. 
So the S.E.C of  Mn³⁺ is: 1s²2s²2p⁶3s²3p⁶3d⁴.
4. The S.E.C of  Mn is: 1s²2s²2p⁶3s²3p⁶4s²3d⁵ . 
Mn loses seven electrons and becomes Mn⁷⁺. 
First, two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
Then five electrons are lost from the d subshell in the penultimate main shell. That is., the 3d subshell. 
So the S.E.C of  Mn⁴⁺ is: 1s²2s²2p⁶3s²3p⁶.
• We see that electrons in the d subshell in the penultimate main shell also takes partin the reactions. 
• This is because, the difference in energy level between the following two is very low:
    ♦ s subshell in the outermost main shell
    ♦ d subshell in the penultimate main shell
• So transition elements in general show variable oxidation states
■ Metallic property
The transition elements are metals
■ Coloured compounds
Most of the compounds formed by transition elements are coloured. Some examples are given below:
• Copper sulphate is a compound of copper, which is a transition element. This compound is blue in colour  
• Cobalt nitrate is a compound of cobalt, which is a transition element. This compound is light pink in colour
• Potassium permanganate is a compound of manganese, which is a transition element. This compound is violet in colour    
• Ferrous sulphate is a compound of iron, which is a transition element. This compound is light green in colour  
• Ammonium dichromate is a compound of chromium, which is a transition element. This compound is orange in colour  
• Compounds of transition elements are used to give colour to glass, prepare some dyes etc.,
■ Uses of transition elements
• Copper is used for making electric cables
• Nickel is used for making coins
• Titanium is used aircraft parts

Properties of f-block elements 

■ Position: The f-block elements are those elements which are placed in two rows at the bottom of the periodic table. We have seen the reason for placing them separately in an earlier chapter. Details here.
The elements of the first row are called Lanthanoids and those in the second row are called Actinoids. They belong to the sixth and seventh periods respectively.
■ S.E.C:
• We know that, the last electron of each element in the f-block will be filled in the f subshell. 
Let us examine the S.E.C of two elements:
• First we will consider ₅₈Ce (Cerium). It is a Lanthanoid 
It has the S.E.C: [Xe]4f¹5d¹6s². 
From the S.E.C it is clear that, cerium falls in the 6^th period.
The last main shell is 6. But the last electron falls in the f subshell of the main shell 4. That is., 4f    
• Next we will consider ₉₀Th (Thorium). It is an Actinoid 
It has the S.E.C: [Rn]5f³6d¹7s². 
From the S.E.C it is clear that, thorium falls in the 7^th period.
The last main shell is 7. But the last electron falls in the f subshell of the main shell 5. That is., 5f

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So we can write:

• f-block elements Lanthanoids belong to the 6^th period. Their last electrons are filled in the 4f subshell 
• f-block elements Actinoids belong to the 7^th period. Their last electrons are filled in the 5f subshell 

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■ Characteristics and uses of f-block elements

• Most of the f-block elements show variable oxidation states like the d-block elements

• Most of the actinoids are radioactive and are artificial elements

• Uranium, thorium, plutonium etc., are used as fuels in nuclear reactors

• Many of them are used as catalysts in the petroleum industry

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Now we will see some solved examples

Solved example 9.7

The element X in group 17 has 3 shells. If so,

(i) Write the S.E.C of the element

(ii) Write the period number

(iii)What will be the chemical formula of the compound formed if the element X reacts with element Y of the third period which contains one electron in the p subshell?

Solution:
Part (i):
1. We have two clues:
• The element belongs to group 17
    ♦ From this it is clear that it belongs to the p-block
• It has 3 shells
    ♦ From this it is clear that it belongs to the 3^rd period
2. For any element in the p-block, the group number is obtained as follows:
• Number of electrons in the p subshell of the last main shell + 12
3. So we can write: 17 =  Number of electrons in the p subshell of the 3^rd main shell + 12
• Thus we get: Number of electrons in the p subshell of the 3^rd main shell = 5
4. For any element in the p-block, the last electron is filled in the p subshell of the last main shell
So we can write:
• The last term of the S.E.C is 3p⁵.
5. If the last term is 3p⁵, the preceding terms will be:
1s²2s²2p⁶3s².
• So the S.E.C is: 1s²2s²2p⁶3s²3p⁵. 
Part (ii):
The period number is 3
Part (iii):
1. First we will write the S.E.C of element Y
We have two clues:
• The element belongs to the third period
    ♦ From this it is clear that the suffix of the last term of the S.E.C is 3
• It has one electron in the p subshell
    ♦ The p subshell can hold 6 electrons. So, if there is only one electron, it will be the last electron
    ♦ So the last term is 3p¹
If the last term is 3p¹, the preceding terms will be:
1s²2s²2p⁶3s²
• So the S.E.C is: 1s²2s²2p⁶3s²3p¹. 
2. The last main shell has 3 electrons. So the valency of element Y is 3
From the S.E.C of element X, it's valency is 8 -7 = 1
3. We have seen how to write the chemical formula from valency. Details here.
4. We have to find the element with lower electronegativity between X and Y 
From the S.E.C of Y, the group number of Y is 13
(Number of electrons in the p subshell of the 3^rd main shell + 12 = 1 + 12 = 13)
So Y is in the group 13 and period 3. X is in group 17 and same period
Y is on the left of X in the same period
Thus Y is less electronegative than X
So Y is to be written first
5. So the required chemical formula is YX₃

Solved example 9.8

The element Cu with atomic number 29 undergoes chemical reaction to form an ion with oxidation state +2

(i) Write down the S.E.C of this ion

(ii) Can Cu show variable valency? Why?

(iii) Write down the chemical formula of one compound formed when Cu reacts with ₁₇Cl. 
Solution:
Part (i):
1. The S.E.C of Cu (copper) is 1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰  OR [Ar]4s¹3d¹⁰.
2. There is only a small energy difference between 4s and 3d subshells. So if conditions are favourable,  electrons in the 3d subshell may also take part in chemical reactions
3. When two electrons are lost to form +2 oxidation state, it is clear that, one electron from the 3d is also lost in addition to the last electon in 4s
4. So the S.E.C of the ion is: 1s²2s²2p⁶3s²3p⁶3d⁹.
Part (ii)
Cu can show variable valency because, if conditions are favourable,  electrons in the 3d subshell may also take part in chemical reactions  
Part (iii):
1. The S.E.C of Chlorine is: 1s²2s²2p⁶3s²3p⁵.
It requires 1 more electron to complete octect. So it's valency is 1
2. Cu is ready to give two electrons. One from the 4s subshell and the other from the 3d subshell. So it's valency is 2
3. We have seen how to write the chemical formula from valency. Details here. 
4. Thus the chemical formula of the compound is CuCl₂. 
Note that Cu, which has a lower electronegativity is written first.

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We have completed our present discussion about Subshell electronic configuration and properties of various blocks. In the next section, we will see Mol concept. 

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Chapter 9.5 - Properties of s-block and p-block elements

In the previous section, we saw how to determine the group number of an element from it's S.E.C. We also saw which all groups fall under each block.  In this section we will see the properties of s-block and p-block elements.

Properties of s-block elements 

■ Position: The s-block consists of the groups I and II of the periodic table
■ Common names:
• Group I elements have the common name Alkali metals 
• Group II elements have the common name Alkaline earth metals (Details here)
■ Acceptance or Donation of electrons during chemical reactions:
Consider the elements Li, Na and K of Group I. Let us write their S.E.C:
1. Li - 1s²2s¹ 
• The outer most main shell is 2.
• This main shell has one electron, which is in 2s subshell
• During chemical reactions, this one electron is donated. This is because, donating one electron is easier than accepting 7 electrons to attain octet 
2. Na - 1s²2s²2p⁶3s¹
• The outer most main shell is 3.
• This main shell has one electron, which is in 3s subshell
• During chemical reactions, this one electron is donated. This is because, donating one electron is easier than accepting 7 electrons to attain octet
3. K - 1s²2s²2p⁶3s²3p⁶4s¹
• The outer most main shell is 4.
• This main shell has one electron, which is in 4s subshell
• During chemical reactions, this one electron is donated. This is because, donating one electron is easier than accepting 7 electrons to attain octet

Consider the elements Be, Mg and Ca of Group II. Let us write their S.E.C:  
1. Be - 1s²2s²
• The outer most main shell is 2.
• This main shell has two electrons, which is in 2s subshell
• During chemical reactions, these two electrons are donated. This is because, donating two electrons is easier than accepting 6 electrons to attain octet 
2. Mg - 1s²2s²2p⁶3s²
• The outer most main shell is 3.
• This main shell has two electrons, which is in 3s subshell
• During chemical reactions, these two electrons are donated. This is because, donating two electrons is easier than accepting 6 electrons to attain octet 
3. Ca - 1s²2s²2p⁶3s²3p⁶4s²
• The outer most main shell is 4.
• This main shell has two electrons, which is in 4s subshell
• During chemical reactions, these two electrons are donated. This is because, donating two electrons is easier than accepting 6 electrons to attain octet 

■ Oxidation state
• All the elements in group I always donate 1 electron. So they always show an oxidation state of +1
• All the elements in group II always donate 2 electrons. So they always show an oxidation state of +2 
• We have seen that some elements show fixed oxidation states which ever be the reaction. While some others show variable oxidation states. (We have seen details about oxidation number or oxidation state here) 
    ♦ Groups I and II elements always show definite fixed oxidation states. 
■ Atomic radius
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the atomic radius decreases as we move from left to right in a period. 
• The s-block elements are at the left most end of the various periods. So we can say that, the s-block elements have high atomic radius
• Also, when we move from top to bottom in any group in the s-block, the atomic radius increases
■ Ionization energy
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the ionization energy increases as we move from left to right in a period. 
• The s-block elements are at the left most end of the various periods. So we can say that, the s-block elements have low ionization energy
• That means lower energies are sufficient to remove electrons from the s-block elements
• Also these elements form mostly ionic compounds
■ Electronegativity
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the electronegativity increases as we move from left to right in a period. 
• The s-block elements are at the left most end of the various periods. So we can say that, the s-block elements have low electronegativity
■ Metallic nature
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the metallic nature decreases as we move from left to right in a period. 
• The s-block elements are at the left most end of the various periods. So we can say that, the s-block elements have high metallic nature
■ Reactivity
• The Group I elements will be the first element in respective periods. 
• These first elements have the greatest reactivity in the respective periods
• Also, the reactivity increases as we move from top to bottom in the groups
■ Oxides and hydroxides
• Magnesium oxide (MgO) is an example of an oxide of an s-block element. It is formed when magnesium ribbon burns in air. 
    ♦ Other examples of oxides from this s-block are: Calcium oxide (CaO) and Barium oxide (BaO) 
• Sodium hydroxide (NaOH) and Potassium hydroxide (KOH) are examples of hydroxides formed from this block. They are alkaline in nature.
• Metallic oxides give alkalies and non-metallic oxides give acids. We have seen the details in an earlier chapter here.
    ♦ Above we have seen that s-block elements are metallic in nature. So we can say that the oxides and hydroxides formed from this block are alkaline in nature.

Properties of p-block elements 

■ Position: The p-block consists of the groups 13 to 18 of the periodic table
■ S.E.C:
• We know that, the last electron of each element in the p-block will be filled in the p subshell. That means, there will always be a p subshell in the last main shell of any p-block element
• Now, p subshell will be filled only after completing the s subshell. So, the last main shell will contain a s subshell also.
• Thus, The general form of the last main shell configuration is: ns²np^{(1 to 6)}.
• Where n is the last main shell. The superscript for s will always be 2
• The superscript for p varies from 1 to 6. This is shown below:
The superscript for p for all elements in group 13 will be 1 (Add 12 to 1, and we get the group no.13) 
The superscript for p for all elements in group 14 will be 2 (Add 12 to 2, and we get the group no.14) 
The superscript for p for all elements in group 15 will be 3 (Add 12 to 3, and we get the group no.15) 
The superscript for p for all elements in group 16 will be 4 (Add 12 to 4, and we get the group no.16) 
The superscript for p for all elements in group 17 will be 5 (Add 12 to 5, and we get the group no.17) 
The superscript for p for all elements in group 18 will be 6 (Add 12 to 6, and we get the group no.18)
An example:
• ₃₄Se    - 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁴  OR [Ar]3d¹⁰4s²4p⁴.
    ♦ Add 12 to the subscript 4 of the 4p subshell. We get 16, which is indeed the group number of Selenium

■ Common names:
• Group 13 elements have the common name Boron family
• Group 14 elements have the common name Carbon family
• Group 15 elements have the common name Nitrogen family
• Group 16 elements have the common name Oxygen family
• Group 17 elements have the common name Halogens
• Group 18 elements have the common name Noble gases

■ Oxidation state
• Some of the elements in the p-block shows variable oxidation states. But in general, we can write the values as follows:
• +3 oxidation state for group 13 elements 
    ♦ donating 3 electrons in the outermost main shell
• +4 oxidation state for group 14 elements 
    ♦ sharing 4 pairs of electrons in the outermost main shell. But all the 4 electrons will move away because of low electronegativity
• -3 oxidation state for group 15 elements 
    ♦ there are 5 electrons in the outermost main shell. Will accept 3 electrons to attain octet 
• -2 oxidation state for group 16 elements 
    ♦ there are 6 electrons in the outermost main shell. Will accept 2 electrons to attain octet  
• -1 oxidation state for group 17 elements 
    ♦ there are 7 electrons in the outermost main shell. Will accept 1 electron to attain octet 
• 0 oxidation state for group 18 elements 
    ♦ Already have an octet configuration
• The above values should be used only as general guide lines. Accurate values should be carefully calculated  for each compound.

■ Atomic radius
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the atomic radius decreases as we move from left to right in a period.
• So, as we move from left to right in any period in the p-block, the atomic radius decreases
• Also, when we move from top to bottom in any group in the p-block, the atomic radius increases
■ Ionization energy
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the ionization energy increases as we move from left to right in a period. 
• So, as we move from left to right in any period in the p-block, the ionization energy increases
• Also, when we move from top to bottom in any group in the p-block, the ionisation energy decreases
■ Electronegativity
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the electronegativity increases as we move from left to right in a period. 
• So, as we move from left to right in any period in the p-block, the electronegativity increases
• Also, when we move from top to bottom in any group in the p-block, the electronegativity decreases

■ Metallic nature
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the metallic nature decreases as we move from left to right in a period. 
• So, as we move from left to right in any period in the p-block, the metallic nature decreases
• Also, when we move from top to bottom in any group in the p-block, the metallic nature increases
• Among the p-block elements, both metals and non-metals are present

■ Reactivity
• Consider the Group 17 elements. 
• They have the smallest atomic size among the p-block elements
• They have the highest electronegativity among the p-block elements
• They need only one more electron to attain octet
• So they have the highest reactivity among the p-block elements
• These first elements have the greatest reactivity in the respective periods
• Also, the reactivity increases as we move from top to bottom in the groups
■ The Group 18 elements needs special mention
• They have eight electrons in the outermost shell
• Their S.E.C ends with ns²np⁶  
• They already have octet. So they does not show any reactivity.
• All the elements in this group are gases
• They are mono atomic because they do not need to combine with other atoms for stability

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In the next section, we will see the properties of d-block and f-block elements. 

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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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