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 4.1 - Families and Periodic trends in the Periodic table

We have seen how the elements are arranged in the periodic table. The arrangement is based on rules 7 and 8 that we saw in the previous section. When the elements are arranged in this way, we obtain some favourable 'side effects'. What are those side effects? Let us examine:

Look at the group I. The elements in this group are: Li, Na, K, Rb. Cs and Fr

• All of them have 1 electron in their outer most shells.

• These elements have similar properties. All of them form alkalies when reacting with water. For example, When sodium (Na) reacts with water, sodium hydroxide (NaOH) is formed.

• Because of the similarities in the properties, these elements can be regarded as 'members of a single family'. This family is given a special name: Alkali metals.

• Similarly, all elements in group II belong to the family called Alkaline Earth metals. The properties of all the elements in this family are also similar. Just like Alkali metals, this family also form alkalies. Then why the term 'earth'? 

• Because, in early days, these elements were seen only as compounds, in 'minerals obtained from earth'. They do not occur in free form in nature. Later, scientists were able to isolate them from the minerals. But the name stuck.

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The table below gives the full list of the families in the periodic table:

Group I: Alkali metals

Group II: Alkaline earth metals

Group III: Boron family

Group IV: Carbon family

Group V: Nitrogen family

Group VI: Oxygen family

Group VII: Halogens

Group VIII: Noble gases

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The elements helium, neon, argon, krypton, xenon and radon, which belong to group VIII are called Noble gases. They are monoatomic molecules. Because they do not need to combine with another atom to attain stability. (We have seen the details here also see here). They already have octet configuration. Normally, they do not combine with other elements also. Hence they are called inert gases. As they are found only in very small quantities, they are also called rare gases.

• Helium is used in weather balloons.

• Neon is used in discharge lamps to obtain orange colour

• Argon is used in electrical bulbs to prevent the evaporation of the filament

• Radon is a radioactive gas

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So we find that elements with similar properties fall together in the periodic table. This is one of the many 'favourable side effects' that we obtain, while making a systematic arrangement of elements. We will see more as we continue our discussion.

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Representative elements

The first 10 elements show a periodicity in electron filling. This is shown in the table below:

Look at the column for L-shell. The electron filling is in sequential order from 1 to 8. In the periodic table, these first 10 elements reside at the top most portions in their respective groups. They are the first elements in their families, and are called Representative elements.

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• The position of hydrogen is still under debate. Some of it's properties makes it eligible to be a member of the Alkali metal family (group I)

• Some of it's other properties makes it eligible to be a member of the halogen family (group VII). Let us examine:
1. Hydrogen loses one electron in some chemical reactions. Alkali metals also lose one electron in chemical reactions

2(a) Hydrogen is a non metal. Halogens are also non metals

(b) Hydrogen is diatomic. Halogens are also diatomic

(c) Hydrogen is not a solid. Halogens are also not solid. (Alkali metals are solid)

(d) Hydrogen is not a metal. Halogens are also not metals. (Alkali 'metals' are metals)

(e) Hydrogen has high ionisation energy. Halogens also have high ionisation energy. (Alkali metals have low ionisation energy). We will learn about ionisation energy later in the next section.

(g) Hydrogen gains one electron in some chemical reactions. Halogens also gain one electron in chemical reactions

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Transition elements

Consider the groups I and II. The elements in these two groups are metals. Consider the groups III to VIII. They are non metals. [However, some elements (inside green squares coming under groups III, IV, V and VI in fig.4.8) show properties of both metals and non metals. They are classified as Metalloids] The elements in the space between group II and group III are called Transition elements. Because, they form a transition from metals to non metals. The following are some of the properties of transition elements:

• They are metals

• They form coloured compounds

• They show similarities in chemical properties in groups as well as in periods. That is:

    ♦ If we take any single vertical group from within the transition elements, all the elements in that group will have similar chemical properties

    ♦ If we take any single horizontal period from within the transition elements, all the elements in that period will have similar chemical properties

• In compounds, they exhibit different oxidation states. For example: Fe²⁺ and Fe³⁺

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Periodic trends in the Periodic table

The word 'period' means an interval of time. An example for the usage of the word 'periodic' is:

Periodic checkup of a car. It means, that, the car has to undergo regular checkup at the ends of definite periods. So the event 'checkup of the car' occurs at the end of regular periods. In other word, it occur periodically.

In the same way, some of the properties of elements occur periodically. Let us see an example:

■ Point your finger on Lithium No.3

• It has 1 electron in it's outer most shell.

• Now add 8 to it's number. We get 3 + 8 = 11. Element no. 11 is sodium.

• It also has 1 electron in it's outer most shell

■ Point your finger on Fluorine No.9

• It has 7 electrons in it's outer most shell.

• Now add 8 to it's number. We get 9 + 8 = 17. Element no. 17 is Chlorine.

• It also has 7 electrons in it's outer most shell

■ So 8 is the period. This period '8' works for elements upto magnesium No.12. 
Let us see the period for the elements after magnesium:

■ Point your finger on Potassium No. 19.

• It has 1 electron in it's outer most shell.

• Now add 18 to it's number. We get 19 + 18 = 37. Element no. 37 is Rubidium.

• It also has 1 electron in it's outer most shell

■ Point your finger on germanium No. 32.

• It has 4 electrons in it's outer most shell.

• Now add 18 to it's number. We get 32 + 18 = 50. Element no. 50 is Tin.

• It also has 4 electron in it's outer most shell

■ So 18 is the period for the elements after 12.

What is the difference between the two periods '8' and '18'?

The difference is 18 – 8 = 10. Note that 10 is the 'number of groups' of the transition elements
It is not just the 'number of electrons in the outer most shell'. Chemical properties of the element obtained by adding the appropriate period 8 or 18 will also be similar.

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• We have not checked the period for any transition elements.

• For the transition elements, we may not get the same number of electrons in the outer most shell after the period 18. 
• But still, the element that we get by adding 18, will have similar chemical properties.

• Within the transition elements, after barium No.56, the period will become '32' instead of '18'. This is because of the presence of 14 Lanthanides and Actinides. [18 + 14 = 32] 

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Now let us see some important properties that vary periodically:

Size of atoms in a group

When we move down any group, the size of atom increases. This is because, the number of shells increases.

Example: In group II, Ca has 4 shells. Sr, which is just below Ca, has 5 shells. Naturally, the atom with 5 shells will be larger than that with 4 shells. Some Bohr models showing the details of shells can be seen here.

Size of atoms in a period

We want to know how the size of atoms vary when we move from left to right in any period. Let us write an analysis:

1. In any period, the 'number of shells' is same for all the elements in that period

2. In any period, when we move from left to right, the number of electrons get increased by '1'

3. The number of electrons is same as the number of protons in the nucleus. So the number of protons also increase as we move from left to right.

4. Greater number of protons and electrons means that there will be greater force of attraction between the positively charged protons and the negatively charged electrons.

5. The greater force of attraction pulls the shells more and more towards the nucleus.

6. As a result, the overall size of the atom decreases as we move from left to right in a period.

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In the next section, we will discuss more Periodic properties. 

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