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 9.4 - Group number from Subshell Electronic Configuration

In the previous section, we saw how to determine the block name and period number of any element using the S.E.C. In this section we will see the method to determine the group number.

Arrangement of elements into Groups

The rule for determining the group number of a given element will depend on the block in which that element resides. That means:
• The elements in the s-block have their own unique rule to determine group numbers
• The elements in the p-block have their own unique rule to determine group numbers
• The elements in the d-block have their own unique rule to determine group numbers
■ However there is basic step which  is common for all blocks. It will be come clear when we write the rule for finding the group for any element.

Rule 9.3
Given any element. We want to know the group in which that element resides in the periodic table. For that, use the following steps:
■ Basic step for any element:
• Apply rule 9.1 and find the block in which the element resides in the periodic table
    ♦ Once we complete the above basic step, we can choose the appropriate one from A, B or C given below:
A. Rule for s-block
1. From the S.E.C, find the outer most Main shell
2. In the outer most main shell, find the s subshell
3. In this s subshell, find the number of electrons
4. This number is the 'group number' of that element

B. Rule for p-block
1. From the S.E.C, find the outer most Main shell
2. In the outer most main shell, find the s and p subshells
3. Find the total number of electrons in those s and p subshells
4. Add '10' to this sum
5. The final sum obtained by adding '10' is the 'group number' of that element
OR
1. From the S.E.C, find the outer most Main shell
2. In the outer most main shell, find the p subshell
3. Find the number of electrons in that p subshell
4. Add '12' to this number
5. The sum obtained by adding '12' is the 'group number' of that element

C. Rule for d-block
1. From the S.E.C, find the outer most Main shell
• In the outer most main shell, find the s subshell
• Write down the number of electrons in this s subshell
2. From the S.E.C, find the second last Main shell
• In the second last main shell, find the d subshell
• Write down the number of electrons in this d subshell
3. Add the number of electrons in (1) and (2)
4. The sum thus obtained is the 'group number' of that element

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Based on the above rules we can determine the group number of any given element. Consider the fig.9.2 that we saw in the previous section. For convenience, it is shown again below: 

Fig.9.2

[Above fig.9.2 is taken from wikipedia. See details here]
■ We get the following information:
• Groups 1 and 2 belong to the s-block  
• Groups 3 to 12 belong to the d-block
• Groups 13 to 18 belong to the f-block
■ We know that, the f-block elements consist of Lanthanides and Actinids. 
• They are kept away from the main body of the periodic table. If we accommodate them, they will also get group numbers. 
• But then,  the periodic table will become very large. We saw those details in a previous chapter here.

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Now we will see some solved examples
Solved example 9.3
The atomic number of an element is 16. Find the following:
(i) Subshell configuration (S.E.C)  (ii) Block (iii) Period  (iv) Group
Solution:
(i) 1s²2s²2p⁶3s²3p⁴  OR [Ne]3s²3p⁴.
(ii) Block: Apply rule 9.1:
• The last electron is filled in the 3p subshell. So the element resides in the p-block
(iii) Period: Apply rule 9.2:
• The highest main shell number in the S.E.C is 3. So the element resides in the third period  
(iv) Group: Apply rule 9.3 that we saw above.
We have already seen that the element belongs to the p-block. So apply Rule.9.3(B)
1. The outermost main shell is 3
2. The p subshell in this main shell is 3p
3. The number of electrons in this subshell is 4
4. Adding 12 to this number, we get: 4+12 = 16
5. So the group number is 16

Solved example 9.4
(a) Which are the subshells in which electrons are filled in the s-block elements of the third period?
(b) Write the subshell electronic configuration (S.E.C) of the last element of this period
Solution:
Part (a):
1. We have two clues:
a. The elements belong to the s-block.
• So the last electron will be filled in the s subshell
b. The elements belong to the third period.
• So the highest 'main shell number' in the S.E.C will be 3
2. Combining the above two clues, we get:
• The last term in the S.E.C is 3s
3. If the last term is 3s, the following subshells would be already filled:
1s, 2s, and 2p
4. So the subshells present are:
1s, 2s, 2p and 3s
Part (b)
1. The clue is:
• The element is the last one of the third period
2. So we can write:
• The element belongs to the last group 18.
• So it belongs to the p-block
• So the last electron will be filled in the 3p subshell
3. Now we take Rule 9.3 (B)
• The group number is 18. It is obtained by adding 12 to the number of electrons in the 3p subshell. 
• Thus, the number of electrons in the 3p subshell is 18-12 = 6
4. So the last term in the S.EC is 3p⁶.
• So the terms before that are: 1s²,2s², 2p⁶ , and 3s²
• Thus the S.E.C is: 1s²2s²2p⁶3s²3p⁶. 

Solved example 9.5

When the last electron of an atom was filled in the 3d subshell, the S.E.C was recorded as 3d⁸. Answer the following questions related to this atom

(i) Write the complete S.E.C

(ii) What is the atomic number, block name, period number and group number of this atom?
Solution:
1. When the last electron is filled, 3d⁸ is obtained. 
2. The filling of 3d can begin only after completely filling up 4s 
3. That means 4s is completely filled up. This gives us 4s².
4. If 4s is completely filled, all the subshells preceding it should be already filled up. Let us write those preceding subshells in order:
1s, 2s, 2p, 3s and 3p 
5. The above subshells are completely filled up. So the S.E.C is:
1s²2s²2p⁶3s²3p⁶4s²3d⁸  OR [Ar]4s²3d⁸

Solved example 9.6

The S.E.C of 5 elements are given below. Pick out the wrong ones from among them
(i) 1s²2s²2p⁷ (ii)1s²2s²2p² (iii)1s²2s²2p⁵3s¹ (iv)1s²2s²2p⁶3s²3p⁶4s¹3d² (v)1s²2s²2p⁶3s²3p⁶4s²3d² 
Solution:
■ (i), (iii) and (iv) are wrong. 
Explanation:
• Consider (i) 1s²2s²2p⁷: The p subshell can carry a maximum of 6 electrons only
• Consider (iii)1s²2s²2p⁵3s¹: The 3s can begin only after completely filling the 2p with 6 electrons
• Consider (iv)1s²2s²2p⁶3s²3p⁶4s¹3d²: The 3d can begin only after completely filling the 4s with 2 electrons

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So we saw how to find the 'Group number' of any given element. We can now proceed to learn the properties of the elements of each block. In the next section, we will see the properties of s-block and p-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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Chapter 4 - The Modern Periodic Table

In the previous section, we completed the discussion on oxidation and reduction. In this section, we will discuss about the Periodic table.

A large number of elements are present in nature. We need to arrange them in order. Proper arrangement is essential in every field. Consider the example of a shop. Goods like soaps, detergents etc., will be kept at one particular place. Goods like fruits, vegetables etc., will be kept at a different place, away from soaps and detergents. In the section for soaps itself, there will be sub-sections. Bathing soaps will be kept at one place, while washing soaps will be kept at another place.

Let us now try to arrange the elements. For that, first, we will make some cards. One card for each element. A sample card (the green square) is shown in fig.4.1 below. It is the card for Iron (Fe).

Fig.4.1

The above card, and the others, that we will use for our present discussion, are taken from Wikimedia commons. Each card will show many important details about the element. For our present discussion, we need only the following details:

• The atomic number Z of the element

• Symbol of the element

• Name of the element

• Electronic configuration of the element

First we need cards upto atomic number Z = 18, that is Argon. Once we prepare those cards, we will put them side by side in sequential order. This is shown in fig.4.2 below (right click, and select 'open in new tab' for an enlarged view):

Fig.4.2

Fig.4.2 shows the cards from 1 to 18 arranged side by side. After argon, we can continue with any number of elements that we like, and arrange like this. But such an arrangement will take up a large horizontal space from left to right. Also, such an arrangement will not serve any special purpose. We want metals to be together at one place, non-metals to be together at another place, gases to be together at yet another place, etc., So let us make some modifications to the above arrangement.

Let us take out the elements after Z= 10. That is., we take out the elements from Sodium with Z = 11, upto Argon with Z = 18. We take them and put them under the first 10 elements. This is shown in fig.4.3 below(right click, and select 'open in new tab' for an enlarged view):

Fig.4.3

• Care must be taken to see that, Sodium No.11 comes under Lithium No.3. 

Now look at the electronic configurations carefully.

♦ Lithium and sodium have 1 electron in their outer most shells, and now, both are grouped together in one column

♦ Beryllium and magnesium have 2 electrons in their outer most shells, and now, both are grouped together in one column

♦ Boron and aluminium have 3 electrons in their outer most shells, and now, both are grouped together in one column

We find the above pattern for all the elements from 3 to 18. Let us write it down:

• All the elements in a column have the same number of electrons in the outer most shells

• Also, when we move from left to right, with the passing of every one column, the number of electrons in the outer most shell increase by 1

We find that hydrogen and helium are left out. Let us give them appropriate positions:

• Take out hydrogen, and put it above lithium No.3

• Take out helium, and put it above neon No.10

The modified table is shown in fig.4.4 below:

Fig.4.4

• Now, the first column has 3 elements. All of them have 1 electron in their outer most shells

• The last column also has 3 elements. But the number of electrons in the outer most shell are not the same. Helium has 2, and the others have 8. We can think about it in this way:

The elements in this column have a stable configuration. That is., maximum number of electrons in the outer most shell. These elements in the last column do not usually take part in reactions because, they are stable.

• So now our table has 3 horizontal rows. Each of these horizontal rows is called a 'period'.
• Also, the table has 7 vertical columns. Each of these vertical columns is called a 'group'.
Do we see any relation ship between the following two:

    ♦ The position of any period

    ♦ The electronic configuration of the elements in that period?

• Indeed there is a relation. Look at the 1^st period. Take a closer look at the electronic configuration of the elements in that period. The configuration has only 1 digit. That means, all the elements in the 1^st period has only 1 shell.

    ♦ In fact, 'all the elements in the world, which have only 1 shell', are included in the 1^st period. Because, after hydrogen and helium, the next element is lithium, which has 2 shells.

• Look at the 2^nd period. Take a closer look at the electronic configuration of the elements in that period. The configuration has 2 digits. That means, all the elements in the 2^nd period has two shells.

    ♦ In fact, 'all the elements in the world, which have 2 shells', are included in the 2^nd period. Because, after neon, the next element is sodium, which has 3 shells.

• Look at the 3^rd period. Take a closer look at the electronic configuration of the elements in that period. The configuration has 3 digits. That means, all the elements in the third period has 3 shells.

    ♦ In fact, 'all the elements in the world, which have 3 shells', are included in the 3^rd period. Because, after argon, the next element is potassium, which has 4 shells.

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Let us write a summary of the discussion that we had so far.

1. The horizontal rows are called Periods

2. The vertical columns are called Groups

3. All the elements in a period will have the same number of shells

4. All the elements in a group will have the same number of electrons in the outer most shell.
5. As we move from top to bottom in the table, with the passing of each period, 1 shell gets added

6. As we move from left to right in the table, with the passing of each new group, 1 electron gets added in the outermost shell
7. The 'name of the period' will indicate:
    ♦ the number of shells present, in each element in that period
• For example, each elements in period 3 will have 3 shells. K, L and M
8. The 'name of the group' will indicate:
    ♦ the number of electrons present in the outer most shell of each element in that group
• For example, each elements in Group V will have 5 electrons in it's outer most shell

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So we have arranged the first 18 elements. Let us now arrange the rest. Rules 7 and 8 written above will help us.

■ The next element is No.19 Potassium. 

• It has 4 shells. K, L, M and N. So, according to rule 7 above, it falls in period 4

• It has 1 electron in the outer most shell. So, according to rule 8, it will fall in group I

• Based on the above 2, potassium is the first element in the 4^th period. So it will fall just below sodium as shown in the fig.4.5 below:

Fig.4.5

■The next element is No.20 calcium. 

• It has 4 shells. K, L, M and N. So, according to rule 7, it falls in period 4

• It has 2 electrons in the outer most shell. So, according to rule 8, it will fall in group II

• Based on the above 2, calcium is the second element in the 4^th period. So it will fall just below magnesium. This is also shown in the fig.4.5 above.

■ The next element is No.21 scandium. 

• It has 4 shells. K, L, M and N. So, according to rule 7, it falls in period 4

• It has 2 electrons in the outer most shell. So, according to rule 8, it will fall in group II

• Based on the above 2, scandium is the second element in the 4^th period. So it will fall just below magnesium in the fig.4.4.

• So here we encounter a problem. We have already assigned the 'position below magnesium' to calcium. Now, scandium is also claiming the same position. Scandium has the claim because, it too has '4 shells', and '2 electrons in the outer most shell'.

■ Let us try the next element. The next element is No.22 titanium. 

• It has 4 shells. K, L, M and N. So, according to rule 7, it falls in period 4

• It has 2 electrons in the outer most shell. So, according to rule 8, it will fall in group II

• Based on the above 2, titanium is the second element in the 4^th period. So it will fall just below magnesium in the fig.4.4.

• Here also we encounter the same problem. We have already assigned the 'position below magnesium' to calcium. Now, scandium and titanium are also claiming the same position. Titanium has the claim because, it too has '4 shells', and '2 electrons in the outer most shell'.

■ Let us try one more element. The next element is No.23 vanadium. 

• It has 4 shells. K, L, M and N. So, according to rule 7, it falls in period 4

• It has 2 electrons in the outer most shell. So, according to rule 8, it will fall in group II

• Based on the above 2, vanadium is the second element in the 4^th period. So it will fall just below magnesium in the fig.4.4.

• Here also we encounter the same problem. We have already assigned the 'position below magnesium' to calcium. Now, scandium, titanium and vanadium are also claiming the same position. Vanadium has the claim because, it too has '4 shells', and '2 electrons in the outer most shell'.

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Why is there more than 1 claim for a single position? Let us analyse:

Consider the electron configuration of the elements that we have seen so far in Period 4:

• No.19: Potassium (K): 2,8,8,1

• No.20: Calcium (Ca): 2,8,8,2

• No.21: Scandium (Sc): 2,8,9,2

• No.22: Titanium (Ti): 2,8,10,2

• No.23: Vanadium (V): 2,8,11,2

We can see that, after No.20, the electrons get added to the second outer most shell. So there is no sequential increase in the number of electrons in the outer most shells of scandium, titanium and vanadium.

We will learn more details about such electron configuration in higher classes. At present, all we need to know is this:

■ The electrons are getting added to the second outer most shell. A phenomenon which creates more than one claim for the same position.

This special situation continues beyond No.23 vanadium. It continues up to No.30 zinc. After that normalcy is restored. That is., after No.30, the electrons get added to the outer most shell. So we have to provide a special place for the elements from 21 to 30. This special place is created in between Groups II and III. This is shown in fig.4.6 below:

Fig.4.6

How much space should be provided between Groups II and III?

• Enough space to accommodate elements from No.21 to No.30. This is shown in fig.4.7 below (right click, and select 'open in new tab' for an enlarged view):

Fig.4.7

A total of 10 elements (from No.21 to No.30) are accommodated in the newly created space. We can see that, the elements which come after No.30, that is., the elements from 31 to 36, strictly follow both the rules 7 and 8.
Now, all the elements in the world, which have 4 shells, are accommodated in the Period 4.

■ Let us move to the next period 5. In this period also, there are some problem causing elements. They have 5 shells, and thus comply with the rule 7. But they do not comply with rule 8. So they are also placed in the portion between Groups II and III. There are exactly 10 'problem causing elements', just as in Period 4

■ Let us move to the next period 6. In this period also, there are some 'problem causing elements'. They have 6 shells, and thus comply with the rule 7. But they do not comply with rule 8. So they are also placed in the portion between Groups II and III. In this period, the number of 'problem causing elements' are more. There are a total of 24 such elements. Note that there are only 10 such elements in Periods 4 and 5. If we include all 24 in between Groups II and III, then the table will become very long from left to right. We will not be able to print or draw it on a single sheet. So, 14 of them (starting from No.57 Lanthanum), are separated from the main table. These 14 are given a special name 'Lanthanides', and are kept at the bottom of the main table. This is indicated by the red arrow in the 'completed periodic table' shown in the fig.4.8 below (right click, and select 'open in new tab' for an enlarged view). This table is obtained from the Wikimedia commons, and can be seen here. All the elements in the world, which have 6 shells come under the red arrow in Period 6.

Fig.4.8

 ■ Let us move to the next period 7. In this period also, there are 24 'problem causing elements'. They have 7 shells, and thus comply with the rule 7. But they do not comply with rule 8. 14 of them (starting from No.89 Actinium) are separated from the main table. These 14 are given a special name 'Actinides', and are kept at the bottom of the main table, below the Lanthanides. This is indicated by the blue arrow in the fig.4.8. All the elements in the world, which have 7 shells come under the blue arrow in Period 7.
• Lanthanides are also known as rare earths
• Actinides are man made artificial elements (except Thorium and Uranium)

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So we have discussed the basics about the arrangement of elements in the tabular form. The Modern Periodic table is based on the works of the Russian scientist Dmitri Ivanovich Mendeleev. Works of the English scientist Henry Moseley contributed to the modifications of the Table.

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In the next section, we will discuss more features of the Periodic table. 

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