Chapter 4.3 - Modern Periodic table - Solved examples

In the previous section, we saw some solved examples related to the position of elements in the periodic table. In this section we will see a few more solved examples.

Solved example 4.3

Electron configuration of elements P, Q, R and S are given below:

P – 2,2    Q – 2,8,2     R – 2,8,5     S – 2,8

(a) Which among these elements are included in the same period?

(b) Which among these elements are included in the same group?

(c) Which among these elements is a noble gas?

(d) To which group and period does the element R belong?

Solution:

(a) Elements in a same period will have the same number of shells

• In the given problem, P and S have both 2 digits in their electron configuration. So they have both 2 shells. Thus, they both belong to the Period 2

• Q and R have both 3 digits in their electron configuration. So they have both 3 shells. Thus, they both belong to the Period 3

(b) Elements in the same group will have the same number of electrons in their outermost shells

In the given problem, P and Q have both 2 electrons in their outermost shells. So they both belong to Group II

(c) S has an octet configuration in the outermost shell. So it is a noble gas

(d) • R has 5 electrons in it's outermost shell. So it belongs to Group V

• It has 3 digits in it's electron configuration. So it has 3 shells. Thus it belong to the Period 3

Solved example 4.4

An incomplete form of the periodic table is given below.

Write answers to the questions related to the position of elements in it.

(a) Which is the element with the biggest atom in group I?

(b) Which is the element with the lowest ionisation energy in group I?

(c) Which is the element with the smallest atom in period II?

(d) Which are the transition elements?

(e) Which among the two elements L and M, has the smallest electronegativity?

(f) Which among the two elements B and I, has higher metallic nature?

(g) Which are the elements that belong to the halogen family?
(h) Which type of ion will be formed from an atom of 'M'. Cation or anion?

(i) Which is the element that resembles E the most in it's chemical properties?

Solution:

(a) We have discussed the periodic trend related to the 'size of atoms', above in this section. We can write this:

• In a group, from top to bottom, size of atom increases

• In a period, from left to right, size of atom decreases

• In this problem, we have to consider the variation within a group only. The bottom most element will be the biggest. So, in group I, D is the biggest.

(b) We have discussed the periodic trend related to the 'ionisation energy', above in this section. We can write this:

• In a group, from top to bottom, the ionisation energy decreases

• In a period, from left to right, the ionisation energy increases

• In this problem, we have to consider the variation within a group only. The bottom most element will have the lowest ionisation energy. So, in group I, D will be having the lowest ionisation energy.

(c) The variation of size was written in problem (a) above. In this problem, we have to consider the variation within a period only. The element at the extreme right will be the smallest. So, in period 2, M is the smallest.

(d) The transition elements fall between groups II and III. So G and H are transition elements.

(e) We have discussed the periodic trend related to the 'electronegativity', above in this section. We can write this:

• In a group, from top to bottom, the electronegativity decreases

• In a period, from left to right, the electronegativity increases

• In this problem, we have to consider the variation within a period only. This is because, L and M fall within a period. The left most element will have the lower electronegativity. So, among L and M, the element L will be having the lower electronegativity.

(f) We have discussed the periodic trend related to the 'metallic nature'. We can write this:

• In a group, from top to bottom, the metallic nature increases

• In a period, from left to right, the metallic nature decreases

• In this problem, we have to consider the variation within a period only. This is because, B and I fall within a period. The left most element will have the higher metallic nature. So, among B and I, the element B will be having the higher metallic nature.

(g) We have learned that, halogen family is the group VII. So, among all the given elements, M and N are the halogens.
(h) • We have lerned about cations and anions here. Positive ions are Cations and negative ions are anions.
• As we move from left to right in a period, the metallic character decreases.
• That means, the tendency to lose electrons decreases. That means, tendency to gain electron increases
• As M is further to the right in it's period, it will have a tendency to gain electrons. So it will become a negative ion. That is., it will become an anion.

(i) Both B and F are close to E. But F will resemble E more because, F, like E has two electrons in the outermost shell
Solved example 4.5

Give examples for the following:

(a) Two elements that have 1 electron in their outermost shells

(b) Three elements that have 2 electrons in their outermost shells

(c) Three elements that have filled outermost shells

Solution:

(a) All elements in the group I have 1 electron in their outermost shells. We can pick any two from this group. Say lithium (Li) and sodium (Na)

(b) All elements in the group II have 2 electrons in their outermost shells. We can pick any three from this group. Say beryllium (Be), magnesium (Mg) and calcium (Ca)

(c) All elements in the group VIII have filled electrons in their outermost shells. We can pick any three from this group. Say neon (Ne), argon (Ar) and krypton (Kr)

Solved example 4.6

Lithium, sodium and potassium are all metals which react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?

Solution: All the three elements have one electron in the outermost shell. During chemical reactions, this single electron is readily donated by these metals

Solved example 4.7

In the modern periodic table, which are the metals among the first 10 elements?

Solution:

Lithium (Li) and Beryllium (Be).

Solved example 4.8

By considering their position in the periodic table, which one of the following elements would you expect to have the maximum metallic character?

Ga, Ge, As, Se, Br

Solution:

All the given elements belong to a same period. Among them, Ga is the one which is at the left most position. So It is the one with the maximum metallic character

Solved example 4.9

Which of the following statements is not a correct statement about the trends when going from left to right in any period

(i) The elements become less metallic in nature

(ii) The number of valence electrons increase

(iii) The atoms lose their electrons more easily

Solution:

Statement (iii) is false. As we move from left to right in any period, the metallic character decreases. That is., tendency to lose electrons decreases.

Solved example 4.10

Which element has

(i) Two shells, both of which are completely filled with electrons

(ii) Electron configuration 2,8,2

(iii) A total of 3 shells, with 4 electrons in the valence shell

(iv) A total of 2 shells, with 3 electrons in the valence shell

(v) twice as many electrons in it's second shell as in it's first shell

Solution:

(i) • Given that the element has two shells. So they are K and L

• Given that they are completely filled up. K-shell can hold a maximum of 2×1² = 2×1 = 2 electrons

• L-shell can hold a maximum of 2×2² = 2×4 = 8 electrons

• So the total number of electrons = total number of protons = 2 + 8 = 10

• Thus the atomic number Z = 10, and the element with this atomic number is Neon (Ne)

(ii) • Given that electron configuration is 2,8,2

• So number of electrons = number of protons = 2 + 8 + 2 = 12

• Thus the atomic number Z = 12, and the element with this atomic number is magnesium (Mg)

(iii) • Given that there are a total of 3 shells. So the shells are K,L and M

• Also given that there are 4 electrons in the valence shell. That means K and L are completely filled up. So number of electrons = 2 + 8 + 4 = 14

• Thus the atomic number Z = 14, and the element with this atomic number is silicon (Si)

(iv) • Given that there are a total of 2 shells. So the shells are K and L

• Also given that there are 3 electrons in the valence shell. That means K is completely filled up. So number of electrons = 2 + 3 = 5

• Thus the atomic number Z = 5, and the element with this atomic number is boron (B)

(v) • Here the total number of shells is not given. But that is not necessary to solve the problem

• Given that number of electrons in the second shell is twice that in the first. It is clear that there are some electrons in the second shell

• The second shell will begin to fill only if the first shell is completely filled up. The maximum capacity of the first shell is 2.

• So the number of electrons in the second shell = 2×2 = 4. The second shell is capable to carry more than this 4. So it follows that there are no more shells than K and L.

• So there are no more electrons than the 2 in first shell, and the 4 in the second shell.

• Thus the atomic number Z = 2 + 4 = 6, and the element with this atomic number is carbon (C)

Solved example 4.11

Nitrogen (Z=7) and Phosphorus (Z=15) belong to group V of the periodic table. Write the electron configuration of these elements. Which will be more electronegative? Why?

Solution:

• Electron configuration of nitrogen is 2,5

• Electron configuration of phosphorus is 2,8,5

• Both of them belong to the same group, with nitrogen above phosphorus. We have seen that while moving from top to bottom in a group, electropositive character increases. That is., electronegative character decreases. So, nitrogen, which is above, is more electronegative.

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We have completed the basic discussion on the Modern periodic table. We will learn more details in higher classes. In the next section, we will discuss about Non-metals. 

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Chapter 4.2 - More periodic trends and Solved examples

In the previous section, we saw the periodic trend in the size of atoms. In this section we will see more properties.

Ionisation energy

We have seen reactions in which ions are formed. Consider for example, the formation of NaCl. We have seen the details here. The sodium (Na) atom loses one electron, and becomes Na+ ion. The chlorine (Cl) atom gains one electron, and become Cl- ion.

• Consider the formation of Na⁺ ion. One electron has to be removed from the Na atom, for the formation of Na⁺. Is such a removal easy?

• The negatively charged electrons are attracted towards the central nucleus. This is because of the positively charged protons present in the nucleus.

• So the electron must over come this attractive force, in order to move away from the atom.

• To over come this attractive force, some energy is required.

• This energy is called ionisation energy

• But, for the same element, this energy may be different under different conditions. For example, the presence of some other element can cause an increase or decrease in the required energy. Also, more energy is required for an electron in an inner shell, than that in an outer shell. So the conditions are to be specified. Thus the definition of ionisation energy is as follows:

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The amount of energy required to liberate the most loosely bound electron from the outer most shell of an isolated gaseous atom of an element is called it's ionisation energy.

[Note that several conditions are specified:

• The electron must be the one which is most loosely bound

• It must be situated in the outer most shell

• The atom must be isolated. That is., there must not be any presence of any other atoms

• The atom must be in the gaseous state]

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So let us analyse the variation of ionisation energy at different parts of the periodic table:

I Moving from top to bottom in a group:

1. When we move from top to bottom in a group, the size of the atom increases. We have already seen the reason here.

2. When the size increases, the distance from the nucleus to the outer most electron also increases

3. When the distance increases, the force exerted by the nucleus (on the outer most electron) decreases

4. When this force is low, low energy is sufficient to liberate the electron.

5. So we can write: The ionisation energy decreases as we move from top to bottom in a group

II Moving from left to right in a period

1. When we move from left to right in a period, the size of the atom decreases. We have already seen the reason.

2. We have seen that the decrease in size is due to the increase in attractive force from the central nucleus

3. From (2) we can readily reach a conclusion: As the attractive force increases, more energy is required to liberate the electron.

4. So we can write: The ionisation energy increases as we move from left to right in a period

Electronegativity

We have already learned about electronegativity. It is the ability of an atom to attract the shared pairs of electrons in covalent bonds. We have seen the details here. Now we want to understand it's periodic trend.

• Electronegativity is closely related to the size of atom. Let us write an analysis:

I Moving from top to bottom in a group:

1. When we move from top to bottom in a group, the size of the atom increases. We have already seen the reason here.

2. When the size increases, the distance from the nucleus to the outer most electron also increases

3. When the distance increases, the force exerted by the nucleus (on the outer most electron) decreases

4. From (3), we see that, when size increases, distance also increases, and the atom becomes lesser and lesser capable to attract 'it's own outer most electrons'.

5. Then there is nothing more to tell about the 'external shared electrons'. We can straight away say: The atom is even less capable to attract the shared pairs of electrons.

6. So we can write: The electronegativity decreases as we move from top to bottom in a group

II Moving from left to right in a period

1. When we move from left to right in a period, the size of the atom decreases. We have already seen the reason.

2. We have seen that the decrease in size is due to the increase in attractive force from the central nucleus

3. So there is an increase in attraction on the 'outer most electrons'. This increase will be felt by the 'shared pairs of electrons' also. We can straight away say: The atom is capable to exert significant force of attraction on the shared pairs of electrons

4. So we can write: The electronegativity increases as we move from left to right in a period

Metallic and Non-metallic Nature

• Elements with a metallic nature, generally loses electrons in chemical reactions.

• Metals are called electropositive because, in chemical reactions, they lose electrons to become positive ions. (Metals are good conductors of electric current because, they readily lose electrons, and hence have lots of free electrons)

• Non-metals are called electronegative because, in chemical reactions, they gain electrons to become negative ions.

• The metallic or non-metallic nature of an element is closely related to it's ionisation energy. Why?

The reason is simple:

    ♦ If the ionisation energy is low, electrons will be easily liberated. If the electrons are easily liberated, it shows a metallic nature.

   ♦ If the ionisation energy is high, electrons will not be easily liberated. If the electrons are not easily liberated, it shows a non-metallic nature.

• So, from the periodic trend of ionisation energy, we will be able to write the periodic trend for metallic and non-metallic nature also. Thus we get:

I Moving from top to bottom in a group:

1. When we move from top to bottom in a group, ionisation energy decreases

2. Decrease in ionisation energy indicates increase in metallic nature. So we can write:

3. The metallic nature increases (consequently, non-metallic nature decreases) as we move from top to bottom in a group

II Moving from left to right in a period

1. When we move from left to right in a period, ionisation energy increases

2. Increasein ionisation energy indicates increase in non-metallic nature. So we can write:

3. The non-metallic nature increases (consequently, metallic nature decreases) as we move from left to right in a period

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Based on the results in I and II above, we can answer an interesting question:

■ Which is the element with the most metallic character in the periodic table?

Solution: 1. From II above we have: In any period, the left most element is the most metallic in nature

2. So the element that we are seeking, is some where in the first group

3. From I above we have: In any group, the bottom most element is the most metallic in nature
4. So the bottom most element in the first group (francium) is the one which is most metallic in character

• However, francium is a man made element. Among the natural elements, caesium is the one which is the most metallic in character. Look at the position of caesium. It is just above francium.

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■ In the above discussion, we derived the periodic trend of metallic and non-metallic character based on the 'trend of ionisation energy'
■ We can derive the periodic trend of metallic and non-metallic character based on the 'trend of electronegativity' also. The reader is advised to write such an analysis by him/herself.

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Now we will see some solved examples

Solved example 4.1

Complete the table given below:

Solution:

1. Consider lithium. It’s electron configuration is given as 2,1. So the atomic number Z = 2 + 1 = 3

2. Consider oxygen. Z is given as 8. So the electron configuration is 2,6

• From the electron configuration, we get: No. of electrons in the outer most shell = 6. So oxygen belongs to Group VI

• There are 2 digits in the electron configuration. So No. of shells = 2. Thus, oxygen belongs to the Period 2

See rules 7 and 8

3. Consider argon. Z is given as 18. So the electron configuration is 2,8,8

• From the electron configuration, we get: No. of electrons in the outer most shell = 8. So argon belongs to Group VIII

• There are 3 digits in the electron configuration. So No. of shells = 3. Thus, argon belongs to the Period 3

4. Consider calcium. It’s electron configuration is given as 2,8,8,2. So Z = 2 + 8 + 8 + 2 = 20

• From the electron configuration, we get: No. of electrons in the outer most shell = 2. So calcium belongs to Group II

• There are 4 digits in the electron configuration. So No. of shells = 4. Thus, oxygen belongs to the Period 4

The completed table is given below:

Solved example 4.2

There are 3 shells in the atom of element 'X'. 6 electrons are present in it's outermost shell.

(a) Write the electron configuration of the element

(b) What is it's atomic number Z?

(c) In which period does this element belong?

(d) In which group does this element belong?

(e) Write the name and symbol of this element

(f) To which family of elements does this element belong?

Solution:

(a) Given that the element has 3 shells. So the element has K, L, and M shells

• Given that the M shell has 6 electrons. So the L and M would be filled up to their full capacities.

• That means there are 2 electrons in K, and 8 electrons in L

• So the total number of electrons = 2 + 8 + 6 = 16

The electron configuration is 2,8,6

(b) Total number of electrons = 16 = Total number of protons

So atomic number Z = 16

(c) There are 3 shells for the atom. So element belongs to the Period 3

(d) There are 6 electrons in the outermost shell. So the element belongs to Group VI

(e) The name of the element with Z = 16 is Sulfur. It's symbol is S

(f) The family is Oxygen family

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In the next section, we will see a few more solved examples. 

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