In the previous section, we saw the periodic trend in the size of atoms. In this section we will see more properties.
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.
Based on the results in I and II above, we can answer an interesting question:
■ 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.
Now we will see some solved examples
In the next section, we will see a few more solved examples.
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:
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]
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
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
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.
■ 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.
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
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
In the next section, we will see a few more solved examples.
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