In the previous section 3.3, we saw how the elements of the periodic table are classified into the s, p, d and f blocks. We also saw some basics about representative elements and families. In this section, we will see how elements are classified as metals and non-metals
1. The elements of the periodic table can be broadly classified into metals and non-metals
• The elements on the left side are metals and the elements on the right side are non-metals
• There is a line which separates the left side from the right side
2. But this line is not at the middle of the periodic table
• The line is far to the right
• That means, left side is very much bigger than the right side
• That means, there are more metals than non-metals
3. Consider all the known elements in the periodic table
• 78% of them are metals
• Only 22% are non-metals
4. Let us write a comparison between the properties of metals and non-metals:
• Metals have high melting and boiling points
♦ Most non-metals are either solids or gases
♦ Non-metallic solids have low melting and boiling points
• Metals are good conductors of heat and electricity
♦ Non-metals are poor conductors of heat and electricity
• Metals are malleable and ductile
♦ Non-metals are brittle. They are neither malleable nor ductile
5. Now let us see the 'boundary line' which separates the metals from the non-metals
• Looking at the periodic table, we see that, the line is a thick stair-step line
♦ It starts from the left side of Si
♦ Passes through the middle of Ge and As
♦ Passes through the middle of Sb and Te
♦ Ends at the middle of Po and At
6. Note that, it is a stair-step line
• It is not a vertical line
• That means, the change from metals to non-metals is not abrupt
♦ If it was abrupt, the line would have been a perfect vertical line
■ In fact, if we move from left to right along a period, we will see that the metallic character decreases
7. Moreover, the elements which touch the stair-step line, have both metallic and non-metallic properties
• The elements which touch the line were mentioned in (5) above
• Those metals are called semi-metals or metalloids
• In an earlier section 3.1 we saw this:
■ Using the electronic configuration, we can find 'the row (period) in which a given element is situated' in the periodic table
The steps are:
1. Write the electronic configuration of that element
2. Note the highest principal quantum number n
3. Let this highest n be denoted as nh
4. Then that element is situated in the nhth period
■ After completing the discussion in the previous section 3.3, we can find block and group also
■ If we know the three items (period, block and group) of an element, we can locate it's position in the periodic table
1. Write the electronic configuration of that element
2. Examine that configuration carefully
(i) If that configuration is in the form [core]ns(1-2), that element belongs to the s-block
(ii) If that configuration is in the form [core]ns2np(1-6) OR [core](n-1)d10ns2np(1-6), that element belongs to the p-block
(iii) If that configuration is in the form [core](n-1)d(1-10)ns(0-2), that element belongs to the d-block
(iv) If that configuration is in the form [core](n-2)f(1-14)(n-1)d(0-1)ns2, that element belongs to the f-block
■ Always remember the method to find the [core]. Only then we would be able to use the above 2 steps. We saw the method in the previous section. Here we will write it again. It consists of 3 steps:
(i) If we know the atomic number of the given element, we can easily spot the just previous noble gas
(ii) The [core] is the electronic configuration of that noble gas
(iii) Instead of writing the 'electronic configuration of that noble gas', it is enough to write it's symbol
1. Write the electronic configuration of that element
2. Examine that configuration carefully and determine the block to which the element belongs
(i) If it is the s-block, obviously, we will be having [core]ns(1-2) before us
• Then the 'superscript of s' is the 'group number' that we are seeking
(ii) If it is the p-block, obviously, we will be having [core]ns2np(1-6) OR [core](n-1)d10ns2np(1-6) before us
• In either case, the sum '(12 + superscript of p)' is the 'group number' that we are seeking
(iii) If it is the d-block, obviously, we will be having [core](n-1)d(1-10)ns(0-2) before us
• Then the sum '(superscript of s + superscript of d)' is the 'group number' that we are seeking
(iv) If it is the f-block, obviously, we will be having [core](n-2)f(1-14)(n-1)d(0-1)ns2 before us
• But for the f-block elements, 'finding the group number' is not very important
♦ All the f-block elements are considered to belong to the 3rd group
• We can locate such elements some where in the two rows below the main body of the periodic table
♦ If n= 6, it is the 1st row of lanthanides
♦ If n= 6, it is the 2nd row of actinides
1. Consider 3Li in Group 1
• It has the general configuration ns1
2. The element below 3Li is 11Na
• It also has the general configuration of ns1
• We know that Li and Na has similar physical and chemical properties
• We have: (11-3) = 8
• We can write:
• After a period of ‘8’, there is a repetition of three items:
♦ General configuration
♦ Physical properties
♦ Chemical properties
3. The element below 11Na is 19K
• It also has the general configuration of ns1
• We know that Na and K has similar physical and chemical properties
• We have: (19-11) = 8
• We can write:
• After a period of ‘8’, there is a repetition of three items:
♦ General configuration
♦ Physical properties
♦ Chemical properties
4. The element below 19K is 37Rb
• It also has the general configuration of ns1
• We know that K and Rb has similar physical and chemical properties
• We have: (37-19) = 18
• We can write:
• After a period of ‘18’, there is a repetition of three items:
♦ General configuration
♦ Physical properties
♦ Chemical properties
5. The element below 37Rb is 55Cs
• It also has the general configuration of ns1
• We know that Rb and Cs has similar physical and chemical properties
• We have: (55-37) = 18
• We can write:
• After a period of ‘18’, there is a repetition of three items:
♦ General configuration
♦ Physical properties
♦ Chemical properties
6. So we see that, after definite periods like 8 and 18, there is repetition of configuration, physical properties and chemical properties
• We can site such repetitions in other groups also
• So it is indeed appropriate to call it the periodic table
Solved example 3.3
The outside electronic structure of an atom is s2p4. Determine the group to which that atom belongs
Solution:
1. The given structure can be written as: [core]ns2np(1-6)
• So the atom belongs to the p block
2. So group number = (12 + superscript of p) = (12+4) = 16
Solved example 3.4
The outside electronic structure of an atom is d5s2. Determine the group to which that atom belongs
Solution:
1. The given structure can be written as: [core](n-1)d(1-10)ns(0-2)
• So the atom belongs to the d block
2. So group number = (superscript of s + superscript of d) = (5+2) = 7
Solved example 3.5
Assign the position of the element having outer electronic configuration (i) ns2np4 for n=3 (ii) (n-1)d2ns2 for n=4, and (iii) (n-2) f7(n-1)d1ns2 for n=6, in the periodic table
Solution:
Part (i):
Given that n = 3
1. So the element belongs to the 3rd period
2. ns2np4 is same as the general form for the p-block: [core]ns2np(1-6)
• So the element belongs to the p-block
3. Group number = (12 + superscript of p) = (12+4) = 16
Part (ii):
Given that n = 4
1. So the element belongs to the 4th period
2. (n-1)d2ns2 is same as the general form for the d-block: [core](n-1)d(1-10)ns(0-2)
• So the element belongs to the d-block
3. Group number = (superscript of s + superscript of d) = (2+2) = 4
Part (iii):
Given that n = 6
1. So the element belongs to the 6th period
2. (n-2) f7(n-1)d1ns2 is same as the general form for the f-block: [core](n-2)f(1-14)(n-1)d(0-1)ns2
• So the element belongs to the f-block
3. If the element belongs to the 6th period and the f-block, it is obviously a lanthanide
• All lanthanides and actinides have the same group number 3
Solved example 3.6
In terms of period and group where would you locate the element with Z =114?
Solution:
1. Given that Z = 114
• The just previous noble gas is Radon (Rn)
• Rn has an atomic number Z = 86
2. The difference is (114-86) = 28
• So we need to distribute the last 28 electrons carefully
3. Rn is the last electron of the 6th period
• After the 6th period, there is only one more period: The 7th period
♦ So the 28 electrons will be distributed in the 7th period
♦ We want to know where the last one of those 28 electrons fall
4. The first two electrons go to the 7s orbital
• The next 14 electrons go to the 5f orbital
♦ This is because, after 7s, the next higher energy level is 5f
• The next 10 electrons go to 6d orbitals
♦ This is because, after 5f, the next higher energy level is 6d
5. So we have successfully distributed (2+14+10) = 26 electrons
• Only 2 electrons are remaining
6. After 6d, the next higher energy level is 7p
• The remaining 2 electrons will go to the 7p
• So the electronic configuration is: [Rn]5f146d107s27p2
7. Thus we can write:
• The last electron goes to the p orbital
• So the element belongs to the p-block
• The superscript of the last p orbital is 2
• Group number = (12 + superscript of p) = (12+2) = 14
Solved example 3.7
Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P.
Solution:
1. We are given 5 elements
• All of them except Be lies in the 3rd period
• Be lies in the 2nd period
2. As we move from left to right along a period, the metallic character decreases
• Also, as we move down a group, the metallic character increases
• We will see the reasons for such variations in later sections
3. The elements belonging to the left side groups will be more metallic
• The elements belonging to the right side groups will be less metallic
• Also, elements belonging to the lower periods will be more metallic
4. We are asked to arrange them in the 'increasing order of metallic character'
• So we must write the lesser metallic elements first
• That means, we must write the 'right side elements' first
5. Thus the required arrangement is:
P < Si < Be < Mg < Na
• Note that, comparing Be and Mg (both belongs to the same group), we get: Be < Mg
♦ This is because, as we move down a group, the metallic character increases
1. The elements of the periodic table can be broadly classified into metals and non-metals
• The elements on the left side are metals and the elements on the right side are non-metals
• There is a line which separates the left side from the right side
2. But this line is not at the middle of the periodic table
• The line is far to the right
• That means, left side is very much bigger than the right side
• That means, there are more metals than non-metals
3. Consider all the known elements in the periodic table
• 78% of them are metals
• Only 22% are non-metals
4. Let us write a comparison between the properties of metals and non-metals:
• Metals have high melting and boiling points
♦ Most non-metals are either solids or gases
♦ Non-metallic solids have low melting and boiling points
• Metals are good conductors of heat and electricity
♦ Non-metals are poor conductors of heat and electricity
• Metals are malleable and ductile
♦ Non-metals are brittle. They are neither malleable nor ductile
5. Now let us see the 'boundary line' which separates the metals from the non-metals
• Looking at the periodic table, we see that, the line is a thick stair-step line
♦ It starts from the left side of Si
♦ Passes through the middle of Ge and As
♦ Passes through the middle of Sb and Te
♦ Ends at the middle of Po and At
6. Note that, it is a stair-step line
• It is not a vertical line
• That means, the change from metals to non-metals is not abrupt
♦ If it was abrupt, the line would have been a perfect vertical line
■ In fact, if we move from left to right along a period, we will see that the metallic character decreases
7. Moreover, the elements which touch the stair-step line, have both metallic and non-metallic properties
• The elements which touch the line were mentioned in (5) above
• Those metals are called semi-metals or metalloids
• We have completed our present discussion on classification of elements in the periodic table
■ Using the electronic configuration, we can find 'the row (period) in which a given element is situated' in the periodic table
The steps are:
1. Write the electronic configuration of that element
2. Note the highest principal quantum number n
3. Let this highest n be denoted as nh
4. Then that element is situated in the nhth period
■ After completing the discussion in the previous section 3.3, we can find block and group also
■ If we know the three items (period, block and group) of an element, we can locate it's position in the periodic table
So next, we will write the steps to find the block:
2. Examine that configuration carefully
(i) If that configuration is in the form [core]ns(1-2), that element belongs to the s-block
(ii) If that configuration is in the form [core]ns2np(1-6) OR [core](n-1)d10ns2np(1-6), that element belongs to the p-block
(iii) If that configuration is in the form [core](n-1)d(1-10)ns(0-2), that element belongs to the d-block
(iv) If that configuration is in the form [core](n-2)f(1-14)(n-1)d(0-1)ns2, that element belongs to the f-block
■ Always remember the method to find the [core]. Only then we would be able to use the above 2 steps. We saw the method in the previous section. Here we will write it again. It consists of 3 steps:
(i) If we know the atomic number of the given element, we can easily spot the just previous noble gas
(ii) The [core] is the electronic configuration of that noble gas
(iii) Instead of writing the 'electronic configuration of that noble gas', it is enough to write it's symbol
Finally, we will write the steps to find the group:
2. Examine that configuration carefully and determine the block to which the element belongs
(i) If it is the s-block, obviously, we will be having [core]ns(1-2) before us
• Then the 'superscript of s' is the 'group number' that we are seeking
(ii) If it is the p-block, obviously, we will be having [core]ns2np(1-6) OR [core](n-1)d10ns2np(1-6) before us
• In either case, the sum '(12 + superscript of p)' is the 'group number' that we are seeking
(iii) If it is the d-block, obviously, we will be having [core](n-1)d(1-10)ns(0-2) before us
• Then the sum '(superscript of s + superscript of d)' is the 'group number' that we are seeking
(iv) If it is the f-block, obviously, we will be having [core](n-2)f(1-14)(n-1)d(0-1)ns2 before us
• But for the f-block elements, 'finding the group number' is not very important
♦ All the f-block elements are considered to belong to the 3rd group
• We can locate such elements some where in the two rows below the main body of the periodic table
♦ If n= 6, it is the 1st row of lanthanides
♦ If n= 6, it is the 2nd row of actinides
We can now write a short note on periodicity. We will write it in steps:
• It has the general configuration ns1
2. The element below 3Li is 11Na
• It also has the general configuration of ns1
• We know that Li and Na has similar physical and chemical properties
• We have: (11-3) = 8
• We can write:
• After a period of ‘8’, there is a repetition of three items:
♦ General configuration
♦ Physical properties
♦ Chemical properties
3. The element below 11Na is 19K
• It also has the general configuration of ns1
• We know that Na and K has similar physical and chemical properties
• We have: (19-11) = 8
• We can write:
• After a period of ‘8’, there is a repetition of three items:
♦ General configuration
♦ Physical properties
♦ Chemical properties
4. The element below 19K is 37Rb
• It also has the general configuration of ns1
• We know that K and Rb has similar physical and chemical properties
• We have: (37-19) = 18
• We can write:
• After a period of ‘18’, there is a repetition of three items:
♦ General configuration
♦ Physical properties
♦ Chemical properties
5. The element below 37Rb is 55Cs
• It also has the general configuration of ns1
• We know that Rb and Cs has similar physical and chemical properties
• We have: (55-37) = 18
• We can write:
• After a period of ‘18’, there is a repetition of three items:
♦ General configuration
♦ Physical properties
♦ Chemical properties
6. So we see that, after definite periods like 8 and 18, there is repetition of configuration, physical properties and chemical properties
• We can site such repetitions in other groups also
• So it is indeed appropriate to call it the periodic table
Now we will see some solved examples:
The outside electronic structure of an atom is s2p4. Determine the group to which that atom belongs
Solution:
1. The given structure can be written as: [core]ns2np(1-6)
• So the atom belongs to the p block
2. So group number = (12 + superscript of p) = (12+4) = 16
Solved example 3.4
The outside electronic structure of an atom is d5s2. Determine the group to which that atom belongs
Solution:
1. The given structure can be written as: [core](n-1)d(1-10)ns(0-2)
• So the atom belongs to the d block
2. So group number = (superscript of s + superscript of d) = (5+2) = 7
Solved example 3.5
Assign the position of the element having outer electronic configuration (i) ns2np4 for n=3 (ii) (n-1)d2ns2 for n=4, and (iii) (n-2) f7(n-1)d1ns2 for n=6, in the periodic table
Solution:
Part (i):
Given that n = 3
1. So the element belongs to the 3rd period
2. ns2np4 is same as the general form for the p-block: [core]ns2np(1-6)
• So the element belongs to the p-block
3. Group number = (12 + superscript of p) = (12+4) = 16
Part (ii):
Given that n = 4
1. So the element belongs to the 4th period
2. (n-1)d2ns2 is same as the general form for the d-block: [core](n-1)d(1-10)ns(0-2)
• So the element belongs to the d-block
3. Group number = (superscript of s + superscript of d) = (2+2) = 4
Part (iii):
Given that n = 6
1. So the element belongs to the 6th period
2. (n-2) f7(n-1)d1ns2 is same as the general form for the f-block: [core](n-2)f(1-14)(n-1)d(0-1)ns2
• So the element belongs to the f-block
3. If the element belongs to the 6th period and the f-block, it is obviously a lanthanide
• All lanthanides and actinides have the same group number 3
Solved example 3.6
In terms of period and group where would you locate the element with Z =114?
Solution:
1. Given that Z = 114
• The just previous noble gas is Radon (Rn)
• Rn has an atomic number Z = 86
2. The difference is (114-86) = 28
• So we need to distribute the last 28 electrons carefully
3. Rn is the last electron of the 6th period
• After the 6th period, there is only one more period: The 7th period
♦ So the 28 electrons will be distributed in the 7th period
♦ We want to know where the last one of those 28 electrons fall
4. The first two electrons go to the 7s orbital
• The next 14 electrons go to the 5f orbital
♦ This is because, after 7s, the next higher energy level is 5f
• The next 10 electrons go to 6d orbitals
♦ This is because, after 5f, the next higher energy level is 6d
5. So we have successfully distributed (2+14+10) = 26 electrons
• Only 2 electrons are remaining
6. After 6d, the next higher energy level is 7p
• The remaining 2 electrons will go to the 7p
• So the electronic configuration is: [Rn]5f146d107s27p2
7. Thus we can write:
• The last electron goes to the p orbital
• So the element belongs to the p-block
• The superscript of the last p orbital is 2
• Group number = (12 + superscript of p) = (12+2) = 14
Solved example 3.7
Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P.
Solution:
1. We are given 5 elements
• All of them except Be lies in the 3rd period
• Be lies in the 2nd period
2. As we move from left to right along a period, the metallic character decreases
• Also, as we move down a group, the metallic character increases
• We will see the reasons for such variations in later sections
3. The elements belonging to the left side groups will be more metallic
• The elements belonging to the right side groups will be less metallic
• Also, elements belonging to the lower periods will be more metallic
4. We are asked to arrange them in the 'increasing order of metallic character'
• So we must write the lesser metallic elements first
• That means, we must write the 'right side elements' first
5. Thus the required arrangement is:
P < Si < Be < Mg < Na
• Note that, comparing Be and Mg (both belongs to the same group), we get: Be < Mg
♦ This is because, as we move down a group, the metallic character increases
In the next section, we will see the periodic trends
No comments:
Post a Comment