In the previous section 3, we completed a discussion on the development of the modern periodic table. In this section, we will see the relationship between the following two items:
• Electronic configuration of an element
• Position of that element in the periodic table
• 'Position of an element' in the Periodic Table depends upon the 'last orbital filled'
• We will now try to understand this in detail:
1. Consider the electronic configuration of any element
2. From that configuration, we can easily write the ‘highest principal quantum number n’ in it’s atom
• Let this highest n value be nh
3. Then that element will be situated in the nhth period
Let us see some examples:
Example 1:
• The electronic configuration of silicon (Si) is [Ne]3s23p2
• Highest n value (nh) = 3
• So Si will be situated in the 3rd period
Example 2:
• The electronic configuration of gold (Au) is [Xe]5d106s1
• Highest n value (nh) = 6
• So Au will be situated in the 6th period
■ From the above discussion, we can write:
♦ All elements having (nh = 1) will be accommodated in the 1st period
♦ All elements having (nh = 2) will be accommodated in the 2nd period
♦ All elements having (nh = 3) will be accommodated in the 3rd period
♦ All elements having (nh = 4) will be accommodated in the 4th period
♦ so on . . .
Let us write a detailed analysis so that, the relation between 'period number' and 'electronic configuration' will become more clear
1. The 1st element of the 1st period is hydrogen (H)
• It has one electron. This electron will be placed in the orbital with the lowest energy, which is the 1s orbital
• So the electronic configuration of 1H is 1s1
• Note that (nh = 1)
♦ H is in the 1st period
(ii) The 2nd element is Helium (He)
• It has two electrons
• It’s first electron can go into the 1s orbital just like in hydrogen
• The 1s orbital can accommodate 2 electrons. So the second electron also goes into the 1s orbital
• Thus the electronic configuration of 2He is: 1s2
• Note that (nh = 1)
♦ He is in the 1st period
■ We can write some general information about the 1st period:
♦ The 1st main shell (K) has one orbital: 1s
♦ That 1 orbital can accommodate 2 electrons
♦ So there will be '2 elements starting from H' in the 1st period
♦ By looking at the periodic table, we find that, this is indeed true
♦ There is no place for a third electron in the 1st main-shell
♦ Such an electron will have to go to the 2nd main-shell
♦ Consequently, such an element will be placed in the 2nd period
2. Now we move on to the 2nd period
(i) The 1st element of the 2nd period is Lithium (Li)
• It has 3 electrons
• It’s first two electrons can go into the 1s orbital just like in He
• We have already seen that, the 1st main-shell is completely filled up in He
• So the 3rd electron in Li will enter the 'smallest energy orbital' in the 2nd main-shell
♦ Obviously, it is the 2s orbital
• Thus the electronic configuration of 3Li is: 1s22s1
♦ This is same as [He]2s1
• Note that (nh = 2)
♦ Li is in the 2nd period
(ii) The 2nd element of the 2nd period is Beryllium (Be)
• It has 4 electrons
• It’s first 3 electrons occupy the same positions as in the previous Li
• the 2s can take one more electron
• So the 4th electron in Be will enter the 2s orbital
• Thus the electronic configuration of 4Be is: 1s22s2
♦ This is same as [He]2s2
• Note that (nh = 2)
♦ Be is in the 2nd period
(iii) The 3rd element of the 2nd period is Boron (B)
• It has 5 electrons
• It’s first 4 electrons occupy the same positions as in the previous Be
• the 2s cannot take more than two electrons
• So the 5th electron in B will enter the next higher orbital, which is 2p
• Thus the electronic configuration of 5B is: [He]2s22p1
• Note that (nh = 2)
♦ B is in the 2nd period
(iv) The 2p orbital can hold 6 electrons
• So starting from Boron, there will be 6 elements. The 6th element from 5B is Neon (Ne)
• When Ne is reached, all the six electrons in 2p will be filled up
• Thus the electronic configuration of 10Ne is: [He]2s22p6
• Note that (nh = 2)
♦ Ne is in the 2nd period
■ We can write some general information about the 2nd period:
♦ The 2nd main shell (L) has two orbitals: 2s and 2p
♦ Those 2 orbitals can accommodate 8 electrons
♦ So there will be '8 elements starting from Li' in the 2nd period
♦ By looking at the periodic table, we find that, this is indeed true
♦ Ne is the last element in this period. It has 10 electrons
♦ There is no place for a 11th electron in the 2nd main-shell
♦ Such an electron will have to go to the 3rd main-shell
♦ Consequently, such an element will be placed in the 3rd period
3. Now we move on to the 3rd period
(i) The 1st element of the 3rd period is Sodium (Na)
• It has 11 electrons
• It’s first 10 electrons occupy the same positions as in the previous Ne
• We have already seen that, the 2nd main-shell is completely filled up in Ne
• So the 11th electron in Na will enter the 'smallest energy orbital' in the 3rd main-shell
♦ Obviously, it is the 3s orbital
• Thus the electronic configuration of 11Na is: [Ne]3s1
• Note that (nh = 3)
♦ Na is in the 3rd period
(ii) The 2nd element of the 3rd period is Magnesium (Mg)
• It has 12 electrons
• It’s first 11 electrons occupy the same positions as in the previous Na
• The 3s can take one more electron
• So the 12th electron in Mg will enter the 2s orbital
• Thus the electronic configuration of 12Mg is: [Ne]3s2
• Note that (nh = 3)
♦ Mg is in the 3rd period
(iii) The 3rd element of the 3rd period is Aluminium (Al)
• It has 13 electrons
• It’s first 12 electrons occupy the same positions as in the previous Mg
• The 3s cannot take more than two electrons
• So the 13th electron in Al will enter the next higher orbital, which is 3p
• Thus the electronic configuration of 13Al is: [Ne]3s23p1
• Note that (nh = 3)
♦ Al is in the 3rd period
(iv) The 3p orbital can hold 6 electrons
• So starting from Al, there will be 6 elements. The 6th element from 13Al is Argon (Ar)
• When Ar is reached, all the six electrons in 3p will be filled up
• Thus the electronic configuration of 18Ar is: [Ne]3s23p6
• Note that (nh = 3)
♦ Ar is in the 3rd period
■ We can write some general information about the 3rd period:
♦ The 3rd main shell (M) has three orbitals: 3s, 3p and 3d
♦ 3d has a higher energy level than 4s
♦ So 3d can begin only after completing 4s
♦ But 4s comes only in the 4th period
♦ So for the 3rd period, we can consider only two orbitals: 3s and 3p
♦ Those 2 orbitals can accommodate 8 electrons
♦ So there will be '8 elements starting from Na' in the 3rd period
♦ By looking at the periodic table, we find that, this is indeed true
♦ Ar is the last element in this period. It has 18 electrons
4. Now we move on to the 4th period
(i) The 1st element in the 4th period is Potassium (K)
• It has 19 electrons
• It’s first 18 electrons occupy the same positions as in the previous Ar
• The 19th electron in K will enter the next higher energy level, which is: 4s
• Thus the electronic configuration of 19K is: [Ar]4s1
• Note that (nh = 4)
♦ K is in the 4th period
(ii) The 2nd element of the 4th period is Calcium (Ca)
• It has 20 electrons
• It’s first 19 electrons occupy the same positions as in the previous K
• The 4s can take one more electron
• So the 20th electron in Ca will enter the 4s orbital
• Thus the electronic configuration of 20Ca is: [Ar]4s2
• Note that (nh = 4)
♦ Ca is in the 4th period
(iii) the 3rd element of the 4th period is Scandium (Sc)
• It has 21 electrons
• It’s first 20 electrons occupy the same positions as in the previous Ca
• The 4s is completely filled in Ca
• The next higher orbital is 3d
• So the 21st electron in Sc will enter the 3d orbital
• Thus the electronic configuration of 21Sc is: [Ar]4s23d1
• Note that (nh = 4)
♦ Sc is in the 4th period
(iv) The 3d orbitals can hold 10 electrons
• There are 10 elements starting from 21Sc and ending at 30Zn
• The 3d orbitals are completely filled up when 30Zn is reached
• This is clear from the configuration of Zn: [Ar]4s23d10
• Note that, in all those 10 elements, we have: (nh = 4)
♦ All of them are in the 4th period
(v) The 10 elements from Sc to Zn are called 3d transition elements
■ Why are they called transition elements?
• The answer can be written in steps:
◦ When the electrons are being filled up in the 3d orbitals, a transition is taking place
◦ That transition is from A to B written below:
A. The completely filled s orbital configuration on the left side of the periodic table
B. The completely filled p orbital configuration on the right side of the periodic table
■ We can also write this:
◦ When the electrons are being filled up in the 3d orbitals, a transition is taking place
◦ That transition is from A to B written below:
A. The elements having 'highly metallic' characteristics on the left side of the periodic table
B. The elements having 'highly non-metallic' characteristics on the right side of the periodic table
• We see that, these 10 elements are indeed in between the left and right blocks of the periodic table
(vi) After the transition elements, comes Gallium (Ga)
• Filling of 4p begins with 31Ga
• The 6 electrons in 4p will be filled up when 36Kr is reached
• There are 6 elements from 31Ga to 36Kr
• 31Ga has the configuration [Ar]4s23d104p1
• 36Kr has the configuration [Ar]4s23d104p6
• Note that, in all those 6 elements, we have: (nh = 4)
♦ All of them are in the 4th period
■ We can write some general information about the 4th period:
♦ The 4th main shell (N) has four orbitals: 4s, 4p, 4d and 4f
♦ 4d and 4f have higher energies than 3d
♦ So even 4d can begin only after completing 3d
♦ So for the 4th period, we can consider only three orbitals: 4s, 3d and 4p
♦ Those 3 orbitals together can accommodate 18 electrons
♦ So there will be '18 elements starting from K' in the 4th period
♦ By looking at the periodic table, we find that, this is indeed true
♦ Kr is the last element in this period. It has 36 electrons
5. Now we move on to the 5th period
(i) The 1st element in the 5th period is Rubidium (Rb)
• It has 37 electrons
• It’s first 36 electrons occupy the same positions as in the previous Kr
• The 37th electron in Rb will enter the next energy level, which is: 5s
• Thus the electronic configuration of 37Rb is: [Kr]5s1
• Note that (nh = 5)
♦ Rb is in the 5th period
(ii) The 2nd element of the 5th period is Strontium (Sr)
• It has 38 electrons
• It’s first 37 electrons occupy the same positions as in the previous Rb
• The 5s can take one more electron
• So the 38th electron in Sr will enter the 5s orbital
• Thus the electronic configuration of 38Sr is: [Kr]5s2
• Note that (nh = 5)
♦ Sr is in the 5th period
(iii) the 3rd element of the 5th period is Yttrium (Y)
• It has 39 electrons
• It’s first 38 electrons occupy the same positions as in the previous Sr
• The 5s is completely filled in Sr
• The next higher orbital is 4d
• So the 39th electron in Y will enter the 4d orbital
• Thus the electronic configuration of 39Y is: [Kr]5s24d1
• Note that (nh = 5)
♦ Y is in the 5th period
(iv) The 4d orbitals can hold 10 electrons
• There are 10 elements starting from 39Y and ending at 48Cd
• The 4d orbitals are completely filled up when 48Cd is reached
• This is clear from the configuration of Cd: [Kr]5s24d10
• Note that, in all those 10 elements, we have: (nh = 5)
♦ All of them are in the 5th period
(v) The 10 elements from Y to Cd are called 4d transition elements
We saw the reason when we discussed the 4th period
(vi) After the transition elements, comes Indium (In)
• Filling of 5p begins with 49In
• The 6 electrons in 5p will be filled up when 54Xe is reached
• There are 6 elements from 49In to 54Xe
• 49In has the configuration [Kr]5s24d105p1
• 54Xe has the configuration [Kr]5s24d105p6
• Note that, in all those 6 elements, we have: (nh = 5)
♦ All of them are in the 5th period
■ We can write some general information about the 5th period:
♦ The 5th main shell (O) has five orbitals: 5s, 5p, 5d, 5f and 5g
♦ 5d, 5f and 5g have higher energies than 4d
♦ So even 5d can begin only after completing 4d
♦ So for the 5th period, we can consider only three orbitals: 5s, 4d and 5p
♦ Those 3 orbitals together can accommodate 18 electrons
♦ So there will be '18 elements starting from Rb' in the 5th period
♦ By looking at the periodic table, we find that, this is indeed true
♦ Xe is the last element in this period. It has 54 electrons
6. Now we move on to the 6th period
(i) The 1st element in the 6th period is Cesium (Cs)
• It has 55 electrons
• It’s first 54 electrons occupy the same positions as in the previous Xe
• The 54th electron in Cs will enter the next energy level, which is: 6s
• Thus the electronic configuration of 37Cs is: [Xe]6s1
• Note that (nh = 6)
♦ Cs is in the 6th period
(ii) The 2nd element of the 6th period is Barium (Ba)
• It has 56 electrons
• It’s first 55 electrons occupy the same positions as in the previous Cs
• The 6s can take one more electron
• So the 56th electron in Ba will enter the 6s orbital
• Thus the electronic configuration of 56Ba is: [Xe]6s2
• Note that (nh = 6)
♦ Ba is in the 6th period
(iii) The element after 56Ba is 57La (Lanthanum)
• After 6s, the next higher energy level is 4f
• So the 14 electrons in the 4f orbitals get progressively filled up
• This gives rise to a new set of elements: The 4f inner transition elements
■ These elements are called Lanthanides
• They are placed as a separate block below the main body of the periodic table
• This block is called the f-block of the periodic table
• The Lanthanides form the first row of the f block
(iv) After completely filling up the 4f orbitals, the next element is 72Hf
• The 72nd electron in Hf goes to the 5d orbital because, 5d has lesser energy than 6p
• The 5d orbitals can hold 10 electrons
• There are 9 elements starting from 72Hf and ending at 80Hg
• The small 'difference of 1' is due to the change in energy levels. We will learn about it in higher classes
• The 5d orbitals are completely filled up when 80Hg is reached
(v) Filling of 6p begins with 81Tl
• The 6 electrons in 6p will be filled up when 86Rn is reached
■ We can write some general information about the 6th period:
♦ The 6th main shell (P) has six orbitals: 6s, 6p, 6d, 6f, . . .
♦ 6p, 6d etc., have higher energies than 4f
♦ Even 5d has a greater energy than 4f
♦ 4f is the next higher energy level after 6s
♦ So for the 6th period, we can consider only four orbitals: 6s, 4f, 5d and 6p
♦ Those 4 orbitals together can accommodate (2+14+10+6) = 32 electrons
♦ So there will be '32 elements starting from Cs' in the 6th period
♦ But the 14 elements corresponding to 4f are placed below the main body of the periodic table
♦ Only the remaining (32-14) = 18 are placed in the proper 6th period
♦ Radon (Rn) is the last element in this period. It has 86 electrons
7. Now we move on to the 7th period
(i) The 1st element in the 7th period is Francium (Fr)
• It has 87 electrons
• It’s first 86 electrons occupy the same positions as in the previous Rn
• The 87th electron in Fr will enter the next energy level, which is: 7s
• Thus the electronic configuration of 37Fr is: [Rn]7s1
• Note that (nh = 7)
♦ Fr is in the 7th period
(ii) The 2nd element of the 7th period is Radium (Ra)
• It has 88 electrons
• It’s first 87 electrons occupy the same positions as in the previous Fr
• The 7s can take one more electron
• So the 88th electron in Ra will enter the 7s orbital
• Thus the electronic configuration of 88Ra is: [Rn]7s2
• Note that (nh = 7)
♦ Ra is in the 7th period
(iii) The element after 88Ra is 89Ac (Actinium)
• After 7s, the next higher energy level is 5f
• So the 14 electrons in the 5f orbitals get progressively filled up
• This gives rise to a new set of elements: The 5f inner transition elements
■ These elements are called Actinides
♦ They are placed below the Lanthanides in the f block
(iv) After completely filling up the 5f orbitals, the next element is 104Rf
• The 104th electron in Rf goes to the 6d orbital because, 6d has lesser energy than 7p
• The 6d orbitals can hold 10 electrons
• There are 9 elements starting from 104Rf and ending at 112Cn
• The small 'difference of 1' is due to the change in energy levels. We will learn about it in higher classes
• The 6d orbitals are completely filled up when 112Cn is reached
(iv) Filling of 7p begins with 113Uut
• The 6 electrons in 7p will be filled up when 118Uuo is reached
■ We can write some general information about the 7th period:
♦ The 7th main shell (Q) has seven orbitals: 7s, 7p, 7d, 7f, . . .
♦ 7p, 7d etc., have higher energies than 5f
♦ Even 6d has a greater energy than 5f
♦ 5f is the next higher energy level after 7s
♦ So for the 7th period, we can consider only four orbitals: 7s, 5f, 6d and 7p
♦ Those 4 orbitals together can accommodate (2+14+10+6) = 32 electrons
♦ So there will be '32 elements starting from Fr' in the 7th period
♦ But the 14 elements corresponding to 5f are placed below the main body of the periodic table
♦ Only the remaining (32-14) = 18 are placed in the proper 7th period
♦ Ununoctium (Uuo) is the last element in this period. It has 118 electrons
■ We see that, the filling up of 4f and 5f orbitals give rise to the inner transition elements
■Those two sets are place away from the main body So that, the structure of the table does not become too long
• Electronic configuration of an element
• Position of that element in the periodic table
Now we are able to find 'the row (period) in which a given element is situated' in the periodic table
The steps are simple:
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
Solved example 3.1
How would you justify the presence of 18 elements in the 5th period of the Periodic table?
Solution:
1. In the 5th period, we see that the filling of 5s orbitals begins
• So we would expect all 'orbitals in the 5th main-shell' to be filled up in the 5th period
• 'Orbitals in the 5th main-shell' are: 5s, 5p, 5d and 5f
• After filling up 5s, we would expect the filling of 5p
2. But 5p has a greater energy level than 4d
• So the 10 electrons of 4d will be filled up before 5p
• So the order of filling is: 5s < 4d < 5p
3. After filling up 5p, we would expect 5d
• But 5d has a greater energy than 6s
• So we cannot expect 5d until 6s is completed
4. But 6s can begin only in the 6th period
• So the only orbitals we can expect in the 5th period are: 5s, 4d and 5p
♦ 5s gives 2 electrons
♦ 4d gives 10 electrons
♦ 5p gives 6 electrons
5. So the total number of electrons = (2 +10 +6) = 18
• Thus we get 18 elements in the 5th period
Solved example 3.2
Atomic number of a few elements are given below:
10, 20, 7, 14
Identify the period to which these elements belong?
Solution:
1. Atomic number: 10
• First we write the electronic configuration: 1s22s22p6
• We get: (nh = 2)
• So the element 10X belongs to the 2nd period
2. Atomic number: 20
• First we write the electronic configuration:
After 18, we can put [Ar] at the beginning. So we get: [Ar]4s2
• We get: (nh = 4)
• So the element 20X belongs to the 4th period
3. Atomic number: 7
• First we write the electronic configuration: 1s22s22p3
• We get: (nh = 2)
• So the element 7X belongs to the 2nd period
4. Atomic number: 14
• First we write the electronic configuration:
After 10, we can put [Ne] at the beginning. So we get: [Ne]3s23p2
• We get: (nh = 3)
• So the element 14X belongs to the 3rd period
• Electronic configuration of an element
• Position of that element in the periodic table
Relation between electronic configuration and period
■ The distribution of electrons into orbitals of an atom is called its electronic configuration
■ The distribution of electrons into orbitals of an atom is called its electronic configuration
• We will now try to understand this in detail:
1. Consider the electronic configuration of any element
2. From that configuration, we can easily write the ‘highest principal quantum number n’ in it’s atom
• Let this highest n value be nh
3. Then that element will be situated in the nhth period
Let us see some examples:
Example 1:
• The electronic configuration of silicon (Si) is [Ne]3s23p2
• Highest n value (nh) = 3
• So Si will be situated in the 3rd period
Example 2:
• The electronic configuration of gold (Au) is [Xe]5d106s1
• Highest n value (nh) = 6
• So Au will be situated in the 6th period
■ From the above discussion, we can write:
♦ All elements having (nh = 1) will be accommodated in the 1st period
♦ All elements having (nh = 2) will be accommodated in the 2nd period
♦ All elements having (nh = 3) will be accommodated in the 3rd period
♦ All elements having (nh = 4) will be accommodated in the 4th period
♦ so on . . .
Let us write a detailed analysis so that, the relation between 'period number' and 'electronic configuration' will become more clear
1. The 1st element of the 1st period is hydrogen (H)
• It has one electron. This electron will be placed in the orbital with the lowest energy, which is the 1s orbital
• So the electronic configuration of 1H is 1s1
• Note that (nh = 1)
♦ H is in the 1st period
(ii) The 2nd element is Helium (He)
• It has two electrons
• It’s first electron can go into the 1s orbital just like in hydrogen
• The 1s orbital can accommodate 2 electrons. So the second electron also goes into the 1s orbital
• Thus the electronic configuration of 2He is: 1s2
• Note that (nh = 1)
♦ He is in the 1st period
■ We can write some general information about the 1st period:
♦ The 1st main shell (K) has one orbital: 1s
♦ That 1 orbital can accommodate 2 electrons
♦ So there will be '2 elements starting from H' in the 1st period
♦ By looking at the periodic table, we find that, this is indeed true
♦ There is no place for a third electron in the 1st main-shell
♦ Such an electron will have to go to the 2nd main-shell
♦ Consequently, such an element will be placed in the 2nd period
2. Now we move on to the 2nd period
(i) The 1st element of the 2nd period is Lithium (Li)
• It has 3 electrons
• It’s first two electrons can go into the 1s orbital just like in He
• We have already seen that, the 1st main-shell is completely filled up in He
• So the 3rd electron in Li will enter the 'smallest energy orbital' in the 2nd main-shell
♦ Obviously, it is the 2s orbital
• Thus the electronic configuration of 3Li is: 1s22s1
♦ This is same as [He]2s1
• Note that (nh = 2)
♦ Li is in the 2nd period
(ii) The 2nd element of the 2nd period is Beryllium (Be)
• It has 4 electrons
• It’s first 3 electrons occupy the same positions as in the previous Li
• the 2s can take one more electron
• So the 4th electron in Be will enter the 2s orbital
• Thus the electronic configuration of 4Be is: 1s22s2
♦ This is same as [He]2s2
• Note that (nh = 2)
♦ Be is in the 2nd period
(iii) The 3rd element of the 2nd period is Boron (B)
• It has 5 electrons
• It’s first 4 electrons occupy the same positions as in the previous Be
• the 2s cannot take more than two electrons
• So the 5th electron in B will enter the next higher orbital, which is 2p
• Thus the electronic configuration of 5B is: [He]2s22p1
• Note that (nh = 2)
♦ B is in the 2nd period
(iv) The 2p orbital can hold 6 electrons
• So starting from Boron, there will be 6 elements. The 6th element from 5B is Neon (Ne)
• When Ne is reached, all the six electrons in 2p will be filled up
• Thus the electronic configuration of 10Ne is: [He]2s22p6
• Note that (nh = 2)
♦ Ne is in the 2nd period
■ We can write some general information about the 2nd period:
♦ The 2nd main shell (L) has two orbitals: 2s and 2p
♦ Those 2 orbitals can accommodate 8 electrons
♦ So there will be '8 elements starting from Li' in the 2nd period
♦ By looking at the periodic table, we find that, this is indeed true
♦ Ne is the last element in this period. It has 10 electrons
♦ There is no place for a 11th electron in the 2nd main-shell
♦ Such an electron will have to go to the 3rd main-shell
♦ Consequently, such an element will be placed in the 3rd period
3. Now we move on to the 3rd period
(i) The 1st element of the 3rd period is Sodium (Na)
• It has 11 electrons
• It’s first 10 electrons occupy the same positions as in the previous Ne
• We have already seen that, the 2nd main-shell is completely filled up in Ne
• So the 11th electron in Na will enter the 'smallest energy orbital' in the 3rd main-shell
♦ Obviously, it is the 3s orbital
• Thus the electronic configuration of 11Na is: [Ne]3s1
• Note that (nh = 3)
♦ Na is in the 3rd period
(ii) The 2nd element of the 3rd period is Magnesium (Mg)
• It has 12 electrons
• It’s first 11 electrons occupy the same positions as in the previous Na
• The 3s can take one more electron
• So the 12th electron in Mg will enter the 2s orbital
• Thus the electronic configuration of 12Mg is: [Ne]3s2
• Note that (nh = 3)
♦ Mg is in the 3rd period
(iii) The 3rd element of the 3rd period is Aluminium (Al)
• It has 13 electrons
• It’s first 12 electrons occupy the same positions as in the previous Mg
• The 3s cannot take more than two electrons
• So the 13th electron in Al will enter the next higher orbital, which is 3p
• Thus the electronic configuration of 13Al is: [Ne]3s23p1
• Note that (nh = 3)
♦ Al is in the 3rd period
(iv) The 3p orbital can hold 6 electrons
• So starting from Al, there will be 6 elements. The 6th element from 13Al is Argon (Ar)
• When Ar is reached, all the six electrons in 3p will be filled up
• Thus the electronic configuration of 18Ar is: [Ne]3s23p6
• Note that (nh = 3)
♦ Ar is in the 3rd period
■ We can write some general information about the 3rd period:
♦ The 3rd main shell (M) has three orbitals: 3s, 3p and 3d
♦ 3d has a higher energy level than 4s
♦ So 3d can begin only after completing 4s
♦ But 4s comes only in the 4th period
♦ So for the 3rd period, we can consider only two orbitals: 3s and 3p
♦ Those 2 orbitals can accommodate 8 electrons
♦ So there will be '8 elements starting from Na' in the 3rd period
♦ By looking at the periodic table, we find that, this is indeed true
♦ Ar is the last element in this period. It has 18 electrons
4. Now we move on to the 4th period
(i) The 1st element in the 4th period is Potassium (K)
• It has 19 electrons
• It’s first 18 electrons occupy the same positions as in the previous Ar
• The 19th electron in K will enter the next higher energy level, which is: 4s
• Thus the electronic configuration of 19K is: [Ar]4s1
• Note that (nh = 4)
♦ K is in the 4th period
(ii) The 2nd element of the 4th period is Calcium (Ca)
• It has 20 electrons
• It’s first 19 electrons occupy the same positions as in the previous K
• The 4s can take one more electron
• So the 20th electron in Ca will enter the 4s orbital
• Thus the electronic configuration of 20Ca is: [Ar]4s2
• Note that (nh = 4)
♦ Ca is in the 4th period
(iii) the 3rd element of the 4th period is Scandium (Sc)
• It has 21 electrons
• It’s first 20 electrons occupy the same positions as in the previous Ca
• The 4s is completely filled in Ca
• The next higher orbital is 3d
• So the 21st electron in Sc will enter the 3d orbital
• Thus the electronic configuration of 21Sc is: [Ar]4s23d1
• Note that (nh = 4)
♦ Sc is in the 4th period
(iv) The 3d orbitals can hold 10 electrons
• There are 10 elements starting from 21Sc and ending at 30Zn
• The 3d orbitals are completely filled up when 30Zn is reached
• This is clear from the configuration of Zn: [Ar]4s23d10
• Note that, in all those 10 elements, we have: (nh = 4)
♦ All of them are in the 4th period
(v) The 10 elements from Sc to Zn are called 3d transition elements
■ Why are they called transition elements?
• The answer can be written in steps:
◦ When the electrons are being filled up in the 3d orbitals, a transition is taking place
◦ That transition is from A to B written below:
A. The completely filled s orbital configuration on the left side of the periodic table
B. The completely filled p orbital configuration on the right side of the periodic table
■ We can also write this:
◦ When the electrons are being filled up in the 3d orbitals, a transition is taking place
◦ That transition is from A to B written below:
A. The elements having 'highly metallic' characteristics on the left side of the periodic table
B. The elements having 'highly non-metallic' characteristics on the right side of the periodic table
• We see that, these 10 elements are indeed in between the left and right blocks of the periodic table
(vi) After the transition elements, comes Gallium (Ga)
• Filling of 4p begins with 31Ga
• The 6 electrons in 4p will be filled up when 36Kr is reached
• There are 6 elements from 31Ga to 36Kr
• 31Ga has the configuration [Ar]4s23d104p1
• 36Kr has the configuration [Ar]4s23d104p6
• Note that, in all those 6 elements, we have: (nh = 4)
♦ All of them are in the 4th period
■ We can write some general information about the 4th period:
♦ The 4th main shell (N) has four orbitals: 4s, 4p, 4d and 4f
♦ 4d and 4f have higher energies than 3d
♦ So even 4d can begin only after completing 3d
♦ So for the 4th period, we can consider only three orbitals: 4s, 3d and 4p
♦ Those 3 orbitals together can accommodate 18 electrons
♦ So there will be '18 elements starting from K' in the 4th period
♦ By looking at the periodic table, we find that, this is indeed true
♦ Kr is the last element in this period. It has 36 electrons
5. Now we move on to the 5th period
(i) The 1st element in the 5th period is Rubidium (Rb)
• It has 37 electrons
• It’s first 36 electrons occupy the same positions as in the previous Kr
• The 37th electron in Rb will enter the next energy level, which is: 5s
• Thus the electronic configuration of 37Rb is: [Kr]5s1
• Note that (nh = 5)
♦ Rb is in the 5th period
(ii) The 2nd element of the 5th period is Strontium (Sr)
• It has 38 electrons
• It’s first 37 electrons occupy the same positions as in the previous Rb
• The 5s can take one more electron
• So the 38th electron in Sr will enter the 5s orbital
• Thus the electronic configuration of 38Sr is: [Kr]5s2
• Note that (nh = 5)
♦ Sr is in the 5th period
(iii) the 3rd element of the 5th period is Yttrium (Y)
• It has 39 electrons
• It’s first 38 electrons occupy the same positions as in the previous Sr
• The 5s is completely filled in Sr
• The next higher orbital is 4d
• So the 39th electron in Y will enter the 4d orbital
• Thus the electronic configuration of 39Y is: [Kr]5s24d1
• Note that (nh = 5)
♦ Y is in the 5th period
(iv) The 4d orbitals can hold 10 electrons
• There are 10 elements starting from 39Y and ending at 48Cd
• The 4d orbitals are completely filled up when 48Cd is reached
• This is clear from the configuration of Cd: [Kr]5s24d10
• Note that, in all those 10 elements, we have: (nh = 5)
♦ All of them are in the 5th period
(v) The 10 elements from Y to Cd are called 4d transition elements
We saw the reason when we discussed the 4th period
(vi) After the transition elements, comes Indium (In)
• Filling of 5p begins with 49In
• The 6 electrons in 5p will be filled up when 54Xe is reached
• There are 6 elements from 49In to 54Xe
• 49In has the configuration [Kr]5s24d105p1
• 54Xe has the configuration [Kr]5s24d105p6
• Note that, in all those 6 elements, we have: (nh = 5)
♦ All of them are in the 5th period
■ We can write some general information about the 5th period:
♦ The 5th main shell (O) has five orbitals: 5s, 5p, 5d, 5f and 5g
♦ 5d, 5f and 5g have higher energies than 4d
♦ So even 5d can begin only after completing 4d
♦ So for the 5th period, we can consider only three orbitals: 5s, 4d and 5p
♦ Those 3 orbitals together can accommodate 18 electrons
♦ So there will be '18 elements starting from Rb' in the 5th period
♦ By looking at the periodic table, we find that, this is indeed true
♦ Xe is the last element in this period. It has 54 electrons
6. Now we move on to the 6th period
(i) The 1st element in the 6th period is Cesium (Cs)
• It has 55 electrons
• It’s first 54 electrons occupy the same positions as in the previous Xe
• The 54th electron in Cs will enter the next energy level, which is: 6s
• Thus the electronic configuration of 37Cs is: [Xe]6s1
• Note that (nh = 6)
♦ Cs is in the 6th period
(ii) The 2nd element of the 6th period is Barium (Ba)
• It has 56 electrons
• It’s first 55 electrons occupy the same positions as in the previous Cs
• The 6s can take one more electron
• So the 56th electron in Ba will enter the 6s orbital
• Thus the electronic configuration of 56Ba is: [Xe]6s2
• Note that (nh = 6)
♦ Ba is in the 6th period
(iii) The element after 56Ba is 57La (Lanthanum)
• After 6s, the next higher energy level is 4f
• So the 14 electrons in the 4f orbitals get progressively filled up
• This gives rise to a new set of elements: The 4f inner transition elements
■ These elements are called Lanthanides
• They are placed as a separate block below the main body of the periodic table
• This block is called the f-block of the periodic table
• The Lanthanides form the first row of the f block
(iv) After completely filling up the 4f orbitals, the next element is 72Hf
• The 72nd electron in Hf goes to the 5d orbital because, 5d has lesser energy than 6p
• The 5d orbitals can hold 10 electrons
• There are 9 elements starting from 72Hf and ending at 80Hg
• The small 'difference of 1' is due to the change in energy levels. We will learn about it in higher classes
• The 5d orbitals are completely filled up when 80Hg is reached
(v) Filling of 6p begins with 81Tl
• The 6 electrons in 6p will be filled up when 86Rn is reached
■ We can write some general information about the 6th period:
♦ The 6th main shell (P) has six orbitals: 6s, 6p, 6d, 6f, . . .
♦ 6p, 6d etc., have higher energies than 4f
♦ Even 5d has a greater energy than 4f
♦ 4f is the next higher energy level after 6s
♦ So for the 6th period, we can consider only four orbitals: 6s, 4f, 5d and 6p
♦ Those 4 orbitals together can accommodate (2+14+10+6) = 32 electrons
♦ So there will be '32 elements starting from Cs' in the 6th period
♦ But the 14 elements corresponding to 4f are placed below the main body of the periodic table
♦ Only the remaining (32-14) = 18 are placed in the proper 6th period
♦ Radon (Rn) is the last element in this period. It has 86 electrons
7. Now we move on to the 7th period
(i) The 1st element in the 7th period is Francium (Fr)
• It has 87 electrons
• It’s first 86 electrons occupy the same positions as in the previous Rn
• The 87th electron in Fr will enter the next energy level, which is: 7s
• Thus the electronic configuration of 37Fr is: [Rn]7s1
• Note that (nh = 7)
♦ Fr is in the 7th period
(ii) The 2nd element of the 7th period is Radium (Ra)
• It has 88 electrons
• It’s first 87 electrons occupy the same positions as in the previous Fr
• The 7s can take one more electron
• So the 88th electron in Ra will enter the 7s orbital
• Thus the electronic configuration of 88Ra is: [Rn]7s2
• Note that (nh = 7)
♦ Ra is in the 7th period
(iii) The element after 88Ra is 89Ac (Actinium)
• After 7s, the next higher energy level is 5f
• So the 14 electrons in the 5f orbitals get progressively filled up
• This gives rise to a new set of elements: The 5f inner transition elements
■ These elements are called Actinides
♦ They are placed below the Lanthanides in the f block
(iv) After completely filling up the 5f orbitals, the next element is 104Rf
• The 104th electron in Rf goes to the 6d orbital because, 6d has lesser energy than 7p
• The 6d orbitals can hold 10 electrons
• There are 9 elements starting from 104Rf and ending at 112Cn
• The small 'difference of 1' is due to the change in energy levels. We will learn about it in higher classes
• The 6d orbitals are completely filled up when 112Cn is reached
(iv) Filling of 7p begins with 113Uut
• The 6 electrons in 7p will be filled up when 118Uuo is reached
■ We can write some general information about the 7th period:
♦ The 7th main shell (Q) has seven orbitals: 7s, 7p, 7d, 7f, . . .
♦ 7p, 7d etc., have higher energies than 5f
♦ Even 6d has a greater energy than 5f
♦ 5f is the next higher energy level after 7s
♦ So for the 7th period, we can consider only four orbitals: 7s, 5f, 6d and 7p
♦ Those 4 orbitals together can accommodate (2+14+10+6) = 32 electrons
♦ So there will be '32 elements starting from Fr' in the 7th period
♦ But the 14 elements corresponding to 5f are placed below the main body of the periodic table
♦ Only the remaining (32-14) = 18 are placed in the proper 7th period
♦ Ununoctium (Uuo) is the last element in this period. It has 118 electrons
■ We see that, the filling up of 4f and 5f orbitals give rise to the inner transition elements
■Those two sets are place away from the main body So that, the structure of the table does not become too long
This completes our present discussion on the relation between two items:
• Position of that element in the periodic table
Now we are able to find 'the row (period) in which a given element is situated' in the periodic table
The steps are simple:
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
Solved example 3.1
How would you justify the presence of 18 elements in the 5th period of the Periodic table?
Solution:
1. In the 5th period, we see that the filling of 5s orbitals begins
• So we would expect all 'orbitals in the 5th main-shell' to be filled up in the 5th period
• 'Orbitals in the 5th main-shell' are: 5s, 5p, 5d and 5f
• After filling up 5s, we would expect the filling of 5p
2. But 5p has a greater energy level than 4d
• So the 10 electrons of 4d will be filled up before 5p
• So the order of filling is: 5s < 4d < 5p
3. After filling up 5p, we would expect 5d
• But 5d has a greater energy than 6s
• So we cannot expect 5d until 6s is completed
4. But 6s can begin only in the 6th period
• So the only orbitals we can expect in the 5th period are: 5s, 4d and 5p
♦ 5s gives 2 electrons
♦ 4d gives 10 electrons
♦ 5p gives 6 electrons
5. So the total number of electrons = (2 +10 +6) = 18
• Thus we get 18 elements in the 5th period
Solved example 3.2
Atomic number of a few elements are given below:
10, 20, 7, 14
Identify the period to which these elements belong?
Solution:
1. Atomic number: 10
• First we write the electronic configuration: 1s22s22p6
• We get: (nh = 2)
• So the element 10X belongs to the 2nd period
2. Atomic number: 20
• First we write the electronic configuration:
After 18, we can put [Ar] at the beginning. So we get: [Ar]4s2
• We get: (nh = 4)
• So the element 20X belongs to the 4th period
3. Atomic number: 7
• First we write the electronic configuration: 1s22s22p3
• We get: (nh = 2)
• So the element 7X belongs to the 2nd period
4. Atomic number: 14
• First we write the electronic configuration:
After 10, we can put [Ne] at the beginning. So we get: [Ne]3s23p2
• We get: (nh = 3)
• So the element 14X belongs to the 3rd period
We have completed a basic discussion on how the electronic configuration is related to periods. In the next section, we will see how the configuration is related to groups
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