In the previous section 2.19, we saw the electronic configuration of atoms up to no.36 krypton (Kr). In this section, we will see higher elements. For the higher elements, we will write only a 'basic pattern'. This is because, for the higher elements from no.37, some 'deviations from the pattern' are observed. We will see the reasons for those deviations in higher classes. At present, we need to be familiar with the 'basic pattern' only
First we will write a summary of the work we did until now
First we will write a summary of the work we did until now
1. We start with 1H
• When 2He is reached, the 1s orbital is completely filled up
• This we saw in fig.2.53
■ 2He is a milestone because, all the orbitals in it are completely filled. And also the 1st main-shell is completely filled
2. After 1s, the next higher orbital is 2s
• Filling of 2s begins in 3Li
• When 4Be is reached, the 2s orbital is completely filled up
• This we saw in fig.2.55
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table. We will see more details about this in the next chapter
3. After 2s, the next higher orbital is 2p
• Filling of 2p begins in 5B
• When 10Ne is reached, the 2p orbitals are completely filled up
• This we saw in fig.2.55
• Note that, there are 6 electrons in the 2p orbitals
♦ There are 6 elements from 5B to 10Ne
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table. We will see more details about this in the next chapter
• Filling of 3s begins in 11Na
• When 12Mg is reached, the 3s orbital is completely filled up
• This we saw in fig.2.56
• Note that, there are 2 electrons in the 3s orbital
♦ There are 2 elements from 11Na to 12Mg
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
5. After 3s, the next higher orbital is 3p
• Filling of 3p begins in 13Al
• When 18Ar is reached, the 3p orbitals are completely filled up
• Note that, there are 6 elements from 13Al to 18Ar
♦ Thus the 6 electrons in the 3p orbitals are filled up
• This we saw in fig.2.58
• Note that, there are 6 electrons in the 3p orbitals
♦ There are 6 elements from 13Al to 18Ar
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table
■ 18Ar is a milestone because, all the orbitals in it are completely filled
6. After 3p, the next higher orbital is 4s
• Filling of 4s begins in 19K
• When 20Ca is reached, the 4s orbital is completely filled up
• This we saw in fig.2.58
• Note that, there are 2 electrons in the 4s orbital
♦ There are 2 elements from 19K to 20Ca
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
7. After 4s, the next higher orbital is 3d
• Filling of 3d begins in 21Sc
• When 30Zn is reached, the 3d orbitals are completely filled up
• This we saw in fig.2.59
• Note that, there are 10 electrons in the 3d orbitals
♦ There are 10 elements from 21Sc to 30Zn
• The last electron in each of these elements fall in the d orbital. So these elements belong to the d-block of the periodic table
■ Also keep in mind the unusual situations in 24Cr and 29Cu
8. After 3d, the next higher orbital is 4p
• Filling of 4p begins in 31Ga
• When 36Kr is reached, the 4p orbitals are completely filled up
• This we saw in fig.2.60
• Note that, there are 6 electrons in the 4p orbitals
♦ There are 6 elements from 31Ga to 31Ga
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table
■ 36Kr is a milestone because, all the orbitals in it are completely filled
• When 2He is reached, the 1s orbital is completely filled up
• This we saw in fig.2.53
■ 2He is a milestone because, all the orbitals in it are completely filled. And also the 1st main-shell is completely filled
2. After 1s, the next higher orbital is 2s
• Filling of 2s begins in 3Li
• When 4Be is reached, the 2s orbital is completely filled up
• This we saw in fig.2.55
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table. We will see more details about this in the next chapter
3. After 2s, the next higher orbital is 2p
• Filling of 2p begins in 5B
• When 10Ne is reached, the 2p orbitals are completely filled up
• This we saw in fig.2.55
• Note that, there are 6 electrons in the 2p orbitals
♦ There are 6 elements from 5B to 10Ne
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table. We will see more details about this in the next chapter
■ 10Ne is a milestone because, all the orbitals in it are completely filled. And also the 2nd main-shell is completely filled
4. After 2p, the next higher orbital is 3s• Filling of 3s begins in 11Na
• When 12Mg is reached, the 3s orbital is completely filled up
• This we saw in fig.2.56
• Note that, there are 2 electrons in the 3s orbital
♦ There are 2 elements from 11Na to 12Mg
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
5. After 3s, the next higher orbital is 3p
• Filling of 3p begins in 13Al
• When 18Ar is reached, the 3p orbitals are completely filled up
• Note that, there are 6 elements from 13Al to 18Ar
♦ Thus the 6 electrons in the 3p orbitals are filled up
• This we saw in fig.2.58
• Note that, there are 6 electrons in the 3p orbitals
♦ There are 6 elements from 13Al to 18Ar
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table
■ 18Ar is a milestone because, all the orbitals in it are completely filled
6. After 3p, the next higher orbital is 4s
• Filling of 4s begins in 19K
• When 20Ca is reached, the 4s orbital is completely filled up
• This we saw in fig.2.58
• Note that, there are 2 electrons in the 4s orbital
♦ There are 2 elements from 19K to 20Ca
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
7. After 4s, the next higher orbital is 3d
• Filling of 3d begins in 21Sc
• When 30Zn is reached, the 3d orbitals are completely filled up
• This we saw in fig.2.59
• Note that, there are 10 electrons in the 3d orbitals
♦ There are 10 elements from 21Sc to 30Zn
• The last electron in each of these elements fall in the d orbital. So these elements belong to the d-block of the periodic table
■ Also keep in mind the unusual situations in 24Cr and 29Cu
8. After 3d, the next higher orbital is 4p
• Filling of 4p begins in 31Ga
• When 36Kr is reached, the 4p orbitals are completely filled up
• This we saw in fig.2.60
• Note that, there are 6 electrons in the 4p orbitals
♦ There are 6 elements from 31Ga to 31Ga
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table
■ 36Kr is a milestone because, all the orbitals in it are completely filled
• So we have completed up to no.36 krypton. We can continue the above steps for the remaining elements. Those steps will help us to become familiar with the 'basic pattern'
• Filling of 5s begins in 37Rb
• When 38Sr is reached, the 5s orbital is completely filled up
• Note that, there are 2 electrons in the 5s orbital
♦ There are 2 elements from 37Rb to 38Sr
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
10. After 5s, the next higher orbital is 4d
• Filling of 4d begins in 39Y
• When 48Cd is reached, the 4d orbitals are completely filled up
• Note that, there are 10 electrons in the 4d orbitals
♦ There are 10 elements from 39Y to 48Cd
• The last electron in each of these elements fall in the d orbital. So these elements belong to the d-block of the periodic table
11. After 4d, the next higher orbital is 5p
• Filling of 5p begins in 49In
• When 54Xe is reached, the 4p orbitals are completely filled up
• Note that, there are 6 electrons in the 4p orbitals
♦ There are 6 elements from 49In to 54Xe
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table
■ 54Xe is a milestone because, all the orbitals in it are completely filled
12. After 5p, the next higher orbital is 6s
• Filling of 6s begins in 55Cs
• When 56Ba is reached, the 6s orbital is completely filled up
• Note that, there are 2 electrons in the 6s orbital
♦ There are 2 elements from 55Cs to 56Ba
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
13. After 6s, the next higher orbital is 4f
• Filling of 4f begins in 57La
• When 71Lu is reached, the 4f orbital is completely filled up
• Note that, there are 14 electrons in the 4f orbital
♦ There are 15 elements from 57La to 71Lu
♦ This small difference is due to the differences in energy levels of orbitals
♦ We will learn about it in higher classes
• The last electron in each of these elements fall in the f orbital. So these elements belong to the f-block of the periodic table
14. After 4f, the next higher orbital is 5d
• Filling of 5d begins in 72Hf
• When 80Hg is reached, the 5d orbitals are completely filled up
• Note that, there are 10 electrons in the 5d orbitals
♦ There are 9 elements from 72Hf to 80Hg
♦ This small difference is due to the differences in energy levels of orbitals
♦ We will learn about it in higher classes
• The last electron in each of these elements fall in the d orbital. So these elements belong to the d-block of the periodic table
15. After 5d, the next higher orbital is 6p
• Filling of 6p begins in 81Tl
• When 86Rn is reached, the 6p orbitals are completely filled up
• Note that, there are 6 electrons in the 6p orbitals
♦ There are 6 elements from 81Tl to 86Rn
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table
16. After 6p, the next higher orbital is 7s
• Filling of 7s begins in 87Fr
• When 88Ra is reached, the 7s orbital is completely filled up
• Note that, there are 2 electrons in the 7s orbital
♦ There are 2 elements from 87Fr to 88Ra
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
17. After 7s, the next higher orbital is 5f
• Filling of 5f begins in 89Ac
• When 103Lr is reached, the 5f orbital is completely filled up
• Note that, there are 14 electrons in the 5f orbital
♦ There are 15 elements from 89Ac to 103Lr
♦ This small difference is due to the differences in energy levels of orbitals
♦ We will learn about it in higher classes
• The last electron in each of these elements fall in the f orbital. So these elements belong to the f-block of the periodic table
18. After 5f, the next higher orbital is 6d
• Filling of 6d begins in 104Rf
• When 112Cn is reached, the 6d orbitals are completely filled up
• Note that, there are 10 electrons in the 6d orbitals
♦ There are 9 elements from 104Rf to 112Cn
♦ This small difference is due to the differences in energy levels of orbitals
♦ We will learn about it in higher classes
• The last electron in each of these elements fall in the d orbital. So these elements belong to the d-block of the periodic table
19. So we have reached up to element no.112. Details about newly discovered elements after 112Cn can be seen here.
9. After 4p, the next higher orbital is 5s
• When 38Sr is reached, the 5s orbital is completely filled up
• Note that, there are 2 electrons in the 5s orbital
♦ There are 2 elements from 37Rb to 38Sr
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
10. After 5s, the next higher orbital is 4d
• Filling of 4d begins in 39Y
• When 48Cd is reached, the 4d orbitals are completely filled up
• Note that, there are 10 electrons in the 4d orbitals
♦ There are 10 elements from 39Y to 48Cd
• The last electron in each of these elements fall in the d orbital. So these elements belong to the d-block of the periodic table
11. After 4d, the next higher orbital is 5p
• Filling of 5p begins in 49In
• When 54Xe is reached, the 4p orbitals are completely filled up
• Note that, there are 6 electrons in the 4p orbitals
♦ There are 6 elements from 49In to 54Xe
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table
■ 54Xe is a milestone because, all the orbitals in it are completely filled
12. After 5p, the next higher orbital is 6s
• Filling of 6s begins in 55Cs
• When 56Ba is reached, the 6s orbital is completely filled up
• Note that, there are 2 electrons in the 6s orbital
♦ There are 2 elements from 55Cs to 56Ba
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
13. After 6s, the next higher orbital is 4f
• Filling of 4f begins in 57La
• When 71Lu is reached, the 4f orbital is completely filled up
• Note that, there are 14 electrons in the 4f orbital
♦ There are 15 elements from 57La to 71Lu
♦ This small difference is due to the differences in energy levels of orbitals
♦ We will learn about it in higher classes
• The last electron in each of these elements fall in the f orbital. So these elements belong to the f-block of the periodic table
14. After 4f, the next higher orbital is 5d
• Filling of 5d begins in 72Hf
• When 80Hg is reached, the 5d orbitals are completely filled up
• Note that, there are 10 electrons in the 5d orbitals
♦ There are 9 elements from 72Hf to 80Hg
♦ This small difference is due to the differences in energy levels of orbitals
♦ We will learn about it in higher classes
• The last electron in each of these elements fall in the d orbital. So these elements belong to the d-block of the periodic table
15. After 5d, the next higher orbital is 6p
• Filling of 6p begins in 81Tl
• When 86Rn is reached, the 6p orbitals are completely filled up
• Note that, there are 6 electrons in the 6p orbitals
♦ There are 6 elements from 81Tl to 86Rn
• The last electron in each of these elements fall in the p orbital. So these elements belong to the p-block of the periodic table
16. After 6p, the next higher orbital is 7s
• Filling of 7s begins in 87Fr
• When 88Ra is reached, the 7s orbital is completely filled up
• Note that, there are 2 electrons in the 7s orbital
♦ There are 2 elements from 87Fr to 88Ra
• The last electron in each of these elements fall in the s orbital. So these elements belong to the s-block of the periodic table
17. After 7s, the next higher orbital is 5f
• Filling of 5f begins in 89Ac
• When 103Lr is reached, the 5f orbital is completely filled up
• Note that, there are 14 electrons in the 5f orbital
♦ There are 15 elements from 89Ac to 103Lr
♦ This small difference is due to the differences in energy levels of orbitals
♦ We will learn about it in higher classes
• The last electron in each of these elements fall in the f orbital. So these elements belong to the f-block of the periodic table
18. After 5f, the next higher orbital is 6d
• Filling of 6d begins in 104Rf
• When 112Cn is reached, the 6d orbitals are completely filled up
• Note that, there are 10 electrons in the 6d orbitals
♦ There are 9 elements from 104Rf to 112Cn
♦ This small difference is due to the differences in energy levels of orbitals
♦ We will learn about it in higher classes
• The last electron in each of these elements fall in the d orbital. So these elements belong to the d-block of the periodic table
19. So we have reached up to element no.112. Details about newly discovered elements after 112Cn can be seen here.
Now we can see the definitions for core electrons and valence electrons
• Every element will have core electrons and valence electrons
• It is easier to define valence electrons first
1. Consider the electronic configuration of an element
2. Note the highest value of the principal quantum number 'n' in that configuration
3. The electrons which have that highest 'n' value are called valence electrons
• In short, the electrons in the outermost main-shell are called valence electrons
4. All electrons which are not valence electrons are called core electrons
■ However, this definition is appropriate only for s-block and p-block elements. We will see the definition related to other block elements in later chapters
Now we will see some solved examples
Solved example 2.81
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energies, list them also
(1) n = 4, l = 2, ml = -2, ms = 1⁄2
(2) n = 3, l = 2, ml = 1, ms = 1⁄2
(3) n = 4, l = 1, ml = 0, ms = 1⁄2
(4) n = 3, l = 2, ml = -2, ms = -1⁄2
(5) n = 3, l = 1, ml = -1, ms = 1⁄2
(6) n = 4, l = 1, ml = 0, ms = 1⁄2
Solution:
• To compare energies, we compare n and l values
1. The least energy will be possessed by the electron in the 'main-shell closest to the nucleus'
• That means, the main-shell which has the lowest n value
2. In our present problem the 2nd, 4th and 5th electrons possess the same smallest quantum number n=3
• So to find the one from among those three, we consider the next quantum number l
3. When we consider l, the 5th electron has the lowest value
So the 5th electron has the lowest energy
4. When we consider the other two, both have the same l value of 2
So 2nd and 4th electrons have the same energy
5. Up to this point, we can write: 5th < [2nd = 4th] <
Three more electrons remain to be compared. They are: 1st, 3rd and 6th
6. Out of those three, all have the same n value of 4
So we consider the next quantum number l
7. 3rd and 6th have the same l value 1
So the result in (5) becomes: 5th < [2nd = 4th] < [3rd = 6th]<
8. Only the 1st remains. So the result becomes:
5th < [2nd = 4th] < [3rd = 6th] < 1st
Solved example 2.82
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ?
Solution:
1. Let us write the (n+l) values:
• For the 2p orbital, the (n+l) value is (2+1) = 3
• For the 3p orbital, the (n+l) value is (3+1) = 4
• For the 4p orbital, the (n+l) value is (4+1) = 5
2. The highest (n+l) value is for the 4p
3. That means, the 4p will be having the greatest energy among the three given orbitals
• This implies that, the 4p will be experiencing the lowest effective nuclear charge
(Recall that, lesser the attraction, greater the energy. Details here)
Solved example 2.83
Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p
Solution:
Part (i):
1. Let us write the (n+l) values:
• For the 2s orbital, the (n+l) value is (2+0) = 2
• For the 3s orbital, the (n+l) value is (3+0) = 3
2. The lower (n+l) value is for the 2s
3. That means, the 2s will be having the lower energy among the two given orbitals
• This implies that, the 2s will be experiencing the larger effective nuclear charge
(Recall that, greater the attraction, lesser the energy. Details here)
Part (ii):
1. Let us write the (n+l) values:
• For the 4d orbital, the (n+l) value is (4+2) = 6
• For the 4f orbital, the (n+l) value is (4+3) = 7
2. The lower (n+l) value is for the 4d
3. That means, the 4d will be having the lower energy among the two given orbitals
• This implies that, the 4d will be experiencing the larger effective nuclear charge
(Recall that, greater the attraction, lesser the energy. Details here)
Part (iii):
1. Let us write the (n+l) values:
• For the 3d orbital, the (n+l) value is (3+2) = 5
• For the 3p orbital, the (n+l) value is (3+1) = 4
2. The lower (n+l) value is for the 3d
3. That means, the 3p will be having the lower energy among the two given orbitals
• This implies that, the 3p will be experiencing the larger effective nuclear charge
(Recall that, greater the attraction, lesser the energy. Details here)
Solved example 2.84
The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
Solution:
1. The unpaired electrons in both Al and Si are in the 3p orbital
So the 'distances from the respective nuclei' are same in both the cases
2. Atomic number of Al is 13 and that of Si is 14
So, even though the distances are the same, the electrons in Si will experience an additional pull from the nucleus (due to the one extra proton)
3. That means, the unpaired electrons in Si will experience more effective nuclear charge from the nucleus
Solved example 2.85
Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Solution:
(a) The electronic configuration of P is [Ne]3s23p3
• The 3p orbitals are being filled
• Applying Hund's rule, it is clear that, the number of unpaired electrons is 3
(b) The electronic configuration of Si is [Ne]3s23p2
• The 3p orbitals are being filled
• Applying Hund's rule, it is clear that, the number of unpaired electrons is 2
(c) The electronic configuration of Cr is [Ar]4s13d5
• The 4s and 3d orbitals are being filled
• Applying Hund's rule, it is clear that, the number of unpaired electrons is 6
(d) The electronic configuration of Fe is [Ar]3d64s2
• The 3d orbitals are being filled
• Applying Hund's rule, it is clear that, the number of unpaired electrons is 4
(e) The electronic configuration of Kr is [Ar]4s24p6
• The element Krypton is a noble gas. It will not have any unpaired electrons
• This fact is clear from the electronic configuration
Solved example 2.86
(a) How many sub-shells are associated with n = 4 ?
(b) How many electrons will be present in the sub-shells having ms value of –1⁄2 for n = 4 ?
Solution:
Part (a):
1. The number of sub-shells in a main-shell will be equal to the n value of that main-shell
2. In our present case, the n value is 4
• So there will be 4 main shells. They are: s, p, d and f
Part (b):
1. The quantum numbers of all electrons in the main-shell with (n=4) will be of the form:
{4, l, ml, ms}
2. Half of those electrons will have the quantum numbers of the form {4, l, ml, 1⁄2}
The other half of those electrons will have the quantum numbers of the form {4, l, ml, –1⁄2}
3. So we have to find the total number of electrons in the main-shell with (n=4)
• In the s sub-shell there is only one orbital. so there will be only 2 electrons
• In the p sub-shell there are 3 orbitals. So there will be 6 electrons
• In the d sub-shell there are 5 orbitals. So there will be 10 electrons
• In the f sub-shell there are 7 orbitals. So there will be 14 electrons
Thus the total number of electrons = (2 +6 +10 +14) = 32
4. So the number of electrons with (ms= –1⁄2) is (32⁄2) = 16
• Every element will have core electrons and valence electrons
• It is easier to define valence electrons first
1. Consider the electronic configuration of an element
2. Note the highest value of the principal quantum number 'n' in that configuration
3. The electrons which have that highest 'n' value are called valence electrons
• In short, the electrons in the outermost main-shell are called valence electrons
4. All electrons which are not valence electrons are called core electrons
■ However, this definition is appropriate only for s-block and p-block elements. We will see the definition related to other block elements in later chapters
Now we will see some solved examples
Solved example 2.81
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energies, list them also
(1) n = 4, l = 2, ml = -2, ms = 1⁄2
(2) n = 3, l = 2, ml = 1, ms = 1⁄2
(3) n = 4, l = 1, ml = 0, ms = 1⁄2
(4) n = 3, l = 2, ml = -2, ms = -1⁄2
(5) n = 3, l = 1, ml = -1, ms = 1⁄2
(6) n = 4, l = 1, ml = 0, ms = 1⁄2
Solution:
• To compare energies, we compare n and l values
1. The least energy will be possessed by the electron in the 'main-shell closest to the nucleus'
• That means, the main-shell which has the lowest n value
2. In our present problem the 2nd, 4th and 5th electrons possess the same smallest quantum number n=3
• So to find the one from among those three, we consider the next quantum number l
3. When we consider l, the 5th electron has the lowest value
So the 5th electron has the lowest energy
4. When we consider the other two, both have the same l value of 2
So 2nd and 4th electrons have the same energy
5. Up to this point, we can write: 5th < [2nd = 4th] <
Three more electrons remain to be compared. They are: 1st, 3rd and 6th
6. Out of those three, all have the same n value of 4
So we consider the next quantum number l
7. 3rd and 6th have the same l value 1
So the result in (5) becomes: 5th < [2nd = 4th] < [3rd = 6th]<
8. Only the 1st remains. So the result becomes:
5th < [2nd = 4th] < [3rd = 6th] < 1st
Solved example 2.82
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ?
Solution:
1. Let us write the (n+l) values:
• For the 2p orbital, the (n+l) value is (2+1) = 3
• For the 3p orbital, the (n+l) value is (3+1) = 4
• For the 4p orbital, the (n+l) value is (4+1) = 5
2. The highest (n+l) value is for the 4p
3. That means, the 4p will be having the greatest energy among the three given orbitals
• This implies that, the 4p will be experiencing the lowest effective nuclear charge
(Recall that, lesser the attraction, greater the energy. Details here)
Solved example 2.83
Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p
Solution:
Part (i):
1. Let us write the (n+l) values:
• For the 2s orbital, the (n+l) value is (2+0) = 2
• For the 3s orbital, the (n+l) value is (3+0) = 3
2. The lower (n+l) value is for the 2s
3. That means, the 2s will be having the lower energy among the two given orbitals
• This implies that, the 2s will be experiencing the larger effective nuclear charge
(Recall that, greater the attraction, lesser the energy. Details here)
Part (ii):
1. Let us write the (n+l) values:
• For the 4d orbital, the (n+l) value is (4+2) = 6
• For the 4f orbital, the (n+l) value is (4+3) = 7
2. The lower (n+l) value is for the 4d
3. That means, the 4d will be having the lower energy among the two given orbitals
• This implies that, the 4d will be experiencing the larger effective nuclear charge
(Recall that, greater the attraction, lesser the energy. Details here)
Part (iii):
1. Let us write the (n+l) values:
• For the 3d orbital, the (n+l) value is (3+2) = 5
• For the 3p orbital, the (n+l) value is (3+1) = 4
2. The lower (n+l) value is for the 3d
3. That means, the 3p will be having the lower energy among the two given orbitals
• This implies that, the 3p will be experiencing the larger effective nuclear charge
(Recall that, greater the attraction, lesser the energy. Details here)
Solved example 2.84
The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
Solution:
1. The unpaired electrons in both Al and Si are in the 3p orbital
So the 'distances from the respective nuclei' are same in both the cases
2. Atomic number of Al is 13 and that of Si is 14
So, even though the distances are the same, the electrons in Si will experience an additional pull from the nucleus (due to the one extra proton)
3. That means, the unpaired electrons in Si will experience more effective nuclear charge from the nucleus
Solved example 2.85
Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Solution:
(a) The electronic configuration of P is [Ne]3s23p3
• The 3p orbitals are being filled
• Applying Hund's rule, it is clear that, the number of unpaired electrons is 3
(b) The electronic configuration of Si is [Ne]3s23p2
• The 3p orbitals are being filled
• Applying Hund's rule, it is clear that, the number of unpaired electrons is 2
(c) The electronic configuration of Cr is [Ar]4s13d5
• The 4s and 3d orbitals are being filled
• Applying Hund's rule, it is clear that, the number of unpaired electrons is 6
(d) The electronic configuration of Fe is [Ar]3d64s2
• The 3d orbitals are being filled
• Applying Hund's rule, it is clear that, the number of unpaired electrons is 4
(e) The electronic configuration of Kr is [Ar]4s24p6
• The element Krypton is a noble gas. It will not have any unpaired electrons
• This fact is clear from the electronic configuration
Solved example 2.86
(a) How many sub-shells are associated with n = 4 ?
(b) How many electrons will be present in the sub-shells having ms value of –1⁄2 for n = 4 ?
Solution:
Part (a):
1. The number of sub-shells in a main-shell will be equal to the n value of that main-shell
2. In our present case, the n value is 4
• So there will be 4 main shells. They are: s, p, d and f
Part (b):
1. The quantum numbers of all electrons in the main-shell with (n=4) will be of the form:
{4, l, ml, ms}
2. Half of those electrons will have the quantum numbers of the form {4, l, ml, 1⁄2}
The other half of those electrons will have the quantum numbers of the form {4, l, ml, –1⁄2}
3. So we have to find the total number of electrons in the main-shell with (n=4)
• In the s sub-shell there is only one orbital. so there will be only 2 electrons
• In the p sub-shell there are 3 orbitals. So there will be 6 electrons
• In the d sub-shell there are 5 orbitals. So there will be 10 electrons
• In the f sub-shell there are 7 orbitals. So there will be 14 electrons
Thus the total number of electrons = (2 +6 +10 +14) = 32
4. So the number of electrons with (ms= –1⁄2) is (32⁄2) = 16
So we have completed a discussion on structure of atom. In the next chapter, we will see Classification of elements and periodicity of properties
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