In the previous section 2.13, we saw the basics about the quantum mechanical model of the atom. We were discussing how the four quantum numbers can be used to find the location of the electron. We wrote 23 steps. In this section we will see the rest of the steps
24. So the postman has reached the correct house
• In a house, there will be several occupants. The postman has to find the correct person from among them
• In the same way, in our present case, we have to find the correct electron from among the various electrons present in the orbital
25. For the postman, the task is easy because, every person has a name. The name would be written on the envelope
• But electrons do not have names. So what do we do?
■ Luckily for us, there will be only two electrons in any orbital
(We will see the reason in a later section)
• So our next task is to distinguish between those two electrons
26. In 1925, George Uhlenbeck and Samuel Goudsmith found out that, the two electrons in an orbital have opposite spins
• 'Opposite spins' can be explained using two steps as follows:
(i) We know that the earth spins around it’s own axis
• In a similar way, an electron spins around it’s own axis
(ii) One of the two electrons spin in the clock-wise direction
• The other spins in the anti clock-wise direction
27. The 'opposite spins' gives us a method to distinguish between the two electrons:
(i) We can name one electron as:
'The electron with the clock-wise spin'
(ii) We can name the other electron as:
'The electron with the anti clock-wise spin'
28. But these are long names. To make things short, we adopt the following method:
(i) We represent one of the two electrons with an up arrow (↑)
• So '↑' is an electron
• Whenever we see '↑', we read it as: The spin up electron
(ii) We represent the other electron with a down arrow (↓)
• So ‘↓’ is also an electron
• Whenever we see '↓', we read it as: The spin down electron
29. In addition to those names, they have numbers also
• We have seen these:
♦ Each of the main-shells possess a unique principal quantum number (n)
♦ Each of the sub-shells possess a unique azimuthal quantum number (l)
♦ Each of the orbitals possess a unique magnetic orbital quantum number (ml)
• In the same way:
♦ Each of the two electrons in an orbital, possess a unique electron spin quantum number
♦ They are denoted by: ms
30. The numbers are:
♦ The ↑ has the ms number: 1⁄2
♦ The ↓ has the ms number: -1⁄2
31. Now we have the correct 'address' of any electron. We can complete the table. It is shown in fig.2.41 below:
Solved example 2.65
An electron has the address: (n, l, ml, ms) = (2, 1, 0, -1⁄2)
Where is this electron situated?
Solution:
1. Given: n = 2
• So the electron is in the 2nd main-shell
2. Given: l = 1
• So the electron is in the p sub-shell
3. Given: ml = 0
• So the electron is in the py orbital
(Recall that, in the p sub-shell, the ml values of -1, 0, 1 denote px , py and pz)
4. Given: ms = -1⁄2
• So it is the spin down electron (↓)
5. We can write the proper address:
• The given electron is:
♦ The ↓ electron
♦ In the py orbital
♦ In the p sub-shell
♦ In the 2nd main shell
Solved example 2.66
An electron has the address: (n, l, ml, ms) = (4, 2, -1, 1⁄2)
Where is this electron situated?
Solution:
1. Given: n = 4
• So the electron is in the 4th main-shell
2. Given: l = 2
• So the electron is in the d sub-shell
3. Given: ml = -1
• So the electron is in the dxz orbital
(Recall that, in the d sub-shell, the ml values of -2, -1, 0, 1, 2 denote $\mathbf\small{d_{xy},d_{xz},d_{yz},d_{x^2- y^2}\;\text{and}\;d_{z^2}}$)
4. Given: ms = 1⁄2
• So it is the spin up electron (↑)
5. We can write the proper address:
• The given electron is:
♦ The ↑ electron
♦ In the dxz orbital
♦ In the d sub-shell
♦ In the 4th main shell
Solved example 2.67
What is the total number of orbitals associated with the principal quantum number n=3?
Solution:
1. 'Principal quantum number' indicates main-shell
• So let us first go to the main-shell with n=3
2. At n=3, we will see 3 sub-shells
• They are: s, p and d
3. Number of orbitals in each sub-shell are:
♦ The s sub-shell has 1 orbital
♦ The p sub-shell has 3 orbitals
♦ The d sub-shell has 5 orbitals
4. So the total number of orbitals associated with n=3 is (1+3+5) = 9
Solved example 2.68
Write the address of the following sub-shells in the s, p, d, f notation
(i) The sub-shell has the quantum number l =1
• It is in the 2nd main-shell
(ii) The sub-shell has the quantum number l =0
• It is in the 4th main-shell
(iii) The sub-shell has the quantum number l =3
• It is in the 5th main-shell
(iv) The sub-shell has the quantum number l =2
• It is in the 3rd main-shell
Solution:
s, p, d, f notation is a shortened method to write the address of any sub-shell
It can be written in two steps:
1. Look for the main-shell in which the sub-shell is situated
♦ The n value of that main-shell is written first
2. The name of the orbital is written on the right side
Part (i):
1. The sub-shell is situated in the 2nd main-shell
• So n=2
2. The sub-shell has the quantum number l =1
• So it is the p sub-shell
3. Thus the s, p, d, f notation is: 2p
Part (ii):
1. The sub-shell is situated in the 4th main-shell
• So n=4
2. The sub-shell has the quantum number l =0
• So it is the s sub-shell
3. Thus the s, p, d, f notation is: 4s
Part (iii):
1. The sub-shell is situated in the 5th main-shell
• So n=5
2. The sub-shell has the quantum number l =3
• So it is the f sub-shell
3. Thus the s, p, d, f notation is: 5f
Part (iv):
1. The sub-shell is situated in the 3rd main-shell
• So n=3
2. The sub-shell has the quantum number l =2
• So it is the d sub-shell
3. Thus the s, p, d, f notation is: 3d
Solved example 2.69
What is the lowest value of n that allows g orbitals to exist ?
Solution:
1. Consider any main-shell
• Let the n value of that main-shell be 'x'
• Then the number of sub-shells allowed in that main-shell = x
2. If g sub-shell is to be present in that main-shell, all sub-shells coming before g must already be present
• That means: s, p, d and f must be already present
3. So, if g also becomes available, there will be a total of 5 sub-shells in that main-shell
• That means: x = 5
• That means: If g is to be present, the value of n should be 5 or greater than 5
Solved example 2.70
An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron
Solution:
Possible values of n:
• From '3d', it is clear that, the 'n' value is none other than 3
Possible values of l:
• If that electron is present in the d sub-shell, it's l value can be none other than 2
• This is because, all d sub-shells has the azimuthal quantum number (l) = 2
Possible values of ml:
• The electron is present in the d sub-shell
• It can be present in any one of the d orbitals
• The five ml values of the d orbitals are: -2, -1, 0, 1, 2
• The electron can have any one of those 5 values
Solved example 2.71
Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n=1; l=0 (b) n = 3; l=1 (c) n = 4; l =2 (d) n=4; l=3.
Solution:
(a) l=0 indicates s sub-shell
• So the s, p, d, f notation is: 1s
(b) l=1 indicates p sub-shell
• So the s, p, d, f notation is: 3p
(c) l=2 indicates d sub-shell
• So the s, p, d, f notation is: 4d
(d) l=3 indicates f sub-shell
• So the s, p, d, f notation is: 4f
Solved example 2.72
Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a) n = 0, l = 0, ml = 0, ms = + ½
(b) n = 1, l = 0, ml = 0, ms = – ½
(c) n = 1, l = 1, ml = 0, ms = + ½
(d) n = 2, l = 1, ml = 0, ms = – ½
(e) n = 3, l = 3, ml = –3, ms = + ½
(f) n = 3, l = 1, ml = 0, ms = + ½
Solution:
1. The set (a) is not possible because:
• The smallest principal quantum number is (n=1)
• There is no main-shell with (n=0)
2. The set (c) is not possible because:
• (l=1) indicates the presence of p sub-shell
• If p sub-shell is present, s sub-shell would also be present
• But when n=1, there can only be one sub-shell
2. The set (e) is not possible because:
• (l=3) indicates the presence of f sub-shell
• If f sub-shell is present, s, p and d sub-shells would also be present
• But when n=3, there can only be three sub-shells
Solved example 2.73
How many electrons in an atom may have the following quantum numbers?
(a) n = 4; ms = – ½ (b) n = 3; l = 0
Solution:
Part (a)
1. In n=4, there will be 4 sub-shells: s, p, d and f
• In s sub-shell, there is only 1 orbital
♦ So number of electrons in the s sub-shell = 1 × 2 = 2
♦ One out of those two electrons will have (ms = -1⁄2)
• In p sub-shell, there are 3 orbitals
♦ So number of electrons in the p sub-shell = 3 × 2 = 6
♦ 3 out of those 6 electrons will have (ms = -1⁄2)
• In d sub-shell, there are 5 orbitals
♦ So number of electrons in the d sub-shell = 5 × 2 = 10
♦ 5 out of those 10 electrons will have (ms = -1⁄2)
• In f sub-shell, there are 7 orbitals
♦ So number of electrons in the f sub-shell = 7 × 2 = 14
♦ 7 out of those 14 electrons will have (ms = -1⁄2)
2. So in (n=4), the total number of electrons with (ms = -1⁄2) = 1 + 3 + 5 + 7 = 16
Part (b)
1. In n=3, there will be 3 sub-shells: s, p, d and f
• But given that (l = 0)
• So we need to consider the s sub-shell only
2. In s sub-shell, there is only 1 orbital
♦ So number of electrons in the s sub-shell = 1 × 2 = 2
3. So number of electrons having [n = 3; l = 0] = 2
24. So the postman has reached the correct house
• In a house, there will be several occupants. The postman has to find the correct person from among them
• In the same way, in our present case, we have to find the correct electron from among the various electrons present in the orbital
25. For the postman, the task is easy because, every person has a name. The name would be written on the envelope
• But electrons do not have names. So what do we do?
■ Luckily for us, there will be only two electrons in any orbital
(We will see the reason in a later section)
• So our next task is to distinguish between those two electrons
26. In 1925, George Uhlenbeck and Samuel Goudsmith found out that, the two electrons in an orbital have opposite spins
• 'Opposite spins' can be explained using two steps as follows:
(i) We know that the earth spins around it’s own axis
• In a similar way, an electron spins around it’s own axis
(ii) One of the two electrons spin in the clock-wise direction
• The other spins in the anti clock-wise direction
27. The 'opposite spins' gives us a method to distinguish between the two electrons:
(i) We can name one electron as:
'The electron with the clock-wise spin'
(ii) We can name the other electron as:
'The electron with the anti clock-wise spin'
28. But these are long names. To make things short, we adopt the following method:
(i) We represent one of the two electrons with an up arrow (↑)
• So '↑' is an electron
• Whenever we see '↑', we read it as: The spin up electron
(ii) We represent the other electron with a down arrow (↓)
• So ‘↓’ is also an electron
• Whenever we see '↓', we read it as: The spin down electron
29. In addition to those names, they have numbers also
• We have seen these:
♦ Each of the main-shells possess a unique principal quantum number (n)
♦ Each of the sub-shells possess a unique azimuthal quantum number (l)
♦ Each of the orbitals possess a unique magnetic orbital quantum number (ml)
• In the same way:
♦ Each of the two electrons in an orbital, possess a unique electron spin quantum number
♦ They are denoted by: ms
30. The numbers are:
♦ The ↑ has the ms number: 1⁄2
♦ The ↓ has the ms number: -1⁄2
31. Now we have the correct 'address' of any electron. We can complete the table. It is shown in fig.2.41 below:
 |
| Fig.2.41 |
So we have completed a discussion on quantum numbers. Let us see some solved examples:
An electron has the address: (n, l, ml, ms) = (2, 1, 0, -1⁄2)
Where is this electron situated?
Solution:
1. Given: n = 2
• So the electron is in the 2nd main-shell
2. Given: l = 1
• So the electron is in the p sub-shell
3. Given: ml = 0
• So the electron is in the py orbital
(Recall that, in the p sub-shell, the ml values of -1, 0, 1 denote px , py and pz)
4. Given: ms = -1⁄2
• So it is the spin down electron (↓)
5. We can write the proper address:
• The given electron is:
♦ The ↓ electron
♦ In the py orbital
♦ In the p sub-shell
♦ In the 2nd main shell
Solved example 2.66
An electron has the address: (n, l, ml, ms) = (4, 2, -1, 1⁄2)
Where is this electron situated?
Solution:
1. Given: n = 4
• So the electron is in the 4th main-shell
2. Given: l = 2
• So the electron is in the d sub-shell
3. Given: ml = -1
• So the electron is in the dxz orbital
(Recall that, in the d sub-shell, the ml values of -2, -1, 0, 1, 2 denote $\mathbf\small{d_{xy},d_{xz},d_{yz},d_{x^2- y^2}\;\text{and}\;d_{z^2}}$)
4. Given: ms = 1⁄2
• So it is the spin up electron (↑)
5. We can write the proper address:
• The given electron is:
♦ The ↑ electron
♦ In the dxz orbital
♦ In the d sub-shell
♦ In the 4th main shell
Solved example 2.67
What is the total number of orbitals associated with the principal quantum number n=3?
Solution:
1. 'Principal quantum number' indicates main-shell
• So let us first go to the main-shell with n=3
2. At n=3, we will see 3 sub-shells
• They are: s, p and d
3. Number of orbitals in each sub-shell are:
♦ The s sub-shell has 1 orbital
♦ The p sub-shell has 3 orbitals
♦ The d sub-shell has 5 orbitals
4. So the total number of orbitals associated with n=3 is (1+3+5) = 9
Solved example 2.68
Write the address of the following sub-shells in the s, p, d, f notation
(i) The sub-shell has the quantum number l =1
• It is in the 2nd main-shell
(ii) The sub-shell has the quantum number l =0
• It is in the 4th main-shell
(iii) The sub-shell has the quantum number l =3
• It is in the 5th main-shell
(iv) The sub-shell has the quantum number l =2
• It is in the 3rd main-shell
Solution:
s, p, d, f notation is a shortened method to write the address of any sub-shell
It can be written in two steps:
1. Look for the main-shell in which the sub-shell is situated
♦ The n value of that main-shell is written first
2. The name of the orbital is written on the right side
Part (i):
1. The sub-shell is situated in the 2nd main-shell
• So n=2
2. The sub-shell has the quantum number l =1
• So it is the p sub-shell
3. Thus the s, p, d, f notation is: 2p
Part (ii):
1. The sub-shell is situated in the 4th main-shell
• So n=4
2. The sub-shell has the quantum number l =0
• So it is the s sub-shell
3. Thus the s, p, d, f notation is: 4s
Part (iii):
1. The sub-shell is situated in the 5th main-shell
• So n=5
2. The sub-shell has the quantum number l =3
• So it is the f sub-shell
3. Thus the s, p, d, f notation is: 5f
Part (iv):
1. The sub-shell is situated in the 3rd main-shell
• So n=3
2. The sub-shell has the quantum number l =2
• So it is the d sub-shell
3. Thus the s, p, d, f notation is: 3d
Solved example 2.69
What is the lowest value of n that allows g orbitals to exist ?
Solution:
1. Consider any main-shell
• Let the n value of that main-shell be 'x'
• Then the number of sub-shells allowed in that main-shell = x
2. If g sub-shell is to be present in that main-shell, all sub-shells coming before g must already be present
• That means: s, p, d and f must be already present
3. So, if g also becomes available, there will be a total of 5 sub-shells in that main-shell
• That means: x = 5
• That means: If g is to be present, the value of n should be 5 or greater than 5
Solved example 2.70
An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron
Solution:
Possible values of n:
• From '3d', it is clear that, the 'n' value is none other than 3
Possible values of l:
• If that electron is present in the d sub-shell, it's l value can be none other than 2
• This is because, all d sub-shells has the azimuthal quantum number (l) = 2
Possible values of ml:
• The electron is present in the d sub-shell
• It can be present in any one of the d orbitals
• The five ml values of the d orbitals are: -2, -1, 0, 1, 2
• The electron can have any one of those 5 values
Solved example 2.71
Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n=1; l=0 (b) n = 3; l=1 (c) n = 4; l =2 (d) n=4; l=3.
Solution:
(a) l=0 indicates s sub-shell
• So the s, p, d, f notation is: 1s
(b) l=1 indicates p sub-shell
• So the s, p, d, f notation is: 3p
(c) l=2 indicates d sub-shell
• So the s, p, d, f notation is: 4d
(d) l=3 indicates f sub-shell
• So the s, p, d, f notation is: 4f
Solved example 2.72
Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a) n = 0, l = 0, ml = 0, ms = + ½
(b) n = 1, l = 0, ml = 0, ms = – ½
(c) n = 1, l = 1, ml = 0, ms = + ½
(d) n = 2, l = 1, ml = 0, ms = – ½
(e) n = 3, l = 3, ml = –3, ms = + ½
(f) n = 3, l = 1, ml = 0, ms = + ½
Solution:
1. The set (a) is not possible because:
• The smallest principal quantum number is (n=1)
• There is no main-shell with (n=0)
2. The set (c) is not possible because:
• (l=1) indicates the presence of p sub-shell
• If p sub-shell is present, s sub-shell would also be present
• But when n=1, there can only be one sub-shell
2. The set (e) is not possible because:
• (l=3) indicates the presence of f sub-shell
• If f sub-shell is present, s, p and d sub-shells would also be present
• But when n=3, there can only be three sub-shells
Solved example 2.73
How many electrons in an atom may have the following quantum numbers?
(a) n = 4; ms = – ½ (b) n = 3; l = 0
Solution:
Part (a)
1. In n=4, there will be 4 sub-shells: s, p, d and f
• In s sub-shell, there is only 1 orbital
♦ So number of electrons in the s sub-shell = 1 × 2 = 2
♦ One out of those two electrons will have (ms = -1⁄2)
• In p sub-shell, there are 3 orbitals
♦ So number of electrons in the p sub-shell = 3 × 2 = 6
♦ 3 out of those 6 electrons will have (ms = -1⁄2)
• In d sub-shell, there are 5 orbitals
♦ So number of electrons in the d sub-shell = 5 × 2 = 10
♦ 5 out of those 10 electrons will have (ms = -1⁄2)
• In f sub-shell, there are 7 orbitals
♦ So number of electrons in the f sub-shell = 7 × 2 = 14
♦ 7 out of those 14 electrons will have (ms = -1⁄2)
2. So in (n=4), the total number of electrons with (ms = -1⁄2) = 1 + 3 + 5 + 7 = 16
Part (b)
1. In n=3, there will be 3 sub-shells: s, p, d and f
• But given that (l = 0)
• So we need to consider the s sub-shell only
2. In s sub-shell, there is only 1 orbital
♦ So number of electrons in the s sub-shell = 1 × 2 = 2
3. So number of electrons having [n = 3; l = 0] = 2
This completes our present discussion on quantum numbers. In the next section, we will see shapes of orbitals
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