Quantum numbers and their calculation

Definition:

Quantum numbers are a set of four values that are used to describe the position and energy of the electron in an atom.

Alternatively, quantum numbers are the numbers used to indicate the size, shape, three-dimensional direction, variation of the energy level, and the direction of rotation of an electron around its own axis.

Four quantum numbers are required to fully describe the position of electrons in an atom or in an atomic orbital. They are:

1. Principal quantum number, n
2. Subsidiary or Azimuthal quantum number, l
3. Magnetic quantum number, m
4. Spin quantum number, s

Relation among four quantum numbers:

1. “n” can have values of 1, 2, 3, 4 ………..(only integers)
2. for a given “n” value, l can have values starting from 0 to (n-1), i.e a total of n values.
3. for a given values of l, there can be total of (2l+1) values of “m” ranging from -l to +l
4. Spin quantum number, “s” does not dependent on other quantum numbers. It can have two values + 1/2 (clockwise) and -1/2 (anti-clockwise)
5. For each value of “m”, there will be two values (+1/2 or -1/2) of s. So, there can be two electrons within an orbital with opposite spin.

For Helium:

He(2) = 1s²

Four quantum numbers are given below:

n = 1,

l = 0 ( electron exists in s orbital, for s orbital l = 0)

m = 0 ( because m = – l to +l through 0)

s = -1/2 ( every orbital contains 2 electrons but their spin should be opposite. If you draw box, it will be more clear)

For N (7) = 1s²2s²2p_x¹2p_y¹2p_z¹ ( well understood by box system)

Here, n = 2

l = 1 ( please see the value of l for p orbital)

m = -1

s = +1/2

For O (8) =1s²2s²2p_x²2p_y¹2p_z¹ ( In this case 8th electron of O pairs)

n = 2

l = 1 ( because electron at p orbital)

m = +1

s = -1/2 ( 8^th electron pairs with 5^th electron and spin quantum number for 5^th electron is =1/2)

For F(9) = 1s²2s²2p_x²2p_y²2p_z¹

n =2

l = 1

m = 0

s = -1/2

For Ne(10) = 1s²2s²2p_x²2p_y²2p_z²

Here,

n =2

l = 1

m = -1

s = -1/2 ( pairing occurs)

For Na(11)

Electronic configuration (EC) of Na(11) = 1s²2s²2p_x²2p_y²2p_z² 3s¹ ( We can also write this configuration as [Ne] 3s¹ )

(A)

(B)

Here,

n = 3

l = 0 ( for n =3, l should have 3 values such as 0, 1, and 2. Since, the11^th electron of Na occupies “s” orbital, the value of l should be 0)

m = 0 ( when l = 0, m = 0)

s = +1/2 ( according to A, clockwise/ upward direction) or -1/2 (according to B, anti-clockwise direction)

For Mg(12)

EC of Mg(12) = 1s²2s²2p_x²2p_y²2p_z² 3s²

Here, n = 3

l = 0 ( because electron is in s orbital)

m = 0

s = -1/2 ( pairing occurs, anti-clockwise direction)

Question: Find out the values of four quantum numbers for the 10^th electron of Na(11).

Hints: See the BOX

For Al(13)

EC of Al(13) = 1s²2s²2p_x²2p_y²2p_z² 3s² 3p_x¹ or [Ne] 3s² 3p_x¹

By box system

Here,

n = 3

l = 1 ( ask yourself why l value is 1)

m = +1 (see the relation between l and m)

s = +1/2

For Si(14)

EC of Si(14) = [Ne] 3s²3p_x¹3p_y¹3p_z

Here,

n = 3

l = 1

m = 0

s = +1/2

For P(15)

EC of P(15)= [Ne] 3s²3p_x¹3p_y¹3p_z¹

Here,

n = 3

l = 1

m = -1

s = +1/2

For S (16)

EC of S(16) = [Ne] 3s²3p_x²3p_y¹3p_z¹

Here,

n = 3

l = 1

m = +1 ( for p_x, l = +1, for p_y, l = 0, and for p_z, l = -1)

s = -1/2 ( due to pairing)

For Chlorine, Cl

EC of Cl(17) = [Ne] 3s²3p_x²3p_y²3p_z¹

Here,

n = 3

l = 1

m = 0 ( read the explanation for S)

s = -1/2 ( due to pairing)

For Ar(18)

EC of Ar(18) = [Ne] 3s²3p_x²3p_y²3p_z²

Here,

n = 3

l = 1

m = -1

s = -1/2

For K(19)

EC of K(19) = 1s²2s²2p_x²2p_y²2p_z² 3s² 3p_x² 3p_y²3p_z² 4s¹ or [Ar] 4s¹

Here,

n = 4

l = 0 ( for n = 4, l has four value such as 0, 1, 2 and 3. For “s” orbital l = 0)

m = 0 ( because l = 0)

s = +1/2