# 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) = 1s22s22px12py12pz1 ( 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) =1s22s22px22py12pz1 ( In this case 8th electron of O pairs)

n = 2

l = 1 ( because electron at p orbital)

m = +1

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

For F(9) = 1s22s22px22py22pz1

n =2

l = 1

m = 0

s = -1/2

For Ne(10) = 1s22s22px22py22pz2

Here,

n =2

l = 1

m = -1

s = -1/2 ( pairing occurs)

For Na(11)

Electronic configuration (EC) of Na(11) = 1s22s22px22py22pz2 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, the11th 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) = 1s22s22px22py22pz2 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 10th electron of Na(11).

Hints: See the BOX

For Al(13)

EC of Al(13) = 1s22s22px22py22pz2 3s2 3px1 or [Ne] 3s2 3px1

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] 3s23px13py13pz

Here,

n = 3

l = 1

m = 0

s = +1/2

For P(15)

EC of P(15)= [Ne] 3s23px13py13pz1

Here,

n = 3

l = 1

m = -1

s = +1/2

For S (16)

EC of S(16) = [Ne] 3s23px23py13pz1

Here,

n = 3

l = 1

m = +1 ( for px, l = +1, for py, l = 0, and for pz, l = -1)

s = -1/2 ( due to pairing)

For Chlorine, Cl

EC of Cl(17) = [Ne] 3s23px23py23pz1

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] 3s23px23py23pz2

Here,

n = 3

l = 1

m = -1

s = -1/2

For K(19)

EC of K(19) = 1s22s22px22py22pz2 3s2 3px2 3py23pz2 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

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