Quantum Numbers

QUANTUM NUMBERS

An atom contains large number of shells and subshells. These are distinguished from one another on the basis of their size, shape and orientation (direction) in space. The parameters are expressed in terms of different numbers called quantum numbers.

Quantum numbers may be defined as a set of four numbers with the help of which we can get complete information about all the electrons in an atom. It tells us the address of the electron i.e., location, energy, the type of orbital occupied and orientation of that orbital.

$\bullet$Principal quantum number (n): It tells the main shell in which the electron resides, the approximate distance of the electron from the nucleus and energy of that particular electron. It also tells the maximum number of electrons that a shell can accommodate is 2n², where n is the principal quantum number.

Shell K L M N

Principal quantum number (n) 1 2 3 4

Maximum number of electrons 2 8 18 32

$\bullet$Azimuthal or angular momentum quantum number$\left(\ell \right)$: This represents the number of subshells present in the main shell. These subsidiary orbits within a shell will be denoted as 0, 1, 2, 3, 4,… or s, p, d, f… This tells the shape of the subshells. The orbital angular momentum of the electron is given as $\sqrt{\ell \left(\ell +1\right)}$ ${}^h/{}_{2\pi }$ or $\sqrt{\ell \left(\ell +1\right)}\hslash$for a particular value of 'n'$\left(\text{where }\hslash =\frac{h}{2\pi }\right)$. For a given value of n values of possible $\ell$ vary from 0 to n – 1.

$\bullet$The magnetic quantum number (m): An electron due to its angular motion around the nucleus generates an electric field. This electric field is expected to produce a magnetic field. Under the influence of external magnetic field, the electrons of a subshell can orient themselves in certain preferred regions of space around the nucleus called orbitals. The magnetic quantum number determines the number of preferred orientations of the electron present in a subshell. The values allowed depends on the value of l, the angular momentum quantum number, m can assume all integral values between –$\ell$ to +$\ell$ including zero. Thus m can be –1, 0, +1 for $\ell$ = 1. Total values of m associated with a particular value of $\ell$ is given by 2$\ell$+ 1.

$\bullet$The spin quantum number (s): Just like earth which not only revolves around the sun but also spins about its own axis, an electron in an atom not only revolves around the nucleus but also spins about its own axis. Since an electron can spin either in clockwise direction or in anticlockwise direction, therefore, for any particular value of magnetic quantum number, spin quantum number can have two values, i.e., +1/2 and –1/2 or these are represented by two arrows pointing in the opposite directions, i.e., $\uparrow$ and $\downarrow$. When an electron goes to a vacant orbital, it can have a clockwise or anti clockwise spin i.e., +1/2 or –1/2. This quantum number helps to explain the magnetic properties of the substances.

Illustration 1. If the principal quantum number n has a value of 3, what are permitted values of the quantum numbers '$\ell$ and 'm'?

Solution: If principal quantum number (n) has the value of 3, subsidary quantum number ($\ell$) will also have three values i.e. 0, 1 and 2 and magnetic quantum number (m) will have n², i.e. 3² i.e. 9 values in all and they are designated as under:

n = 3 $\ell$ = 0 m = 0 (s)

$\ell$ = 1 m = – 1 (p)

m = 0

m = + 1

$\ell$ = 2 m = – 2 (d)

m = – 1

m = 0

m = +1

m = +2

Pauli's Exclusion Principle:

According to this principle, an orbital can contain a maximum number of two electrons and these two electrons must be of opposite spin.

Two electrons in an orbital can be represented by

SHAPES AND SIZE OF ORBITALS

An orbital is the region of space around the nucleus within which the probability of finding an electron of given energy is maximum (90–95%). The shape of this region (electron cloud) gives the shape of the orbital. It is basically determined by the azimuthal quantum number$\ell$, while the orientation of orbital depends on the magnetic quantum number (m). Let us now see the shapes of orbitals in the various subshells.

s–orbitals: These orbitals are spherical and symmetrical about the nucleus. The probability of finding the electron is maximum near the nucleus and keep on decreasing as the distance from the nucleus increases. There is vacant space between two successive s–orbitals known as radial node. But there is no radial node for 1s orbital since it is starting from the nucleus.

The size of the orbital depends upon the value of principal quantum number (n). Greater the value of n, larger is the size of the orbital. Therefore, 2s–orbital is larger than 1s orbital but both of them are non-directional and spherically symmetrical in shape.

p–orbitals ($\ell$=1):

The probability of finding the p–electron is maximum in two lobes on the opposite sides of the nucleus. This gives rise to a dumb–bell shape for the p–orbital. For p–orbital $\ell$ = 1. Hence,

m = –1, 0, +1. Thus, p–orbital have three different orientations. These are designated as p_x,p_y & p_z depending upon whether the density of electron is maximum along the x y and z axis respectively. As they are not spherically symmetrical, they have directional character. The two lobes of p–orbitals are separated by a nodal plane, where the probability of finding electron is zero.

The three p-orbitals belonging to a particular energy shell have equal energies and are called degenerate orbitals.

d–orbitals ($\ell$= 2):

For d–orbitals, l = 2. Hence m = –2,–1, 0, +1, +2. Thus there are 5d orbitals. They have relatively complex geometry. Out of the five orbitals, the three (d_xy, d_yz,d_zx) project in between the axis and the other two $d_{z^2}$ and $d_{x^2-y^2}$ lie along the axis.

Illustration 2. An electron has a spin quantum number $+\frac{1}{2}$and a magnetic quantum

number -1. It cannot occupy a/an

(A) d orbital (B) f orbital

(C) p orbital (D) s orbital

Solution: For s orbital m = 0. Hence (D) is correct.

RULES FOR FILLING OF ELECTRONS IN VARIOUS ORBITALS

The atom is built up by filling electrons in various orbitals according to the following rules.

Aufbau Principle: This principle states that the electrons are added one by one to the various orbitals in order of their increasing energy starting with the orbital of lowest energy. The increasing order of energy of various orbital is

1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,5f,6d,7p……………………

How to remember such a big sequence? To make it simple we are giving you the method to write the increasing order of the orbitals. Starting from the top, the direction of the arrow gives the order of filling of orbitals.

Alternatively, the order of increasing energies of the various orbitals can be calculated on the basis of (n +$\ell$) rule.

The energy of an orbital depends upon the sum of values of the principal quantum number (n) and the azimuthal quantum number ($\ell$). This is called (n +$\ell$) rule. According to this rule,

"In neutral isolated atom, the lower the value of (n +$\ell$) for an orbital, lower is its energy. However, if the two different types of orbitals have the same value of (n +$\ell$), the orbitals with lower value of n has lower energy''.

Illustration of (n +$\ell$) Rule:

Hund's Rule of Maximum Multiplicity:

This rule deals with the filling of electrons in the equal energy (degenerate) orbitals of the same sub shell (p,d and f). According to this rule,

"Electron pairing in p,d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied".

This is due to the fact that electrons being identical in charge, repel each other when present in the same orbital. This repulsion can, however, be minimized if two electrons move as far apart as possible by occupying different degenerate orbitals. All the electrons in a degenerate set of orbitals will have same spin.

Electronic Configuration of Elements:

Electronic configuration is the distribution of electrons into different shells, subshells and orbitals of an atom.

Keeping in view the above mentioned rules, electronic configuration of any orbital can be simply represented by the notation]

Alternatively:

Orbital can be represented by a box and an electron with its direction of spin by arrow. To write the electronic configuration, just we need to know (i) the atomic number (ii) the order in which orbitals are to be filled (iii) maximum number of electrons in a shell, sub–shell or orbital.

$\bullet$Each orbital can accommodate two electrons

$\bullet$The number of electrons to be accomodated in a subshell is 2 x numbers of

degenerate orbitals.

$\bullet$The maximum number of electrons in each shell (K, L, M, N…) is given by 2n².

where n is the principal quantum number.

$\bullet$The maximum number of orbitals in a shell is given by n² where n is the principal quantum number.

Importance of Knowing the Electronic Configuration:

The chemical properties of an element are dependent on the relative arrangement of its electrons.

Illustration 3. Write the electronic configuration of nitrogen (atomic number= 7)

Solution:

Exceptional Configurations:

Stability of Half Filled and Completely Filled Orbitals:

Cu has 29 electrons. Its expected electronic configuration is

1s²2s²2p⁶3s²3p⁶4s²3d⁹

But a shift of one electron from lower energy 4s orbital to higher energy 3d orbital will make the distribution of electron symmetrical and hence will impart more stability.

Thus the electronic configuration of Cu is

1s² 2s²2p⁶3s²3p⁶4s¹3d¹⁰

Fully filled and half filled orbitals are more stable.

Illustration 4. Which of the following metals has the highest value of exchange energy required for exchange stabilization?

(A) Mn (B) Cr

(C) Cu (D) Zn

Solution: Cu has full filled d-subshell. Hence (C) is correct.

Illustration 5. The total spin resulting from a d⁷ configuration is

(A) 1 (B) 2

(C) ${}^5/{}_2$ (D) ${}^3/{}_2$

Solution: S = n/2, where n = number of unpaired electrons = 3. Hence (D) is correct.

Illustration 6. The electronic configuration of an element is 1s²2s²2p⁶3s²3p⁶3d⁵4s¹. This represents its

(A) excited state (B) ground state

(C) cationic form (D) anionic form

Solution: It is ground state electronic configuration of Cr. Hence (B) is correct.

Illustration 7. Write the electronic configuration of the following ions:

(a) H⁺ (b) Na⁺ (c)O^2- (d) F^-

Solution: (a) 1s⁰ (b) 1s²2s²2p⁶ (c) 1s²2s²2p⁶ (d) 1s²2s²2p⁶