Samacheer Class 11 Chemistry - Quantum Mechanical Model of Atom

(b) For n – 1, the electron has a more negative energy then it does for n = 6 which means that the eiectron is more loosely bound in the smallest allowed orbit.

When n = 0, l = 0, 1,2 and 3. Hence, there are four subshells namely, s, p, d and f
l = 0, m 1 = 0; one 4s orbital, l = 1, m = -1, 0, +1; three 4p orbitals,
l = 2, m 1 = – 2, -1, 0, +1, +2; five 4d orbitals and
l = 3, m 1 = – 3, -2, -1, 0, 1, 2, 3; seven 4f orbitals. Hence, the number of possible orbitals when n = 4 are sixteen.

The number of radial nodes is equal to (n – l – 1) and angular nodes is l.
Orbital
N
l
Radial node (n – l – 1)
Angular node, l
2s
2
0
1
0
4p
4
1
2
1
5d
5
2
2
2
4f
4
3
0
3
The number of radial nodes for 2s, 4p, 5d, and 4f orbitals are respectively 1,2,2 and 0 and the number of angular nodes for 2s, 4p, 5d, and 4f orbitals respectively are 0, 1, 2, and 3.

The exactly half-filled orbitals have greater stability. The reason for their stability are –
* symmetry
* exchange energy.
(1) Symmetry: The half-filled orbitals are more symmetrical than partially filled orbitals and this symmetry leads to greater stability.
(2) Exchange energy: The electrons with the same spin in the different orbitals of the same sub-shell can exchange their position. Each such exchange releases energy and this is known as exchange energy. Greater the number of exchanges, the greater the exchange energy, and hence greater the stability. In d-orbital, 10 exchanges are possible but in p-orbital 6 exchanges are possible. So, d – orbital with 5 unpaired electrons (10 exchanges)n i.e. half-filled is more stable than p – orbital with 3 unpaired electrons (6 exchanges).

Pauli’s exclusion principle states that “No two electrons in an atom can have the same set of values of all four quantum numbers”.
Illustration: H(Z = 1) 1s 1.
One electron is present in hydrogen atom, the four quantum numbers are n = 1, l = 0, m = 0 and s = + \(\frac {1}{2}\). For helium Z = 2. He: 1s 2. In this one electron has the quantum number same as that of hydrogen, n = 1,l = 0, m = 0 and s = +½ For other electron, fourth quantum number is different, i.e. n = 1, l = 0, m = 0 and s = – ½.

Orbital is a three-dimensional space in which the probability of finding the electron is maximum. The values of ‘n’ and ‘l’ for 3px orbital are n = 3 and l = 1, 4d x 2 – y 2 orbital are n = 4 and l = 2.

Erwin Schrodinger expressed the wave nature of electrons in terms of a differential equation. This equation determines the change of wave function in space depending on the field of force in which the electron moves. The time-independent Schrodinger equation can be expressed as
Hψ = Eψ, where H is called Hamiltonian operator, ψ is the wave function and is a function of position coordinates of the particle and is denoted as ψ(x, y, z), E is the energy of the system.
The above Schrodinger wave equation does not contain time as a variable and is referred to as time-independent Schrodinger wave equation. This equation can be solved only for certain values of E, the total energy, i.e., the energy of the system is quantized. The permitted total energy values are called eigenvalues and corresponding wave functions represent the atomic orbitals.

Heisenberg’s Uncertainty Principle is ∆x. ∆v > h/4πm.
Given:
∆v = 0.1%,
υ = 22 × 10 6 ms -1.
h = 6.626 × 10 -34 kgm 2 s -1.
m = 9.1 X 10 21 kg.
∆v = \(\frac{0.1 \times 2.2 \times 10^{6} \mathrm{~ms}^{-1}}{100}\)
= 2.2 × 10 3 ms -1
Uncertainty in position,
∆x ≥ \(\frac{h}{4 \pi m}\)
∆x ≥ \(\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} \mathrm{~s}^{-1}}{9.1 \times 10^{-31} \mathrm{~kg}. \times 2.2 \times 10^{3} \mathrm{~ms}^{-1}}\)
∆x ≥ 2.64 × 10 -8 m.

(1) O (Z = 8) 1s 2 2s 2 2p x 2 2p y 1 2p z 1
Four quantum numbers for 2p x 1 electron in oxygen atom:
n = principal quantum number = 2
l = azimuthal quantum number =1
m = magnetic quantum number =+1
s = spin quantum number = +\(\frac {1}{2}\)
(2) Cl (Z = 17) 1s 2 2s 2 2p 6 3s 2 3p x 2 3p y 2 3p z 1
Four quantum numbers for 15 th electron in chlorine atom:
n = 3, l = 1, m = 0, s = + ½
(3) Cr (Z = 24) 1s 2 2s 2 2p 2 3s 2 3p 2 3d 2 4s 1
n = 3, l = 2, m = +2, s = + ½

Energy of the electron in the nth orbit is
E n = –\(\frac{-13.6}{n^{2}}\)eV/atom.
When n = 3,
E 3 = \(\frac{-13.6 \mathrm{eV} / \text { atom }=-1.51 \mathrm{eV} / \text { atom }}{9}\)
When n = 4,
E 4 = \(\frac{-13.6 \mathrm{eV} / \text { atom }=-0.85 \mathrm{eV} / \text { atom }}{16}\)
∆E = (E 4 – E 3 ) = (-0.85) – (-1.51) = 0.66 eV/atom
Wavelength corresponding to this transition,

de Broglie wavelength,
λ = \(\frac{h}{m v}\)
Given:
de Broglie wavelength, λ = 5400 Å and mass, m = 54 g.
Velocity of the tennis ball
v = \(\frac{h}{m \lambda}\)
v = \(\frac{6.626 \times 10^{-34} J s}{54 \times 10^{-3} k g \times 5400 \times 10^{-10} m}\)
v = 2.27 × 10 -26 ms -1

1. Mn (Z = 25)
Mn → Mn 2+ + 2e –
Mn 2+ electronic configuration is 1s 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5
2. Cr (Z = 24)
Cr → Cr 3+ + 3e –
Cr 3+ electronic configuration is Is 2 2s 2 2p 6 3s 2 3p 6 3d 3

Aufbau Principle states that “In the ground state of the atoms, the orbitals are filled in the order of their increasing energies”. That is the electrons first occupy the lowest energy orbital available to them. Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals. The order of filling of various orbitals as per the Aufbau principle is given in the figure. which is in accordance with the (n + l) rule.

(i) Atomic Number of the element, z = No. of protons or No. of electrons. = 35
Mass number of the element, A = No. of protons + No of neutrons = 35 + 45 = 80
Number of Protons = 80 – 45 = 35.
(ii) Electronic configuration of the element (z = 35)
1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5
(iii) Quantum number for the last electron (4p z ),
n = 4, l = 1, m = +1, or -1 s = +1/2

According to the de Broglie concept, the electron that revolves around the nucleus exhibits both particle and wave character. In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave, Otherwise, the electron wave is out of phase.
Circumference of the orbit = nλ
2πr = nλ
2πr = \(\frac{n h}{m v}\)
Rearranging,
mvr = \(\frac{n h}{m v}\)
Angular momentum = \(\frac{n h}{2 \pi}\)
The above equation was already predicted by Bohr.
Hence, de Broglie and Bohr’s concepts are in agreement with each other.

He + (g) → He 2+ (g) + e and
E n = – 13.6z 2 /n 2
E l = – \(\frac{13.6(2)^{2}}{(1)^{2}}\) = -56.4 eV
E ∞ = \(\frac{-13.6(2)^{2}}{(\infty)^{2}}\) = 0
Required energy for the given process is,
E ∞ – E l = 0 – (-56.4) = 56.4 eV.

Let the number of electrons in an ion = x
number of neutrons = n = x + \(\frac{11.1}{100}\) eV = 1.111 x
(As the number of neutrons is 11.1% more than the number of electrons)
In the neutral of the atom, a number of electrons.
e – = x – 1 (as the ion carries -1 charge)
Similarly number of protons = P = x – 1
Number of protons + number of neutrons = mass number = 37
(x – 1) + 1.111 x = 37.
2.111 x = 37 +1
2.111 x = 38
x = \(\frac{38}{2.111}\) = 18.009 = 18
∴ Number of protons = atomic number – 1 = 18-1 = 17
∴ The symbol of the ion = \(_{17}^{37} \mathrm{Cl}\).

Given:
velocity, v = 2.85 × 10 8 ms -1.
mass, m = 1.673 × 10 -27 kg
λ = \(\frac{h}{m v}\)
λ = \(\frac{6.626 \times 10^{-34} \mathrm{kgms}^{-1}}{1.673 \times 10-27 \mathrm{~kg} \times 2-85 \times 10^{8} \mathrm{~ms}^{-1}}\)
λ = 1.389 × 10 -8 Å

Given:
velocity, v = 140 km/hr. = \(\frac{140 \times 10^{3}}{60 \times 60 \mathrm{~ms}^{-1}}\)
mass, m = 160g = 160 × 10 -3 kg
λ = \(\frac{h}{m v}\)
λ =\(\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} \mathrm{~s}^{-\mathrm{le}}}{160 \times 10^{-3} \mathrm{~kg} \times 3.88 \mathrm{~ms}^{-1}}\)
λ = 1.605 × 10 -34 m.

Heisenberg Uncertainity Principle is ∆x. ∆p ≥ \(\frac{h}{4 \pi}\)
Given:
∆x = 0.6 Å = 0.6 × 10 -10 m
h = 6.626 × 10 -34 kgm 2 s -1
Uncertainity in momentum,
∆p ≥ \(\frac{h}{4 \pi \Delta x}\)
∆p ≥ \(\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} \mathrm{~s}^{-1}}{4 \times 3.14 \times 0.6 \times 10^{-10} \mathrm{~m}}\)
∆p ≥ 8.8 × 10 -25 kgms -1

Potential difference = 100 V = 100 × 106 × 10 -19 J
λ = \(\frac{h}{\sqrt{2} \text { mev }}\)
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Kgm}^{2} \mathrm{~s}}{\sqrt{2} \times 9.1 \times 10^{-31} \mathrm{~kg} \times 100 \times 1.6 \times 10^{-19} \mathrm{~J}}\)
λ = 1.22 × 10 -10