Electronic configuration of Fe³⁺ 1s²2s²2p⁶3s²3p⁶3d⁶4s²

Electronic configuration of mn²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
Five unpaired electrons
Electronic configuration of Ar is 1s² 2s² 2p⁶ 3s² 3p⁶ 
no unpaired electrons.

\(\triangle x=?\)
\(\triangle v=?\)
\(\triangle x.\triangle p\ge\frac{h}{4\pi}\)
\(\lambda (m\triangle v)\ge\frac{h}{4\pi}\)
\(\triangle v\ge\frac{h}{4\pi(m\lambda)}\)
\(\triangle \ge\frac{h}{4\pi\times m \times\frac{h}{mv}}\)   \(\left[ \because \lambda-={{h}\over{mv}} \right]\)
\(\triangle v \ge \frac{v}{4\pi}\)
therefore, minimum uncertainty in velocity \(=\frac{v}{4\pi}\)

\(\triangle\)x = 0.6\(\mathring{A}\) = 0.6 x 10^-10m
\(\triangle p\) = ?
\(\triangle x.\triangle p\ge\frac{h}{4\pi}\)
\(\triangle x.\triangle p\ge5.28\times10^{-35}kgm^{2}s^{-1}\)
\((0.6 \times10^{-10}) \triangle p \ge5.28\times10^{-35}\)
\(\Rightarrow \triangle \ge \frac{5.28\times10^{-35}kgm^{2}s^{-1}}{0.6510^{-1}m}\)
\(\triangle p \ge8.8\times10^{-25}kgms^{-1}\)

v = 2.85 x 10⁸ ms^-1
m_p = 1.673 x 10^-27Kg
\(\lambda=\frac{h}{mv}\)
\(=\frac{6.626\times10^{-34}kgm^{2}s^{-1}}{1.673\times10^{-27}kg\times\times2.85\times10^{8}ms^{-1}}\)
\(\lambda=1.389\times10^{-15}\Rightarrow\lambda=1.389\times10^{-15}A\) \([\because \mathring{A}={10}^{-10}m]\)

\({ r_{n} }=\frac{(0.529)n^2}{z}\mathring{A}\ \ { E_{n} }=\frac{-13.6(z)^2}{(n)^2}ev atom^{-1}\)
for Li²⁺ z = 3
Bohr radius for the third orbit (r₃)
= \(\frac { (0.529){ (3) }^{ 2 } }{ 3 } \)
= 0.529\(\times\)3
=1.587 \(\mathring{A}\)
Energy of an electron in the fourth orbit
\(({E}_{6})=\frac { -13.6{ (3) }^{ 2 } }{ { (4) }^{ 2 } } \)
=-7.65eV atom^-1

\({ He }^{ + }\longrightarrow { He }^{ 2+ }+{ e }^{ - }\)
\({ E }_{ n }=\frac{-13.6(2)^2}{n^2}\)
\({ E }_{ 1 }=\frac{-13.6(2)^2}{(1)^2}=-54.4\)
\({ E }_{ \infty }=\frac{-13.6(2)^2}{(\infty)^2}=0\)
\(\therefore\) Required Energy for the given process
= E_\(\infty\) - E_I = 0 - (- 54.4) = 56.4 ev.

\(\triangle x.\triangle p\ge \frac { h }{ 4\pi } \)
\(\triangle x.\triangle p\ge 5.28 \times{ 10 }^{ -35 }Kg{ m }^{ 2 }{ s }^{ -1 }\)
\(\triangle x.(m\triangle v)\ge 5.28 \times{ 10 }^{ -35 }Kg{ m }^{ 2 }{ s }^{ -1 }\)
Given \(\triangle\)v = 0.1%
v = 2.2 x 106 ms^-1
m = 9.1 x 10^-31Kg
\(\triangle\)v = \(\frac{0.1}{100}\times2.2\times{10}^{6}ms^{-1}\)
= \(2.2\times{10}^{6}ms^{-1}\)
\(\therefore \triangle x\ge \frac { { 5.28\times 10 }^{ -35 }{ Kgm }^{ 2 }{ s }^{ -1 } }{ 9.1\times { 10 }^{ -31 }Kg\times 2.2\times { 10 }^{ 3 }m{ s }^{ -1 } } \)
\(\\ \triangle x\ge 2.64\times { 10 }^{ -8 }m\)

i) ₂₅Mn - 1s², 2S^2, 2p⁶, 3s², 3p⁶, 4s², 3d⁵
₂₃Mn²⁺ - 1s², 2S^2, 2p⁶, 3s², 3p⁶, 3d⁵
ii) ₂₄Cr - 1s^2, 2S²,2p⁶, 3s², 3p⁶, 4s¹, 3d³
₂₁Cr³⁺ - 1s^2, 2S²,2p⁶, 3s², 3p⁶, 3d³

Erwin Schrodinger expressed the wave nature of electron 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,
\(\overset { \wedge }{ H } \psi =E\psi \)          .........(1)
Where \(\overset { \wedge }{ H } \) is called Hamiltonian operator, \(\psi \) is the wave function and is a function of position coordinates of the particle and is denoted as \(\psi \) (x, y, z) E is the energy of the system
\(\overset { \wedge }{ H } =\left[ \frac { { -h }^{ 2 } }{ 8{ \pi }^{ 2 } } \left( \frac { { \partial }^{ 2 }\psi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial y }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial z }^{ 2 } } \right) +V \right] \)
can be written as
\(\left[ \frac { { -h }^{ 2 } }{ 8{ \pi }^{ 2 }m } \left( \frac { { \partial }^{ 2 }\psi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial y }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial z }^{ 2 } } \right) +V\Psi \right] =E\Psi \)
Multiply by \(\frac { 8{ \pi }^{ 2 }m}{ { -h }^{ 2 } } \)and rearranging
\(\frac { { \partial }^{ 2 }\psi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial y }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial z }^{ 2 } } +\frac { 8{ \pi }^{ 2 }m }{ { -h }^{ 2 } } (E-V)\Psi =0\)          ........(2)
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 quantised. The permitted total energy values are called eigen values and corresponding wave functions represent the atomic orbitals.

(i) n= 2 and 1 =0. The orbital is 2s. Its shape is symmetrically spherical.
(ii) n = 2 and 1 = 1. The orbital is 2p. Its shape is dumbbell.

(i) \({ _{ 30 }^{ 64 }{ Zn } }^{ 2+ }\)
No. of protons = 30
No. of electrons = 30 - 2 = 28
No. of neutrons = 64 - 30 = 34
(ii) ³⁵Cl^-
Atomic number = Mass number - No. of neutrons
= 35 - 18 = 17
Atomic number = No. of protons = 17
(iii) Ar
Mass number = No. of protons + No. of neutrons
= 18 + 22 = 40
Atomic number = No. of protons = 18
No. of electrons =  No. of protons = 18.

m 160 g = 160 x 10^-3 kg
\(v=140 km hr^{-1}=\frac{104\times10^{3}}{60\times60}ms^{-1}\)
\(v=38.88ms^{-1}\)
\(\lambda=\frac{h}{mv}\)
\(=\frac{6.626\times^{-34}Kgm^{2}s^{-1}}{160\times10^{-3}Kg\times38.88ms^{-1}}\)
\(\lambda=1.065\times10^{-34}m\)

The word Aufbau in German means 'building up'. 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 which is in accordance with (n + l) rule.

The solution to Schrodinger equation gives the permitted total energy values called eigen values and the corresponding wave function represent atomic orbitals.

Statement : "No two electrons in an atom can have the same set of values of all four quantum numbers"
Explanation : It means that, each electron must have unique values for the four quantum numbers (n, l, m and s).
For the lone electron present in hydrogen atom, the four quantum numbers are: n = 1; l = 0; m = 0 and s = +1/2. For the two electrons present in helium, one electron has the quantum numbers same as the electron of hydrogen atom, n = 1.
l = 0, m = 0 and s = +1/2. For other electron, the fourth quantum number is different i.e., n = 1, l = 0, m = 0 and s = -1/2.
As we know that the spin quantum number can have only two values +1/2 and - 1/2, only two electrons can be accommodated in a given orbital in accordance with pauli exclusion principle.

n = 4 l = 0,1,2,3
4 sub shells s, p, d & f.
I = 0 m₁ = 0 + one 4s orbital.
I = 1 m₁ = -1, 0, + 1 \(\Rightarrow\) three 4p orbitals.
I = 2 m₁ = -2,:1, 0, +1, +2 \(\Rightarrow\) five 4d orbitals.
I = 3 m₁ = -3, -2, -1,0, +1, +2, +3 \(\Rightarrow\) seven 4f orbitals.
Over all 16 orbitals are possible.

Given \(\triangle\)v = 5.7 x 10⁵ ms^-1. \(\triangle\)x = ?
According to Heisenbergs uncertainty principle \(\Delta x \cdot \Delta p \geq \frac{\mathrm{h}}{4 \pi}\)
\( \frac{\mathrm{h}}{4 \pi}=\frac{6.626 \times 10^{-34}}{4 \times 3.14} \mathrm{kgm}^{2} \mathrm{~s}^{-1}=5.28 \times 10^{-35} \)
\(\Delta x \cdot \Delta \mathrm{p} \geq 5.28 \times 10^{-35} \)
\(\Delta x . \mathrm{m} \Delta \mathrm{v} \geq 5.28 \times 10^{-35} \)
\(\Rightarrow \Delta x \geq \frac{5.28 \times 10^{-35} \mathrm{kgm}^{2} \mathrm{~s}^{-1}}{9.1 \times 10^{-31} \mathrm{~kg} \times 5.7 \times 10^{5} \mathrm{~ms}^{-1}} \Rightarrow \Delta x \geq 1.017 \times 10^{-10} \mathrm{~m}\)

The reasons are:
(i) Symmetrical distribution: 3d⁵(half-filled) and 3d¹⁰ (completely filled) one more symmetrical and hence more stable.
(ii) Exchange energy: Electrons with parallel spins in degenerate orbitals tend to exchange their positions. As a result, energy is released. This energy is called exchange energy. Greater the exchange energy, greater is the stability.

\({ E }_{ n }=\frac { -13.6 }{ { n }_{ 2 } } ev{ \quad atom }^{ -1 }\)
n = 3 E₃ = \(\frac { -13.6 }{ { 3 }^{ 2 } } =\frac { -13.6 }{ 9 } \)
= -1.51 ev atom^-1
n = 4 E₄ =\(\frac { -13.6 }{ { 4 }^{ 2 } } =\frac { -13.6 }{ 16 } \)
= -0.85  ev atom^-1
\(\triangle \)E = (E₄-E₃) = (-0.85) - (-1.51) ev atom^-1
= (-0.85 + 1.51)
= 0.66eV atom^-1
(1eV = 1.6 x 10^-19J)
\(\triangle \)E = 0.66 x 1.6 x 10^-19J
\(\triangle \)E = 1.06 x 10^-19J
hv = 1.06 x 10^-19J
\(\frac { hv }{ \leftthreetimes } \) = 1.06 x 10^-19J
\(\therefore\)\( \leftthreetimes\) = \(\frac { hc }{ 1.06\times { 10 }^{ -19 }J } \)
= \(\frac { 6.626\times { 10 }^{ -34 }JS\times 3\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 1.06\times { 10 }^{ -19 }J } \)
​​​​​​​\(\lambda=1.875\times10^{-6}m\)

(i) ground state :

This type of electronic configuration have ten possible arrangement (i.e. Half filled configuration is More).

a) Principal quantum number defines energy and size of an orbital.
b) Azimuthal quantum number defines shape of an orbital
c) Magnetic quantum number defines spatial orientation (direction) of an orbital.