Shape of s-orbital - sphere
Shape of p-orbital- dumb bell
Shape of d-orbital- clover leaf

n = 3, main shell is m.
Total number of orbitals in 3rd energy level =?

Total number of orbitals = 9.

(i) Albert Einstein proposed that light has dual nature. i.e. like photons behave both like a particle and as a wave.
(ii) Louis de Broglie extended this concept and proposed that all forms of matter showed dual character.
(iii) He combined the following two equations of energy of which one represents wave character (hv) and the other represents the particle nature (mc²).

Cr (Z = 24) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹.
The reason for this is, Cr with 3d⁵ configuration is half filled and it will be more stable.
Chromium has [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s² due to the symmetrical distribution and exchange energies of d electrons.

KE = \(\frac{1}{2}\)mv² = 5.85 x 10^-25 J
By de Broglie equation \(\lambda=\frac{h}{mv}\)
But \(\lambda=\frac{\nu}{V}=\frac{h}{mv}\)
v = \(\frac{mv^2}{h}=\frac{2\times 5.85\times 10^{-25}J}{6.026\times 10^{-34}J}\)
= 1.77 x 10⁹ s^-1.

Angular momentum of electron in any orbital = \(\sqrt{l(l+1)}\times\frac{h}{2\pi}\)
(i) for 2s orbital, l = 0
Therefore angular momentum = \(\sqrt{0(0+1)}\times\frac{h}{2\pi}=0\)
(ii) for 4f orbital, I = 3
Therefore angular momentum = \(\sqrt{3(3+1)}\times\frac{h}{2\pi}\)
= \(2\sqrt{3}\frac{h}{2\pi}\)
\(=\sqrt{3}\frac{h}{\pi}\)

(i) In AI, 2p should be filled before filling 3s, orbital states.
\(\therefore\) correct electronic configuration is 1s² 2s² 2p⁶ 3s² 2p¹
(ii) In B, total electrons = 5. Electronic configuration is 1s² 2s² 2p¹
(iii) In F, total electrons = 9. Electronic configuration is 1s² 2s² 2p⁵.

Similarities:
(i) Both have similar shape.
(ii) Both have same angular momentum = \(\sqrt{l(l+1)} \frac{h}{2\pi}\)
Dissimilarities:
(i) Is orbital has no node while 2s orbital has one node.
(ii) Energy of 2s orbital is greater than Is orbital.
(iii) The size of the 2s orbital is larger than Is orbital.

(i) Mass number is a whole number because it is the sum of number of protons and number of neutrons.
(ii) Atomic mass is fractional because it is the average relative mass of its atom as compared with mass of an atom of C-12 isotope taken as 12.

The charge of an electron is 1.602 x 10^-19 coulomb
The mass of an electron is 9.11 x 10-³¹ kg

Energy electrons symmetry
This is due to the symmetrical distribution and exchange energy of given d- electrons. Symmetry leads to stability.
Exchange energy:
If two or more electrons with the same spin are present in degenerate orbitals, there is a possibility for exchanging their positions. During exchange process, the energy is released and the released energy is called exchange energy. If more number of exchanges are possible, more exchange energy in released. More number of exchanges are possible only in case of half filled and fully filled configurations.
For example, in chromium the electronic configuration is [Ar]3d⁵ 4s¹. The 3d orbital is half filled and there are ten possible exchanges as shown in figure. On the other hand only six exchanges are possible for [Ar]3d⁴ 4s² configuration. Hence, exchange energy for the half filled configuration is more. This increases the stability of half filled 3d orbitals.

The exchange energy is the basis for Hund's rule, which allows maximum multiplicity, that is electron pairing is possible only when all the degenerate orbitals contain one electron each.

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

De Broglie wavelength of the tennis ball equal to 5400 \(\overset { 0 }{ A } \).
m = 54 g 
V = ?
\(\lambda=\frac{h}{mV}\)
\(V=\frac{h}{m\lambda}\)
\(\mathrm{v}=\frac{6.626 \times 10^{-34} \mathrm{JS}}{54 \times 10^{-3} \mathrm{~kg} \times 5400 \times 10^{-10} \mathrm{~m}}=2.27 \times 10^{-26} \mathrm{~ms}^{-1}\)