(i) no. of electrons: 35 (given)
no. of protons : 35
(ii) Electronic configuration
1s² 2S² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵
(iii) Last electron:

4P_x        4P_y     4p_z
last electron present in 4P_y orbital y
n = 4, l = 1 m₁ = either + 1 or -1 and s = -1/2

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.

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.

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

Electronic configuration of oxygen

ஃ 8^th electron present in 2p_x orbital and the quantum numbers are
n = 2,l = 1,m₁ = either + 1 or -1 and s = -1/2
Electronic configuration of chlorine 

15^th electron present in 3P_z orbital and the quantum numbers are n = 3, l = 1, m₁ = either +1 or -1 and ms = +1/2

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