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.

The balanced equation for the formation of HCI is,
H₂(g) + CI₂(g) \(\rightarrow\) 2 HCI (g)
As per the stoichiometric equation, under given conditions,
To produce 2 moles of HCI, 1 mole of chlorine gas is required.
To produce 44.8 litres of HCI, 22.4 litres of chlorine gas are required.
\(\therefore\) To produce 11.2 litres of HCI,

= 5.6 litres of chlorine are required.

Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction
A B \(\rightarrow\) A + B
Example: 2KClO₃ \(\rightarrow\) 2KCl + O₂
PCl₅ \(\rightarrow\) PCl₃ +Cl₂

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.

Molecular mass of H₂0 = 18
gram molecular mass of H₂0 = 18g
1 mol of H₂0 = 18g of H₂0 = 6.023 x 10²³molecules of water
1.8g of H₂0 = \({6.023\times10^{23}\over18}\times1.8 \)
\(=6.023\times10^{22} molecules\)

(i) An element consists of only one type of atom ..An atom is the smallest electrically neutral - particle, being made up of fundamental particles, namely electrons, protons and neutrons.
(ii) Compounds are made up of molecules which contain two or more atoms of different elements. Properties of compounds are different.from those of their constituent elements. The constituents of a compound is present in a fixed ratio by weight.

Molar mass of formic acid = 46 g
Molar volume of 46 g (1 mole) of formic acid = 2.24 \(\times\) 10^-2 m3
Molar volume of 460 g of (10 moles) of formic acid = \(\frac { 2.24\times { 10 }^{ -2 }\times 460 }{ 46 } \)
= 2.24 \(\times\) 10^-2 m³

Phosphorous acid (H₃ P0₃)
equivalent mass of phosphorous acid
= \(\frac{Molar \ mass}{basicity} =\frac{82}{2}=41\)
∴ equivalent mass of H₃P0₃ = 41 

1 x AW of Na = 1 x 22.99 = 22.99 amu
1 x AW of O = 1x 16 = 16.00 amu
1 x AW of H = 1 x1.008 = 1.008 amu
Formula weight of NaOH is = 39.998 amu

6 x AW of C = 6x12.01 = 72.06 amu
12 x AW of H = 12x1.008 = 12.096 amu
6 x AW of O = 6 x16 = 96.0 amu
Formula weight of Glucose is = 180.156 am

Molar mass of NH₃ = (1 x 14 + 3 x 1) = 17g mol^-1
Mass of 0.2 mol of NH₃ = 0.2 mol x 17g mol^-1
= 3.4 g

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.

Electronic configuration of Fe³⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁵
Electronic configuration of Ni²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁸
Fe³⁺ has stable 3d⁵ half filled configuration.

\({ E }_{ n }=\frac { -13.6 }{ { n }_{ 2 } } ev{ \quad atom }^{ -1 }\)
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
\(\therefore\)\( \leftthreetimes\) = \(\frac { hc }{ 1.06\times { 10 }^{ -19 }J } \)
= \(\frac { 6.626\times { 10 }^{ -34 }JS\times 3\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 1.06\times { 10 }^{ -19 }J } \)
​​​​​​​\(\lambda=1.875\times10^{-6}m\)