Rain water normally has a pH of 5.6 due to dissolution of atmospheric CO₂ into it. Oxides of sulphur and nitrogen in the atmosphere may be absorbed by droplets of frater that make up clouds and get chemically converted into sulphuric acid and nitric acid respectively as a results of pH of rain water drops below the level 5.6, hence it is called acid rain. Acid rain is a by-product of a variety of sulphur and nitrogen oxides in the atmosphere. Burning of fossil fuels (coal and oil) in power stations, furnaces and petrol, diesel in motor engines produce sulphur dioxide and nitrogen oxides The main contributors of acid rain are SO₂ and NO₂.They are converted into sulphuric acid and nitric acid respectively by the reaction with oxygen and water.
2SO₂ + O₂ + 2H₂O ⟶ 2H₂SO₄
4NO₂ + O₂ + 2H₂O ⟶ 4HNO₃
Harmful effects of acid rain
Some harmful effects are discussed below :
(i) Acid rain causes extensive damage to buildings and structural materials of marbles. This attack on marble is termed as Stone leprosy.
CaCO₃ + H₂SO₄ ⟶ CaSO₄ + H₂O +CO₂ ↑
(ii) Acid rain affects plants and animal life in aquatic ecosystem.
(iii) It is harmful for agriculture, trees and plants as it dissolves and removes the nutrients needed for their growth.
(iv) It corrodes water pipes resulting in the leaching of heavy metals such as iron, lead and copper into the drinking water which have toxic effects.
(v) It causes respiratory ailment in humans and animals.

compound	Structural formula	Name
A	CH2 = CH2	Ethene
B		1,2 - dichloroethane
C	HCHO	Methanal
D	\(\mathrm{CH} \equiv \mathrm{CH}\)	Ethyne

(k_H)_bonzene = 4.2 x 10^–5 mm Hg
Solubility of methane = ?
P = 750mm Hg           
P = 840 mm Hg
According to Henrys Law,
P = K_H . x_{in solution}.
750 mm Hg = 4.2 x 10^–5 mm Hg . x_{in solution}
\(\Rightarrow X_{insolution}={750\over 4.2\times 10^{-5}}\)
i.e, solubility = 178.5 x 10⁵
similarly at P = 840 mm Hg
solubility = \({840\over 4.2\times 10^{-5}}\)
= 200 x 10^-5.

ΔH_f(Fe₂O₃)= -741 kJ mol^-1
ΔH_f(CO)= -137 kJ mol^-1
ΔH_f(CO₂)= -394.5 kJ mol^-1
Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂ ΔH_r=?
ΔH_r=Σ(ΔH_f)products - Σ(ΔH_f)reactants
ΔH_r=[2ΔH_f=(Fe)+3ΔH_f(CO₂)]-[ΔH_f(Fe₂O₃)+3ΔH_f(CO)]
ΔH_r=[0 + 3 (-394.5)] - [-741 +3 (-137)]
ΔH_r=[-1183.5] - [-1152]
ΔH_r=-1183.5 + 1152
ΔH_r=-31.5 kJ mol^-1

Potential difference = 100V
= 100 x 1.6 x 10-¹⁹J
\(\lambda=\frac{h}{\sqrt{2mev}}\)
\(=\frac { 6.626\times { 10 }^{ -34 }Kg{ m }^{ 2 }{ s }^{ -1 } }{ \sqrt { 2\times 9.1\times { 10 }^{ -31 }Kg\times 100\times 1.6\times { 10 }^{ -19 }J } } \)
\(\lambda=1.22\times10^{-10}m\)

Element	%	Relative number of atoms	Simple Ratio
Na	14.31	\(\frac { 14.31 }{ 23 } =0.62\)	\(\frac { 0.62 }{ 0.31 } =2\)
S	9.97	\(\frac { 9.97 }{ 32 } =0.31\)	\(\frac { 0.31 }{ 0.31 } =1\)
H	6.22	\(\frac { 6.22 }{ 1 } =6.22\)	\(\frac { 6.22 }{ 0.31 } =20\)
O	69.5	\(\frac { 69.5 }{ 16 } =4.34\)	\(\frac { 4.34 }{ 0.31 } =14\)

Empirical formula = Na₂ SH₂₀ O₁₄ 
\(\left[ \begin{matrix} { Na }_{ 2 }{ SH }_{ 20 }{ O }_{ 14 } \\ =(2\times 23)+(1\times 32)+(20\times 1)+14(16) \\ =46+32+20+234 \\ =322 \end{matrix} \right] \)
n = \(\frac { molar\quad mass }{ caluclated\quad empirical\quad formula\quad mass } =\frac { 322 }{ 322 } =1\)
Molecular formula = Na₂ SH₂₀O₁₄
Since all the hydrogen in the compound are present as water
\(\therefore \) The molecular formula is Na₂ SO₄ 10H₂O.

(i) Magnesium and calcium also play a vital role in biological functions. A typical adult human body contains about 25 g of magnesium and 1200 g of calcium. Magnesium plays an important· role in many biochemical reactions catalysed by enzymes. It is the co-factor of all enzymes that utilize ATP in phosphate transfer and energy release. It is also essential for DNA synthesis and is responsible for the stability and proper functioning of DNA.It is also used for balancing electrolytes in our body. Deficiency of magnesium results into convulsion and neuromuscular irritation.
(ii) Calcium is a major component of bones and teeth. It is also present in blood and its concentration is maintained by hormones (calcitonin and parathyroid hormone). Deficiency of calcium in blood causes it to take longer time to clot. It is also important for muscle contraction.
(iii) The main pigment that is responsible for photosynthesis, chlorophyll, contains magnesium which plays an important role in photosynthesis.

Screening effect: The repulsive force between the inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus, the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called shielding effect.
Pauling's scale: Pauling, he assigned arbitrary value of electronegativities for hydrogen and fluorine as 2.2 and 4.0 respectively. Based on this the electronegativity values for other elements can be calculated using the following expression.
\(({ X }_{ A }-{ X }_{ B })=0.182\sqrt { E_{ AB } } -({ E }_{ AA }*{ E }_{ BB })^{ 1/2 }\)
Where E_AB' EAA and E_BB are the bond dissociation energies of AB, A₂ and B₂ molecules respectively. The electronegativity of any given element is not a constant and its value depends on the element to which it is covalently bound. The electronegativity values play an important role in predicting the nature of the bond.

The group metal - 1 which is present in common salt is sodium.
So (A) is sodium
Sodium reacts with hydrogen (B) to give, sodium hydride (C). In sodium hydride the hydrogen is present in -1 oxidation state.
\(2\underset { (A) }{ Na } +\underset { (B) }{ { H }_{ 2 } } \rightarrow 2\underset { (C) }{ NaH } \)
So (B) hydrogen and (C) is sodium hydride
H₂ reacts with oxygen gas (D) to give an universal solvent, water (E) follows:
\(2\underset { (B) }{ { H }_{ 2 } } +{ O }_{ 2 }\rightarrow 2\underset { (D) }{ { H }_{ 2 }O } \)
So (E) is water. Water is the universal solvent
Water (E)  reacts with sodium  (A) follow to give (F), which is a strong base.
\(2\underset { (E) }{ { H }_{ 2 }O } +\underset { (A) }{ 2Na } \rightarrow 2\underset { (F) }{ NaOH } +{ { H }_{ 2 } } \)
So (F) is sodium hydroxide.

Element / Compound	Symbol / Formula	Name
A	Na	Sodium
B	H2	Hydrogen
C	NaH	Sodium hydride
D	H2O	Water
E	NaOH	Sodium hydroxide

The van der equation for a real gas is
\(\left( P+{{{am}^{2}}\over{{V}^{2}}} \right)(V-nb)=nRT\)
Pressure, Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.
Van der Waals found out the forces of attraction experienced by a molecule near the wall are directly proportional to the square of the density of the gas.
\(P^{\prime} \propto \rho^{2} ; \quad \rho=\frac{n}{v}\)
where n is the number of moles of gas and
V is the volume of the container
\( \Rightarrow p^{\prime} \alpha \frac{n^{2}}{V^{2}} \)
\(\Rightarrow p^{\prime}=a \frac{n^{2}}{V^{2}}\)
where a is proportionality constant and depends on the nature of gas
Therefore \(P_{\text {ideal }}=P+\frac{\operatorname{an}^{2}}{V^{2}}\)

Volume Correction
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container,
V. Van der Waals introduced a correction factor V' to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules
\(=\frac{4}{3} \pi(2 r)^{3}=8\left(\frac{4}{3} \pi r^{3}\right)=8 V_{m}\)
Where V_m it a volume of a single molecule
Excluded volume for single molecule = \(\frac{8 \mathrm{~V}_{\mathrm{m}}}{2}=4 \mathrm{~V}_{\mathrm{m}}\)
Excluded volume for n molecule = n(4V_m) = nb
Where b is van der waals constant which is equal to 4V_m
\( \Rightarrow V^{\prime}=n b \)
\(V_{\text {ideal }}=V-n b\)
Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the Van der Waals equation of state for real gases as below,
\(\left(p+\frac{a^{2}}{V^{2}}\right)(V-n b)=n R T\)
The constants a and b are van der Waals constants and their values vary with the nature of the gas. It is an approximate formula for the non-ideal gas.

CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g)
K_C = 4 x 10^–2 mol lit^–2
Volume = 500 ml = 1/2 L
\(\left[\mathrm{CH}_{4}\right]_{\text {in }}=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}} \)
= 2 mol L^-1
\(\left[\mathrm{CS}_{2}\right]_{\text {in }}=\frac{1 \mathrm{~mol}}{1 / 2 \mathrm{~L}}\)
= 2 mol L^-1
\( {\left[\mathrm{H}_{2} \mathrm{~S}\right]_{\text {in }}=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1} \quad\left[\mathrm{H}_{2}\right]=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1}} \)
\(\mathrm{Q}=\frac{\left[\mathrm{CS}_{2}\right]\left[\mathrm{H}_{2}\right]^{4}}{\left[\mathrm{CH}_{4}\right]\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}}=\frac{2 \times(4)^{4}}{(2) \times(4)^{2}}=16 \)
Q > K_c
\(\therefore \) The reaction will proceed in the reverse direction to reach the equilibrium.