(i) Monovalent sodium and potassium ions are found in large proportions in biological fluids.
(ii) These ions perform important biological functions such as maintenance of ion balance and nerve impulse conduction.
(iii) Sodium - Potassium play an important role in transmitting nerve signals.
(iv) A typical 70 kg man has 90 g of Na and 170 g of K.
(v) Sodium ions are found on the outside of cells, being located in blood plasma and in the interstitial fluid which surrounds the cells. These ions .participate in the transmission of nerve signals, in regulating the flow of water across cell membranes and in the transport of sugars and amino acids into cells.
(vi) Potassium ions are the most abundant cations within cell fluids, where they activate many enzymes, participate in the oxidation of glucose to produce ATP and with sodium, are responsible for the transmission of nerve signals.

(i) Andrew plotted isotherms of carbon dioxide at different temperatures. It is then proved that many real gases behave In a similar manner like CO₂.
(ii) At a temperature of 303.98 K, CO₂ remains as a gas. Below this temperature, CO₂ turns into liquid CO₂ at 73 atm. It is called the critical temperature of CO_2.
(iii) At 303.98 K and 73 atm pressure, CO₂ becomes a liquid but remains a gas at higher temperature.
(iv) Below the critical temperature, the behaviour of CO₂ is different. For example, consider an isotherm of CO₂ at 294.5 K, it is a gas until the point B, is reached. At B, a liquid separates along the line BC, both the liquid and gas coexist. At C, the gas is completely condensed.
(v) lf the pressure is higher. than at C, only the liquid is compressed so, a steep rise in pressure is observed. Thus, there exist a continuity of state.
(vi) A gas below the critical temperature can be liquefied by applying pressures.

(i) For 1 mole of an ideal the plot becomes parallel to the pressure axis.
(ii) At low pressures all gases exhibit ideal behaviour.\(\frac{PV}{RT}\) values converges to 1 as P approaches to zero.
(iii) At moderate pressures, \(\frac{PV}{RT}\)becomes less than 1. This means that gases exhibit negative derivation from ideal behaviour. This is because attractive forces operate among molecules at relatively short distances.
(iv) At high pressures, \(\frac{PV}{RT}\) becomes greater than 1. This means that gases exhibit negative deviation from ideal behaviour. Intermolecular force becomes significant effect to affect the motion of the molecules.

Experiment	\(\frac{v_1}{T_1}\)=constant
1	\(\frac{1.54}{293}\) = 0.0053
2	\(\frac{1.65}{313}\) = 0.0053
3	\(\frac{1.95}{373}\) = 0.0053
4	\(\frac{2.07}{393}\) = 0.0053

The average value of the constant is 0.0053.

The distinctive behaviour of Li+ ion is due to its exceptionally small size, high polarising power, high hydration energy and non availability of d-orbitals.

Lithium	Other elements of the family
Hard, high melting and boiling point.	Soft and lower melting and boiling point.
Least reactive (For example it reacts with oxygen to form normal oxide, forms peroxides with great difficulty and its higher oxides are unstable).	More reactive.
Reacts with nitrogen to give Li3N (lithium nitride)	No reaction.
Reacts with bromine slowly.	React violently.
Reacts directly with carbon to form ionic carbides. eg: 2Li + 2C ⟶ Li2C2 (lithium carbide).	Do not react with carbon directly, but can react with carbon compounds. eg: Na + C2H2 ⟶ Na2C2
Compounds are sparingly soluble in water.	Compounds are highly soluble in water.
Lithium nitrate decomposes to give an oxide.	decompose to give nitrites.

(i) An element A belonging to II group and II period is beryllium (A)
(ii)  Beryllium reacts with chlorine to form beryllium chloride (B)
Be +Cl₂ \(\longrightarrow \) BeCl₂
                         (B)
(iii) Beryllium chloride on treatment with LialH Forms beryllium hydride (c)
2BeCl₂ + LiA1H₄ \(\longrightarrow \) 2BeH₂ + LiCl + AlCl₃ 
                                                             (C)

A	Be	Beryllium
B	Becl2	Beryllium Chloride
C	BeH2	Beryllium Hydride

Given data P₁ = 1 atm V₁ = 5.82 dm³
P₁ = 1.92 atm V₂ = ?
According to Boyles law,
P₁V₁ = P₂V₂
\({ V }_{ 2 }=\frac { { P }_{ 1 }{ V }_{ 1 } }{ { P }_{ 2 } } \)
\({ V }_{ 2 }=\frac { 1\ atm\times 5.82\ { dm }^{ 3 } }{ 1.92\ atm } \)
V₂ = 3.031 dm²

a = 3.67 atm lit² mol^-2
b = 0.0408 lit mol^-1
R = 0.0821 atm lit K^-1 mol^-1
(i) \({ T }_{ c }=\frac { 8a }{ 27Rb } \)
\(=\frac { 8\times 3.67 }{ 27\times 0.082\times 0.0408 } =324.7\quad K\)
T_c = 324.7 K
(ii) \({ P }_{ c }=\frac { a }{ 27{ b }^{ 2 } } \)
\(=\frac { 3.67 }{ 27\times { (0.0408) }^{ 2 } } =81.6\)
P_c = 81.6 atm.

Density (d) = 0.9 gm lit^-1
T = 350 K
Mass of neon (M) = 20 gm mol^-1
\(P=\frac { dRT }{ M } \quad \left( \because n=\frac { d }{ M } \right) \)
\(=\frac { 0.9\times 8.314\times 10^{ -2 }\times 350 }{ 20 } =1.309\quad bar\)
Pressure of neon = 1.309 bar

(i) The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT. This is termed as compression factor. Mathematically,
\(Z=\frac { PV }{ nRT } \)
For ideal gases Z=1 at all temperatures and pressures, because PV = nRT.
(ii) When a gas deviates from ideal behaviour, its Z value deviates from unity.
(iii) At high pressure these gases have Z > 1 and are difficult to compress. At intermediate pressures, Z< 1.
(iv) Above the Boyle point, Z > 1 for real gases, ie., the real gases show positive deviation.
(v) Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increase with the increase in pressure.
Hence, the compressibility factor Z can be rewritten as
\(Z=\frac { { PV }_{ real } }{ nRT } \)
\({ V }_{ ideal }=\frac { nRT }{ P } \)
Substituting b in a
\(Z=\frac { { V }_{ real } }{ { V }_{ ideal } } \)
(vi) Where V_real  is the molar volume of the real gas and V_ideal is the molar volume of it when it behaves ideally.

(i) Isotherm:
At any constant temperature when pressure is increased, volume is decreased and vice versa. Such P - V curves at constant temperature are known as isotherms.
(ii) Critical Temperature: (T_c)
It is defined as the characteristic temperature of a gas above which no liquefaction occurs.
(iii) Critical Pressure: (P_c)
It is defined as the minimum pressure required to liquefy 1 mole of a gas present at its critical temperature.
(iv) Critical volume: (V_c)
The volume occupied by 1 mole of a gas at its critical pressure and at critical temperature is the critical volume of the gas
(v) Compressibility factor:
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT. This is termed as compressibility factor.
\(Z=\frac { PV }{ nRT } \)

(i) Sodium hydroxide is prepared commercially by the electrolysis of brine solution in Castner-Kellner cell .
(ii) Cathode- mercury
anode - carbon 
(iii) Sodium metal is discharged at the cathode and combines with mercury to form sodium amalgam.
(iv) Chlorine gas is evolved at the anode.
(v) The sodium amalgam thus obtained is treated with water to give sodium hydroxide
(vi) At cathode: Na⁺ + e^- ⟶ Na(amalgam)
At anode : Cl^- ⟶ 1/2 Cl₂ ↑ + e^-
2Na(amalgam) + 2H₂O ⟶ 2NaOH + 2Hg + H₂↑

S.No	Properties
1	Both lithium and magnesium are harder than other elements in the respective groups
2	Lithium and magnesium react slowely with water. Their oxides and hydroxides are much less soluble and their hydroxides decompose on heating
3	Both form a nitride, Li3N and mg3N2 , by direct combination with nitrogen
4	They do not give any superoxides and form only oxides, Li2O and MgO
5	The carbonates of lithium and magnesium decompose upon heating to form their respective oxides and CO2
6	Lithium and magnesium do not form bicarbonates
7	Both LiCl and MgCl2 are soluble and are deliquescent

(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.

1. This group contains Be, Mg, Ca, Sr, Ba & Ra.
2. Except Be all these elements are called as alkaline earth metals because their oxides and hydroxides are alkaline in nature.
3. Beryllium is the rare element and Radium is the rarest (10% rocks) ; Their Occurrence: Be- Beryl; Mg - carnallite; Dolomite Ca - Fluorapatite; Sr - Celestite Ba - Barytes.
4. Radium is radioactive.
5. The general electronic configuration is :
[Noble gas] ns² eg. Be - [He] 2s²
6. On moving down the group the radii increase. Their atomic radii are smaller than alkali metals.
7. They exhibit +2 oxidation state.
8. The ionisation enthalpies are less than p - block elements due to large size. Down the group the ionisation enthalpy decreases.
9. The IE₁ of group 2 elements are greater than group 1 elements.
10. The IE₂ values are higher than that of alkali metals.
11. They are less electro Positive elements than alkali metals.
12. The hydration enthalpy decreases with increase in ionic radii.
13. MgCl₂ form MgCI₂ .6H₂O and CaCl₂ form CaCl₂.6H₂O
14. The electronegativity value decreases down the group.
15. With concentrated HCl They impart flame colour. Ca - Brick Red ; Sr - Crimson red and Barium - Apple Green.
16. All form metallic halides at elevated temperatures. M + X₂ + MX₂
17. All elements except Beryllium combine with hydrogen to form hydrides of formula MH₂.

(i) Alkaline earth metal (A) belonging to 3rd period is magnesium.
(ii) So A is Magnesium. Magnesium reacts with oxygen and nitrogen as follows.
\(2Mg+O_2⟶\underset{(B)}{2MgO}\)
\(3Mg+N_2⟶\underset{(C)}{Mg_3N_2}\)
So B is Magnesium oxide and C is magnesium nitride.
(iii) Magnesium undergoes metal displacement reaction with AgNO₃ as follow to give D as follows :
\(Mg+2AgNO_3⟶\underset{D}{Mg(NO_3)_2}+2Ag\)
So D is Magnesium nitrate.
Result  :

Compound or Element	Symbol or Formula	Name
A	Mg	Magnesium
B	MgO	Magnesium oxide
C	Mg3N2	Magnesium nitride
D	Mg(NO3)2	Magnesium nitrate

S.No	Properties
1	Beryllium chloride forms a dimeric structure like aluminium chloride with chloride bridges. Beryllium chloride also forms polymeric chain structure In addition to dimer. Both are soluble in organic solvents and are strong lewis acids.
2	Beryllium hydroxide dissolves in excess of alkali and gives beryllate ion and [Be(OH)2]2- and hydrogen as aluminium hydroxide which gives aluminate ion, [Al(OH)4]-
3	Beryllium and aluminum ions have strong tendency to form complexes, \(BeF_4^{2-} AIFt{_6^{3-}}\)
4	Both beryllium and aluminium hydroxides are amphoteric in nature.
5	Carbides of beryllium (Be2C) like aluminum carbide (Al4C3) give methane on hydrolysis
6	Both beryllium and aluminium are rendered passive by nitric acid.

The van der Waals equation for n moles is
\(\left( P+\frac { { an }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-nb \right) =nRT\) ....(1)
For 1 mole
\(\left( P+\frac { { a }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-b \right) =RT\) ...(2)
From the equation we can derive the values of critical constants P_c, V_c and T_c, in terms of a and b, the van der Waals constants, On expanding the above equation
\(pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT=0\)    ...(3)
Multiply equation (3) by V² / P
\(\frac { { v }^{ 2 } }{ p } \left( Pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT \right) =0\)
\({ v }^{ 3 }+\frac { av }{ p } +-{ bv }^{ 2 }-\frac { ab }{ { v }^{ 2 } } -\frac { { RTV }^{ 2 } }{ p } \)    ...(4)
When the above equation is rearranged in powers of Y.
\({ v }^{ 3 }-\left[ \frac { RT }{ P } +b \right] { v }^{ 2 }+\left[ \frac { a }{ p } \right] v-\left[ \frac { ab }{ p } \right] =0\)        ...(5)
The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume V_C. The pressure and temperature becomes Pc and T_c respectively
i.e., V = V_c
V - V_c = 0
(V - V_C)³ = 0
V3 - 3V_CV² + 3V_c²V - V_c³ = 0 .....(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V², V and constant terms in (5) and (6).
\(-3{ v }_{ c }{ v }^{ 2 }=-\left[ \frac { { RT }_{ c } }{ { p }_{ c } } +b \right] { v }^{ 2 }\)
\(3{ v }_{ c }=\frac { { RT }_{ c } }{ { p }_{ c } } +b\)           .....(7)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)           ......(8)
\(3{ v }_{ c }^{ 2 }=\frac { ab }{ { p }_{ c } } \)           ...(9)
Divide equation (9) by equation (8)
\(\frac { { v }_{ c }^{ 3 } }{ 3{ v }_{ c }^{ 2 } } =\frac { ab/{ p }_{ c } }{ a/{ p }_{ c } } \)
\(\frac { { V }_{ c } }{ 3 } =b\)
i.e. V_c = 3b          .....(10)
when equation (10) is substituted in (8)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)
\({ p }_{ c }=\frac { a }{ 3{ v }_{ c }^{ 2 } } =\frac { a }{ 3\left( { 3b }^{ 2 } \right) } =\frac { a }{ { 3\times 9b }^{ 2 } } =\frac { a }{ { 27 }b^{ 2 } } \)
\({ p }_{ c }=\frac { a }{ { 27 }b^{ 2 } } \)            ...(11)
substituting the values of V_c and P_c in equation (7),
\(3vc=b+\frac { { RT }_{ c } }{ p } \)
\(3\left( 3b \right) =b+\frac { { RT }_{ c } }{ \left( \frac { a }{ { 27b }^{ 2 } } \right) } \)
\(9b-b=\left( \frac { { RT }_{ c } }{ a } \right) 27{ b }^{ 2 }\)
\(8b=\frac { { t }_{ c }R{ 27b }^{ 2 } }{ a } \)
\(\therefore { T }_{ c }=\frac { 8ab }{ 27R{ b }^{ 2 } } =\frac { 8a }{ 27Rb } \)
\({ T }_{ c }=\frac { { 8 }_{ a } }{ 27Rb } \)           ......(12)
The critical constants can be calculated using the values of van der Waals constant of a gas and vice versa.
\(a=3{ V }_{ C }^{ 2 }{ P }_{ C }\quad and\quad b=\frac { { V }_{ C } }{ 3 } \)

n_A = 1.5 mol n_B = ?
V_A= 37.6 dm³ V_B = 16.5 dm³
(T = 298 K constant)
\(\frac { { V }_{ A } }{ { n }_{ A } } =\frac { { V }_{ B } }{ { n }_{ B } } \)
\({ n }_{ A }=\left( \frac { { n }_{ A } }{ { V }_{ A } } \right) { V }_{ B }\)

= 0.66 mol.

Pressure of the gas in the tank at its melting point
T₁ = 298 K; P₁ = 2.98 atm; T₂ = 1100 K; P₂ = ?
\(\frac { { P }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 } }{ { T }_{ 2 } } \)
\(\Rightarrow \quad { P }_{ 2 }=\frac { { P }_{ 1 } }{ { T }_{ 1 } } \times { T }_{ 2 }\)
\({ P }_{ 2 }=\frac { 2.98\quad atm }{ 298\ K } \times 1100\ K=11\ atm\)
At 1100 K the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.