We know \(\triangle_r{G}^{\ominus}\) = -2.303 RO log K_p
and R = 8.314 JK^-1 mol^-1
Therefore, \(\triangle_r{G}^{\ominus}\) = -2.303 (8.314 J K^-1 mol^-1) x (log 2.47 x 10^-29)
= 163000 J mol^-1 = 163 mol^-1

According to Hess law,  \(\Delta\)H₁ = \(\Delta\)H₂ + _\(\Delta\)H₃
-393.5 kJ = X - 283.5 kJ
X = -110.5 kJ

(i) At room temperature, HCI is a colourless gas and the solution of HCI in water is called hydrochloric acid and it is in liquid state.
(ii) Sodium hydride NaH is an ionic compound and it is made of sodium cations (Na⁺) and hydride (H^-) anions. It has the octahedral crystal structure. It is an alkali metal hydride.

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.

The balanced chemical equation is,
\({ CaCO }_{ 3 }\left( s \right) \overset { \triangle }{ \longrightarrow } { CaO }_{ (s) }+{ CO }_{ 2 }(g)\)
As per the stoichiometric equation,
1 mole (100g) CaC0₃ on heating produces 1 mole CO₂

At STP, 1 mole of CO₂ occupies a volume of 22.7 liters
At STP, 50 g of CaCO₃ on heating produces,

= 11.35 liters of CO₂

The balanced stoichiometric equation for the formation of ammonia is
N₂(g) + 3H₂ (g) \(\rightarrow\) 2NH₃ (g)
As per the stoichiometric equation,
To produce 2 moles of ammonia, 3 moles of hydrogen are required.
\(\therefore\)  to produce 10 moles of ammonia,

= 15 moles of hydrogen are required.

H₂O \(\rightleftharpoons\) H₂O₍₁₎ \(\triangle_{fus}\) H is 6 kJ mol-1
(a) ΔG=ΔH-TΔS
When the reaction is carried out at 0°K
or ΔS=0
ΔG=ΔH
(b) \(H_2O_{s} \rightleftharpoons H_2O{_{(I)}}\)
\(\Delta_{fus}H= 6\) kJ mol^-1
= 6000 kJ mol^-1
\(\Delta_{fus}H= 6\) kJ mol^-1
= 6000 J mol^-1

The equation representing the enthalpy of formation of methane is:
\(C_{(s)}+2H_{2_{(g)}} \rightarrow CH_{4_{(g)}}\)  ΔH = -78.84 kJ
ΔH = 78.84 kJ; Δ^ng = 1 - 2 = -1 mol
R= \(8.314 \times 10 ^{-3}\)kJ-1 K^-1 mol^-1;  T=300 K
According to the relation,
ΔH = ΔU + Δ^ng RT
ΔU = ΔH + Δ^ng RT
=(-78.84 kJ) - (1 mol) x (\(8.314 \times 10 ^{-3}\)kJ K^-1 mol^-1) х 300 K
= -78.84-2.49 = -81.35 kJ.

The complete ethylene combustion reaction can be written as,
\(C_2H_{4_(g)}+30_{2_(g)} \rightarrow 2CO_{2(g)}+2H_2O_{(I)}\)
ΔU= -1406kJ
Δn=\(n_{p(g)}-n_{r(g)}\)
Δn= 2-4= -2
ΔH=ΔU+RTΔn_g
ΔH=-1406+(8.314x10^-3x300x(-2))
ΔH= -1410.9 kJ

We know  ΔU=n C_V(T₂-T₁)
ΔH=n C_P(T₂-T₁)
Here \(n=\frac{128}{32}=4\)moles;
T₂ = 100°C = 373 K; T₁ = 0°C = 273 K
ΔU = n C_V (T₂ - T₁)
ΔU = 4 x 21 x (373-273)
ΔU = 8400 J
ΔU = 8.4 kJ
ΔH =n C_p(T₂-T₁)
ΔH = 4 x 29 x (373 - 273)
ΔH = 11600 J
ΔH = 11.6 kJ

The standard enthalpy change for the combustion of ethanol can be calculated from the standard enthalpies of formation of \(C_2 H_5 OH_{I}, \)\(CO_{2{g}}\) and \(H_2 O(_{I})\). The enthalpies of formation are -277, - 393.5 and -285.5 kJ mol^-1 respectively.
\(C_2O_5 OH_{(I)}+3O_{2{g}} \rightarrow 2CO_{2{g}}+3 H_2 O_{(I)}\)
\(\Delta H^{0}_{1} = [(\Delta H^{0}_{f})_{products}-(\Delta H^{0}_{f})_{reactants})]\)
\(\Delta H^{0}_{r}=[2(\Delta H^{0}_{f})_{CO_{2}}+3(\Delta H^{0}_{f})_{H_2O}]-[1(\Delta H^{0}_{f})_{C_2H_5OH}+3(\Delta H^{0}_{f})_{O_2}]\)
\(\triangle {H}_{r}^{0}=\begin{bmatrix} 2\ mol(-393.5)\ kJ\ {mol}^{-1} \\ +3\ mol(-285.5)\ kJ\ {mol}^{-1} \end{bmatrix}-\begin{bmatrix} 1\ mol(-227)\ kJ\ {mol}^{-1} \\ +3\ mool(0)\ kJ\ {mol}-1^{} \end{bmatrix}\)
= [-787-856.5]-[-277]
=-1643.5+277
\(\Delta H^{0}_{r}\)= -1366.5 kJ

Given data q = 400 J; V₁= 5L; V₂ = 10L
Δu = q- w (heat is given to the system (+q); work is done by the system(-w)
Δu=q-PdV
=400 J - 1 atm (10 - 5)L
= 400 J - 5 atm L
[∴ 1L atm=101.33 J]
= 400 J - 5 x 10l.33 J
= 400 J - 506.65 J
= - 106.65 J

Given:
S⁰ (urea) = 173.8 J mol^-1 K^-1
S⁰ (H₂O) = 70 J mol^-1 K^-1
S⁰ (CO₂) = 213.5 J mol^-1K^-1
S⁰ (NH₃) = 192.5 J mol^-1K^-1
NH₂ - CO - NH₂ + H₂O ⟶ 2NH₃ + CO₂
ΔS_r⁰ = Σ(S⁰)_products - Σ(S⁰)_reactants
ΔS_r⁰ = [2 S⁰(NH₃) + S⁰(CO₂)] - [S⁰(urea) + S⁰(H₂O)]
ΔS_r⁰ = [2 x 192.5 + 213.5]-[173.8+70]
ΔS_r⁰ = [598.5] - [243.8]
ΔS_r⁰ = 354.7 J mol^-1K^-1.

Given:
T = 25°C = 298K;
ΔU = -3268.12 kJ mol^-1
ΔH = ?
ΔH = ΔU + Δn_gRT
ΔH = (ΔU + (n_p - n_r)RT
ΔH = 3268.12 + \(\left( 6-\frac { 7 }{ 2 } \right) \) x 8.314 x 10^-3 x 298
ΔU = -3268.12 + (1.5 x 8.314 x 10^-3 x 298)
ΔU = -3268.12 - 3.72
ΔU = -3274.84 kJ mol^-1

Given: Number of moles of water n = \(\frac { 180g }{ 18g\ mol^{ -1 } } =10\) mol molar heat capacity of water
C_p = 75.3 J K^-1 mol^-1
T₂ = 100°C = 373 K
T₁ = 25°C = 298K
ΔH = ?
ΔH = nC_p(T₂-T₁)
ΔH = 10 mol x 75.3 J mol^-1K^-1 x (373 - 298)K
ΔH = 56475 J
ΔH = 56.475 KJ.

\(\frac{rate_{unknown}}{rateN_2}=\sqrt{\frac{M_{N_2}}{M_{unknown}}}\)
\(0.5=\sqrt{\frac{28\ g\ mol^{-1}}{M_{unknown}}}\)
Squaring on both sides
\((0.5)^2=\frac{28gmol^{-1}}{M_{unknown}}\)
\(\Rightarrow M_{unknown}=\frac{28}{0.25}\) = 112 g mol^-1

P_Ne = X_Ne P_Total
\(X_{Ne}=\frac{n_{Ne}}{n_{Ne}+n_{Ar}+n_{Xe}}\)
\(=\frac{4.76}{4.76+0.74+2.5}=0.595\)
\(X_{Ar}=\frac{n_{Ar}}{n_{Ne}+n_{Ar}+n_{Xe}}\)
 \(=\frac{0.74}{4.76+0.74+2.5}=0.0093\)
\(X_{Xe}=\frac{n_{Xe}}{n_{Ne}+n_{Ar}+n_{Xe}}\)
\(=\frac{2.5}{4.76+0.74+2.5}=0.312\)
P_Ne = X_Ne P_Total = 0.595 x 2
= 1.19 atm.
P_Ar = X_Ar P_Total = 0.093 x 2
= 0.186 atm.
P_Xe = X_Xe P_Total = 0.312 x 2
= 0.624 atm.

According to Charles' law,
\(\frac{V_1}{T_1}=\frac{V_2}{T_2}=\frac{V_3}{T_3}\)
\(\frac{0.3dm^3}{200K}=\frac{V_2}{300K}=\frac{0.15dm^3}{T_3}\)
\(\frac{V_2}{300K}=\frac{0.3dm^3}{200K}\)

V₂ = 0.45 dm³ and
\(\frac{0.15dm^3}{T_3}=\frac{0.3dm_3}{200K}\)

T₃ = 100 K.

According to Boyle's law, at constant temperature for a given mass of gas at constant temperature,
P₁V_I = P₂V₂ = P₃V₃
1 atm x 1 dm³ = 2 atm x V₂ = P₃ x 0.25 dm³
\(\therefore\) 2 atm x V₂ = 1 atm x 1 dm³

\(\boxed{V_2=0.5\ dm^3}\)
and P₃ x 0.25 dm³ = 1 atm x 1 dm³

\(\boxed{P_3=4\ atm}\)

X³⁺,Y²⁺, Z^- are isoelectronic.
∴ Effective nuclear charge is in the order
(Z_eff)_Z-< (Z_eff)_y2+< (Z_eff)_X3+ and hence 
ionic radius should be in the order r_z- > r_y²⁺ >r_X³⁺
∴ The correct values are

n = 4; f orbital l = 3 \(\Rightarrow\) m₁= - 3, -2,-1, 0, +1, +2, +3
out of two electrons, one electron occupies 4f orbital with m₁ = -3 and another electron occupies 4f orbital with m₁ = -2.
All the four quantum numbers for the two electrons are

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.

Mass of chloride = 0.606 - 0.456
= 0.146 g
0.146g of chlorine combines with 0.456 g of metal
∴ 35.5 g of chlorine will combine with 
 = \(\frac { 35.5\times 0.456 }{ 0.146 } \)
= 110.8 g of metal
∴ equivalent mass of metal = 110.8g equ^-1

Equivalent mass =  \(\frac { Atomic\ mass }{ Valency } \)
Equivalent mass of sodium = \(\frac { 23 }{ 1 } \) = 23

Equivalent mass of an ion = \(\frac { Formula\ mass }{ Change\ of\ ion } \)
Equivalent mass of NO₃^- = \(\frac { 62 }{ 1 } \) = 62

Equivalent mass = \(\frac { Atomic\ mass }{ Valency } \)
=  \(\frac { 65 }{ 2 } \) = 32.5 q eq^-1

6.023 1023 molecules = 1 mole
 3.0115 \(\times\)10²³ molecules = \(\frac { 1 }{ 6.023\times { 10 }^{ 23 } } \times 3.0115\times { 10 }^{ 23 }\)
= 0.5 moles
Molar volumes of 1 mole of SO₂ = 2.24 \(\times\)10^-2m³
Molar volumes of 0.5 moles of SO₂ = 2.24 \(\times\) 10^-2 \(\times\)0.5
= 1.12 \(\times\) 10^-2 m³

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

Molar mass of methane = 16 g
Molar volume of 16g (1 mole) of methane = 2.24 \(\times\) 10-² m³
volume of 5 moles(80 g) of methane = \(\frac { 2.24\times { 10 }^{ -2 }\times 80 }{ 16 } \)
= 11.2 \(\times\) 10^-2 m³

Molar mass of CO₂ = 44 g
Molar volume of 44 g (1 mole) of CO₂
 = 2.24 \(\times\) 10^-2 m³
The volume of 88g (2 moles) = \(\frac { 2.24\times { 10 }^{ -2 }\times 88 }{ 44 } \)
= 4.48 \(\times\) 10^-2 m³

Atomic mass of potassium = 39
No. of moles = \(\frac { Mass }{ Molar\ mass } \)
No.of moles = \(\frac { 19.5 }{ 39 } \) = 0.5 moles

Molar mass of MgO = 40
No .of moles = \(\frac { Mass }{ Molar\ mass } \)
No .of moles = \(\frac { 90 }{ 40 } \)  =  2.25 moles

Molar mass of ethanol = 46
No.of moles = \(\frac { Mass }{ Molar\ mass } \)
No.of moles (n) =  \(\frac { 46 }{ 46 } \) = 1 mole

Molar mass of sodium hydroxide = 40
No.of moles = \(\frac { Mass }{ Molar\ mass } \)
No.of moles (n) = \(\frac { 120 }{ 40 } \)
= 3 moles

Molar mass of calcium chloride = 111
No. of moles =  \(\frac { Mass }{ Molar\ mass } \)
No. of moles = \(\frac { 50 }{ 111 } \)
 = 0.450 moles

Molecular mass of acetic acid = 60
60g of acetic acid contains = 6.023 \(\times\) 10²³
Molecules of acetic acid
∴ 1000g of acetic acid contains
= \(\frac { 100\times 6.023\times { 10 } }{ 60 } \)
 = 100 \(\times\) 10²³
 Molecules of acetic acid

Molecular mass of SO₂ = 64
64 g of sulphur dioxide contains = 6.023 \(\times\) 10²³
Molecules of SO₂
∴ 100g of SO₂ contains = \(\frac { 100\times 6.023\times { 10 }^{ 23 } }{ 64 } \)
 = 9.41 
Molecules of SO₂

1 mole of water = 18 g mol^-1
18 g of water contains 6.023 \(\times\)10²³ molecules of water
1.8 g of water contains = \(\frac { 1.8\times 6.023\times { 10 }^{ 23 } }{ 18 } \)
= 0.602 \(\times\)10²³
= 6.02 \(\times\) 10²⁴
Molecules of water

Atomic mass of Hg = 200 gmol-1
200 g of mercury contains 6.023 \(\times\) 10²³ atoms of mercury.
10g of mercury contains = \(\frac { 10\times 6.023\times { 10 }^{ 23 } }{ 200 } \)
= 0.301 \(\times\)10²³
= 3.01\(\times\) 10²⁴
atoms of mercury

 C ⟶  1 \(\times\)12 =12
H ⟶ 4\(\times\) 1 = 4
\(\underline { \overline { 16 } } \)           
∴ Molecular mass of CH₄ = 16

\(\overset { COOH }{ \underset { COOH }{ |\quad \quad \quad } } .2{ H }_{ 2 }O\)
 C ⟶ 1 \(\times\) 12 = 24
 O ⟶ 4 \(\times\) 16 = 64
H₂ \(\times\) 1 = 2
 \(\underline { \overline { 90 } } \)
4\(\times\) 1 = 4
2 \(\times\)16 = 32
\(\underline { \overline { 126 } } \)
∴ Molecular mass of oxalic acid 126  

1 atomic mass of   K = 1 \(\times\) 39 = 39
Mn = 1\(\times\) 55 = 55
O = 4\(\times\) 16 = 63
\(\overline { \underline { 158 } } \)
∴ Molecular mass of KMnO₄ = 158

NO₃^-
x + 3(-2) = -1
x - 6 = -1
x = -1 + 6
x = +5
Oxidation number of N in NO₃^- is +5

K₂CrO₄
2(1) + x + 4(-2) = 0
2 + x - 8 = 0
x - 6 = 0
x = +6
Oxidation number of Cr in K₂CrO₄ is +6

K₂MnO₄
Oxidation number of Mn be x
2 (1) + x + 4 (-2) = 0
2 + x - 8 = 0
x - 6 = 0
x = 6
Oxidation number of Mn in K₂MnO₄ is +6

Oxygen
Mass of oxygen atom = 2.656 x 10^-23 gram
1 amu = 1.6605 x 10^-24g
The mass of oxygen atom in amu = \(\frac { 2.656\times { 10 }^{ -23 } }{ 1.66056\times { 10 }^{ -24 } } \approx 15.99\quad amu\)

Al₂(SO₄)₃ + 3 BaCl₂ \(\longrightarrow\) 3 BaSO₄ + 2 AlCl₃
When 1 mole of aluminium sulphate reacts with barium chloride, 3 moles of BaSO₄ is precipitated.

(i) Nitrous acid: HNO₂
+ 1 + x - 2 x 2 = 0
x = +3
(ii) Nitric acid: HNO3
+ 1 + x - 2 x 3 =0
x = +5.

1 x AW of Mg = 1 x 24.305 = 24.305 amu
2 x AW of O = 2 x 16 = 32.000 amu
2 x AW of H = 2 x 1.008 = 2.016 amu
Formula weight of Mg(OH)₂ is = 58.321 amu
Formula weight of Mg(OH)₂ is = 58 amu.

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

1 x AW of N = 1 x 14 = 14amu
2 x AW of O = 2 x16 = 32 amu
Formula weight of NO₂ = 46 amu

Molar mass of H₅IO₆ = (5x1 + 1x127 + 6x16)
= 228 g mol^-1
Mass of 5.14 mol of H₅IO₆ =5.14 mol x 228g mol^-1
= 1171.9 g.

Molar mass of CO₂ = (1 x 12 + 2 x 16)
= 44 g mol^-1
Mass of 3 moles of CO₂ = 3 mol x 44g mol^-1
= 132 g

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

No. of silver atoms = 1 million = 1 x 10⁶
Mass of one million Ag atoms = 1.79 x 10^-16g
Mass of 6.023 x 10²³atoms of silver
= \(\frac { 1.79\times { 10 }^{ -16 }g }{ 1\times { 10 }^{ 6 } } \times 6.023\times { 10 }^{ 23 }\)
= 107.8 g.
Atomic mass of silver = 6.023 x 10²³ atoms of Ag
\(\therefore\)  The atomic mass of Ag = 107.8 g

Molecular mass of water = (2 x 1u) + (1 x 16u)
= 18 u
Molar mass of water = 18 g mol^-1
Mass of 1 molecule of water
= \(\frac { Molar\ mass\ of\ water }{ Avogadro's\ number } \)
= \(\frac { 18\ g\ { mol }^{ -1 } }{ 6.023\times { 10 }^{ 23 }\ { mol }^{ -1 } } \)
= 2.99 x 10^-23 g
Mass of 1 molecule of water = 2.99 x 10^-23 g

Molecular mass of benzene (C₆H₆) = (6 x 12.01 u) + (6 x 1 u) = 78.06 u
Molar mass of benzene = 78.06 g mol^-1
Then, mass of [molecule of benzene = \(\frac { Molar\ mass }{ Avogadro's\ number } \)
= \(\frac { 78.06g\ { mol }^{ -1 } }{ 6.023\times { 10 }^{ 23 }\ { mol }^{ -1 } } \)
= 12.96 x 10^-23 g

Molecular mass of silver (Ag) = 107.87 u
Molar mass of Ag = 107.87 g mol^-1
\(\therefore\) Mass of 1 atom of Ag = \(\frac { Molar\ mass }{ Avogadro's\ number } \)
= \(\frac { 107.87g\ { mol }^{ -1 } }{ 6.023\times { 10 }^{ 23 }\ { mol }^{ -1 } } \)
= 17.91 x 10^-23g
Mass of 1 atom of Ag = 17.91 x 10^-23 g.

52g of He contains 7.83 x 10²⁴ He atoms.
1 mol of He contains 6.023 x 10²³ He atoms
\(\therefore\) 52 moles of He contains =  \(\frac { 6.023\times { 10 }^{ 23 }\times 52 }{ 1} \)
= 3.132 x 10²⁵
52 moles of He contains 3.132 x 10²⁵ He atoms

1 mol of He \(\equiv\) 4g \(\equiv\) 6.022 x 10²³ He atoms
(ie) 4g of He contains 6.022 x 10²³ He atoms
\(\therefore\) 52g of He contains = \(\frac { 6.023\times { 10 }^{ 23 }\times 52 }{ 4 } \)
= 7.83 x 10²⁴

(i) Kelvin-Planck statement: It is impossible to construct a machine that absorbs heat from a hot source and converts it completely into work by a cyclic process without transferring a part of heat to a cold sink.
(ii) Clausius statement: It is impossible to transfer heat from a cold reservoir to a hot reservoir without doing some work.
(iii) Entropy statement: The entropy of an isolated system increases during a spontaneous process

Strong acids and strong bases exist in the fully ionised form in aqueous solutions as below:
H₃O⁺ + CI^- + Na⁺ + OH^- ⟶ Na⁺ + CI + 2H₂O
(or)
H₃O⁺_(aq)+OH^-_(aq)⟶2H₂O(1)⁰
ΔH^o =-57.32 KJ.
The H⁺ions produced in water by the acid molecules exist as H₃O⁺. Thus, enthalpy change of neutralisation is essentially due to enthalpy change per mole of water formed from H₃O⁺ and OH- ions. Therefore, irrespective of the chemical nature, the enthalpy of neutralisation of strong acid by strong base is a constant value which is equal to -57.32 KJ

Given K_eq= 10
Gas constant R = 8.314 JK^-1 mol^-1
T=300K
The relationship between Free energy change ΔG and equilibrium constant K is ΔG^o=-RTlnK
Since K, T and R are positive values, ΔG^o will be negative.
When ΔG is -ve, the process is spontaneous and feasible

(i) State function: A state function is a thermodynamic property of a system, which has a specific value for a given state and does not depend on the path (or manner) by which the particular state is reached.
Example: Pressure (P), Volume (V), Temperature(T)
(ii) Path functions: A path function is a thermodynamic property of the system whose value depends on the path by which the system changes from its initial to final states.
Example: Work (w), Heat (q).

(i) In reactions, (i) & (ii) AgNO₃ and CuSO₄ act as oxidising agents. They oxidise H₃PO₂ (hypophosphorous acid) to (orthophosphoric acid)
(ii) In reaction (iii) [Ag (NH₃)₂]⁺ _(aq) oxidises C₆H₅CHO to C₆H₅COOH.
(iii) In reaction (iv) Cu²⁺ does not oxidise C₆H₅ to C₆H₅COOH.
(iv) This indicates that Ag⁺ is a strong oxidising agent than Cu.

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

Alkali metals have one e^o in their outer most shell. Alkaline earth metals have 2 e^o in their outer most shell. More valence electrons and more positively charged nucleii leads to greater opportunity for metallic bonding.

In Na₂ SO₄.10 H₂O. So the metal will be Na (sodium). The salt is washing soda.

It will be more difficult to drink water with a straw on the top of a Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water up into the straw, below the water surface.
The force that propels water through a straw is atmospheric pressure, which is less at high altitude.

The passenger in a moving bus falls in the forward direction, when brakes are applied suddenly is in accordance with Newton's first law of motion.
When breaks are applied, the automobile comes to rest but the passenger due to inertia of motion tend to continue to move in forward direction hence fall forward.
The movement of helium balloon in the opposite direction is due to the difference in the density of surrounding air. When the car stops suddenly, the air moves forward due to inertia of motion.
As a result the air in the front of the car is more dense that at the rear of the car creating more pressure on the front side of the balloon. This causes the balloon move towards the rear of the car where air is less dense.
The body is at rest but the car is forward motion, due to inertia the passengers are pushed toward the front of the car on forward acceleration.

Circumference of the orbit - 2\(\pi\)r         .... ...(1)
Circumference of the orbit of H - atom (n = 1) - n\(\lambda\),  .........(2)
(1) = (2) \(2 \pi r=\lambda \text { (or) } 2 \pi r=\frac{\mathrm{h}}{\mathrm{mv}}\)

a) Hydrogen is the lightest gas in the atmosphere. So it rises up and other gases which are heavier like O₂ & N₂ come down towards the surface of the earth according to Graham's law of diffusions \(\mathrm{r} \alpha \sqrt{\frac{1}{M}}\). Hydrogen diffuses very fast. Its mean speed in greater than the escape velocity from the earth. As a consequence, H₂ would have escaped from the atmosphere long time ago.
b) The acceleration due to gravity 'g' on moon surface is small. The value of escape velocity is also small. The molecules of the atmospheric gases on the moon's surface have thermal velocities greater than the escape velocity.
All the molecules have escaped. So the atmosphere is so thin.

(i) The order : NH₃ < H₂O < HF
(ii) Strength of H bond depends upon the atomic size and electronegativity of the other atom to which H atom is covalently bonded. 
Since electronegativity of F > O > N, the strength of H-bond is in the order H-F......H > H - O......H > H-N......H.
H-bonds are much weaker than covalent bonds.

Metal Hydride (Hydrogen Sponge):
The best studied binary hydrides are the palladium - hydrogen system. Hydrogen interacts with palladium in a unique way, and forms a limiting monohydride, PdH. Upon heating, H atoms diffuse through the metal to the surface and recombine to form molecular hydrogen. Since no other gas behaves this way with palladium, this process has been used to separate hydrogen gas from other gases.

\(2 \mathrm{Pd}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{PdH}(\mathrm{s})\)
The hydrogen molecule readily adsorbs on the palladium surface, where it dissociates into atomic hydrogen. The dissociated atoms dissolve into the interstices or voids (octahedral/tetrahedral) of the crystal lattice.
Technically, the formation of metal hydride is by chemical reaction but it behaves like a physical storage method, i.e., it is absorbed and released like a water sponge. Such a reversible uptake of hydrogen in metals and alloys is also attractive for hydrogen storage and for rechargeable metal hydride battery applications.

1. Deuterium is used in heavy water moderated fission reactors, usually as liquid D₂O to slow neutrons
2. To prepare D₂O (heavy water)
3. To prepare deuterated halides
4. To prepare compounds like deutero methane, deutero ammonia etc..
5. To manufacture deuterium lamps with quartz or UV glass bulb.

When compounds containing hydrogen are treated with D₂O, hydrogen undergoes an exchange for deuterium
\( 2 \mathrm{NaOH}+\mathrm{D}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOD}+\mathrm{HOD} \)
\(\mathrm{HCl}+\mathrm{D}_{2} \mathrm{O} \rightarrow \mathrm{DCl}+\mathrm{HOD} \)
\(\mathrm{NH}_{4} \mathrm{Cl}+4 \mathrm{D}_{2} \mathrm{O} \rightarrow \mathrm{ND}_{4} \mathrm{Cl}+4 \mathrm{HOD}\)
These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound.
For example, when D₂O is treated with of hypo-phosphorus acid only one hydrogen atom is exchanged with deuterium. It indicates that, it is a monobasic acid.
\(\mathrm{H}_{3} \mathrm{PO}_{2}+\mathrm{D}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{DPO}_{2}+\mathrm{HDO}\)

1. Heavy water is widely used as moderator in nuclear reactors as it can lower the energies of fast neutrons
2. It is commonly used as a tracer to study organic reaction mechanisms and mechanism of metabolic reactions
3. It is also used as a coolant in nuclear reactors as it absorbs the heat generated.

Elements having same atomic number but different mass numbers are called isotopes. These are variant, of an element which differ in neutron numbers, which contain equal number of protons. They differ in relative atomic mass but not in chemical properties.
The isotopes of hydrogen are:
1. Protium \(\left(\mathrm{H}^{1} \text { or } \mathrm{H}\right)\) [Predominant form]
2. Deuterium (or) Heavy hydrogen \(\left({ }_{1} \mathrm{H}^{2} \text { or }{ }_{1} \mathrm{D}^{2}\right)\)
3. Tritium \(\left({ }_{1} \mathrm{H}^{3} \text { or }{ }_{1} \mathrm{T}^{3}\right)\)

The carbon monoxide of the water gas can be converted to carbon dioxide by mixing the gas mixture with more steam at 400^oC and passed over a shift converter containing iron/copper catalyst. This reaction is called as water-gas shift reaction.
CO + H₂O ➝ CO₂ + H₂

a) Oxidation is performed in acidic medium:
Eg : H₂O₂ Oxidises FeSO₄ to Fe (SO₄)₃
H₂O₂ + 2H⁺+ 2e^- ➝ 2H₂O (E⁰ = +1.77 V)
2FeSO₄ + H₂SO₄ + H₂O₂ ➝ Fe₂(SO₄)₃ + 2H₂O
b) Reduction is performed in basic medium:
Eg : H₂O₂ reduces KMnO₄ to MnO₂
HO₂^- + OH^- ➝ H₂O + 2e^- (E⁰ = +0.08 V)
\(2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{MnO}_{2}+2 \mathrm{KOH}+2 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{O}_{2}\)

(i) 2KMnO₄ + 3H₂O₂ ➝ 2MnO₂ + 2KOH + 2H₂O + 3O₂ (Redox reaction)
(ii) CrCl₃ + 6H₂O ➝ [Cr Cl₂ (H₂O)CI. 2H₂O [Hydration reaction]
(iii) CaO + H₂O ➝ Ca(OH)₂ (Hydration reaction)

Gases which obey Boyle's Law and Charle's law or ideal gas equation PV = nRT are called ideal gases.
All gases whose behaviour is consistent with the assumption of kinetic theory of gases under all condition are called ideal gases.

The following table gives the pressure of a gas with increasing temperature at constant volume for two cylinders.

Both the models explain Gay - Lussac's Law. P \(\propto\) T are constant V & n
For 50 L container:
The pressure increases with rise in temperature (at constant volume)
For 75 L container:
The pressure increases with rise in temperature (at constant volume)

Charles law states that "At constant pressure, the volume of a given mass of an ideal gas is directly proportional to its temperature." According to Charles Law, if we were to take a balloon filled with air and increase the temperature of the air inside, the volume of air would increase causing the balloon to expand. This is caused by the heating of the molecules of air inside the balloon causing them to move rapidly. In the same manner if we cooled the balloon in a freezer, the volume of air decreases, making the balloon look partially deflated.

At a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
\(V\alpha \frac { 1 }{ P }\) at constant T& n.
Mathematical form: P₁V₁ = P₂ V₂ = K

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.

(i) Lithium and beryllium form more covalent compounds, unlike the alkali and alkali earth metals which predominantly form ionic compounds.
(ii) The elements of the second period have only four orbitals (2s & 2p) in the valence shell and have a maximum co-valence of 4, whereas the other members of the subsequent periods have more orbitals in their valence shell and shows higher valences. For example, boron forms BF₄^- and aluminium forms AIF₆^3-.

In the periodic table vertical columns are called groups.
Elements in the same vertical column possess similar number of electrons in the outer orbitals.
∴ Elements a and c belongs to group 18
Elements b and d belongs to group 13

(i) According to aufbau's principle 5th period has nine orbital (one 5s, five 4d and three 6p) to be filled.
(ii) Nine orbitals can accommodate a maximum of 18 electrons. Hence fifth period of the periodic table should has 18 elements from rubidium (2 = 37) to Xenon (Z = 54).

Z = 118; [₈₆Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶
In the periodic table the element with Z = 118 is located in p-block,
Period no = 7 (as n = 7 for valence shell)
Group no. = 18 (group no = 10+ ns electrons + np electrons) (n - outer most shell)

(i) Electronic configuration is the arrangement of electrons in an atom. The outermost electron shell is often referred to as the "valence shell"determines the chemical properties.
(ii) Ionization energy and electron affinity is the amount of energy released or required in pulling out or adding an electron to a neutral atom. So both depend on electronic configuration of the element.

Ionisation energy is the amount of energy required to remove the electron from the ground state (EI) to excited state (E^∞)
E₁=-2.18 x 10^-18 J; E_∞=0
ΔE=E_∞-E₁
=0-(-2.18 x 10^-18 J)=2.18 x 10^-18 J
I.E per hydrogen atom = 2.18 x 10^-18 J
I.E per mole of H-atom =2.18 x 10^-18 J x 6.023 x 10²³
=13.13 x 10⁵ J mol^-1

Equate the total no. of electrons in the reactant side by cross multiplying.
\(\Rightarrow 3 \mathrm{As}_{2} \mathrm{~S}_{3}+28 \mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO}\)
Based on reactant side, balance the products
\(\Rightarrow 3 \mathrm{As}_{2} \mathrm{~S}_{3}+28 \mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow 6 \mathrm{H}_{3} \mathrm{AsO}_{4}+9 \mathrm{HSO}_{4}+28 \mathrm{NO}\)
Product side : 36 hydrogen atoms & gg Orygen atoms
Reactant side : 28 hydrogen atoms & 74 Orygen atoms
Difference is 8 hydrogen atoms & 4 oxygen atoms
∴ Add 4 H₂O molecule on the reactant side.
Balanced equation is,
\(3 \mathrm{As}_{2} \mathrm{~S}_{3}+28 \mathrm{HNO}_{3}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow 6 \cdot \mathrm{H}_{3} \mathrm{AsO}_{4}+9 \mathrm{H}_{2} \mathrm{SO}_{4}+28 \mathrm{NO}\)

(a) Isothermal process: An isothermal process is defined as one in which the temperature of the system remains constant, during the change from its initial to final state. The system exchanges heat with its surroundings and the temperature of the system remains constant.  
For an isothermal process dT = 0
(b) Adiabatic process: An adiabatic process is defined as one in which there is no exchange of heat (q) between the system and surrounding during the process. For an adiabatic process q = 0
(c) Isobaric process: An isobaric process is defined as one in which the pressure of the system remains constant during its change from the initial to final state. For an isobaric process dP = 0 .
(d) Isochoric process: An isochoric process IS defined as the one in which the volume of system remains constant during its change from initial to final state. For an isochoric process, dV= 0.

Magnesium oxide in having higher melting point. The lattice energy of MgO & MgF₂ are 3938 and 2957 respectively.
MgO has +2, -2 charges, MgF₂ has +2, -1 charges. When the two charges are multiplied together, MgO results in larger amount of lattice energy since it has a higher charge, The strong attraction cause most ionic material to be hard and brittle and have high melting points.

The word Aufbau in German means 'building up'. In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them.
Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals. The order of filling of various orbitals as per the Aufbau principle which is in accordance with (n + l) rule.

(i) By treatment with catalyst like Pt or Fe.
(ii) By passing an electric discharge
(iii) By heating to 800°C or more.
(iv) By mixing with paramagnetic molecules like O₂,NO,NO₂·
(v) By mixing with nascent hydrogen or atomic hydrogen.

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.

ΔH_f⁰ CO₂ = -393.5 kJ mol^-1
ΔH_f⁰ CO = -111.31 kJ mol^-1
ΔH_f⁰ (H₂O) = - 242 kJ mol^-1
CO₂(g) + H₂(g) - CO(g) + H₂O(g)
ΔH_f⁰ = ?
ΔH_f⁰ = Σ(ΔH_f⁰)_products-Σ(ΔH_f⁰)_reactions
ΔH_f⁰ = [ΔH_f⁰(CO) + ΔH_f⁰(H₂O)] - [ΔH_f⁰(CO₂) + ΔH_f⁰(H₂)]
ΔH_f⁰ = [-111.31 + (-242)] - [-393.5 + (0)]
ΔH_f⁰ = [-353.31] + 393.5
ΔH_f⁰ = 40.19
ΔH_f⁰ = +40.19 kJ mol^-1.

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.

n = 4 l = 0,1,2,3
4 sub shells s, p, d & f.
I = 0 m₁ = 0 + one 4s orbital.
I = 1 m₁ = -1, 0, + 1 \(\Rightarrow\) three 4p orbitals.
I = 2 m₁ = -2,:1, 0, +1, +2 \(\Rightarrow\) five 4d orbitals.
I = 3 m₁ = -3, -2, -1,0, +1, +2, +3 \(\Rightarrow\) seven 4f orbitals.
Over all 16 orbitals are possible.

Heavy water is toxic when taken in large quantities. Heavy water is not radioactive. The deuterium in it is stable; it does not decay. Nobody will be in danger at all from radiation. It is heavier than plain water. D₂O performs little different from H₂O in chemical reactions. One has to drink a lot of D₂O to kill him.

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

2NH₃ + H₂O + CO₂ ⟶ (NH₄)₂ CO₃
(NH₄)₂ CO₃ + H₂O + CO₂ ⟶ 2NH₄ HCO₃
2NH₄HCO₃ + NaCl ⟶ NH₄Cl + NaHCO₃
2NaH CO₃ ⟶ Na₂ CO₃ + CO₂ + H₂O.

The solubility product of NaCl is lower than that of NaOH. The more soluble a substance is, the higher the K_sp value it has In aqueous solution NaOH gives OH^- ions. It can be solvated by establishing H-bonds with water molecules. So it is more water soluble.

Gases are very less denser. They have negligible intermolecular forces of attraction between the molecules. They are in continuous kinetic motion. So they won't settle at the bottom due to gravitational forces. The material that is most dense will sink to the bottom; the less denser will go up.

a) KH, CaH₂, ------ GaH₃, GeH₄, AsH₃, H₂Se, HBr
b) Trend in the formula:

c) First 2 members: KH & CaH₂. KH is alkali hydride (Grey powder) CaH₂ is an alkaline earth hydride.
KH reacts vigorously with water liberating H₂ gas.
\(\mathrm{LiH}_{(\mathrm{s})}+\mathrm{H}_{2} \mathrm{O}_{(h)} \rightarrow \mathrm{Li}(\mathrm{OH})_{a q}+\mathrm{H}_{2}\)
CaH₂ reacts with water liberating H₂ gas
\(\mathrm{CaH}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}+2 \mathrm{H}_{2}\)

For ideal gases, a = 0. The value of 'a' is a measure of the attractive forces between the molecules. There will not be any intermolecular forces of attraction. So it cannot be liquefied.

According to Avogadro's hypothesis: V \(\alpha\) n
\(\therefore \frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}=\text { constant }\)
V₁ & n₂ are the volume and number of moles of a gas and V₂ & n₂ are a different set of values of volume and number of moles of the same gas at same temperature and pressure.