This equation gives the quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change (\(\triangle\)G^O) and equilibrium constant is
\(\Delta { G }^{ 0 }=-RTln\ K\)    ...(1)
We know that
\(\Delta { G }^{ 0 }=\Delta { H }^{ 0 }-T\Delta { S }^{ 0 }\)
Substituting (1) in equation (2)
\(-RTln\ K\ =\Delta { H }^{ 0 }-T\Delta { s }^{ 0 }\)
Rearranging
In \(K=\cfrac { -\Delta H^{ 0 } }{ RT } +\cfrac { { \Delta S }^{ 0 } }{ R } \)     ...(3)
Differentiating equation (3) with respect to temperature
\(\cfrac { d\left( In\quad K \right) }{ dT } =\cfrac { \Delta { H }^{ 0 } }{ { RT }^{ 2 } } \)       ...(4)
Equation 4 is known as differential form of Van't Hoff equation.
On integrating the equation 4, between T₁ and T₂ with their respective equilibrium constants K₁ and K_2.
\(\int _{ { k }_{ 1 } }^{ { K }_{ 2 } }{ d\left( In\ K \right) =\cfrac { \Delta { H }^{ 0 } }{ R } \int _{ { T }_{ 2 } }^{ { { T }_{ 2 } } }{ \cfrac { dT }{ { T }^{ 2 } } } } \)
\(\left[ In\quad K \right] _{ { K }_{ 1 } }^{ { K }_{ 2 } }=\cfrac { \Delta { H }^{ 0 } }{ R } \left[ -\cfrac { 1 }{ T } \right] ^{ { T }_{ 2 } }_{ { T }_{ 1 } }\)
\(In\quad { K }_{ 2 }-In\quad { K }_{ 1 }=\cfrac { \Delta { H }^{ 0 } }{ R } -\left[ \cfrac { 1 }{ { T }_{ 2 } } +\cfrac { 1 }{ { T }_{ 2 } } \right] \)
\(In\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \)
\(log\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ 2.303R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \)  ...(5)
Equation (5) is known as integrated form of Van't Hoff equation.

(i) Over a range of low pressures, the real gases can behave ideally at a particular temperature called as Boyle temperature or Boyle point.
(ii) The Boyle point varies with the nature of the gas.
(iii) Above the Boyle point, the compression point Z > 1 for real gases i.e. real gases show positive deviation.
(iv) Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increase with increase in pressure. So, it is clear that at low pressure and at high temperature, the real gases behave as ideal gases.
(v) Hence, \(Z=\frac{PV_{real}}{nRT}\)
\(PV_{ideal}=\frac{nRT}{P}\)
So, \(Z=\frac{V_{real}}{V_{ideal}}\)

In metallic hydrides hydrogen occupies the interstitial sites (voids) of their crystal lattice. Hence,they are called interstitial hydrides; these hydrides show properties similar to parent metals andhence they are also known as metallic hydrides. Most of the hydrides are non-stoichiometric with variable composition (TiH_{1.5_ 1.8} and PdH_{O.6_0.8}) Some are relatively light, inexpensive and the thermally unstable which make them useful for hydrogen storage applications. Electropositive metals and some other metals form hydrides with the stoichiometry MH or sometimes MH₂(M = Ti, Zr, Hf, V, Zn)

For example, if a balloon is moved from an ice cold water bath to a boiling water bath, the temperature of the gas increases. As a result, the gas molecules inside the balloon move faster and gas expands. Hence, the volume increases.

Metal displacement reaction

(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).

High purity hydrogen (> 99.9%) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
 At anode: 2OH^- ➝ H₂O + 1/2O₂ + 2e^-
At cathode: 2H₂O + 2e^- ➝ 2OH^- + H₂
Overall reaction: H₂O ➝ H₂ + 1/2O₂ 

The lattice energy of LiF is higher due to the smaller size of Li⁺ and F^-. So LiF has lower solubility.

6Li + N₂ ⟶ 2Li₃ N

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

(i) The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is known as metallic bond.
(ii) According to Drude and Lorentz, metallic crystal is an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to ionisation of the valence electrons of the atoms of the metal.
(iii) As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is referred to as electronic bonding.
(iv) The electrostatic attraction between the metal ions and the free electrons yield a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.
(v) As the close packed structure contains many slip planes along which movement can occur during mechanical loading, metal acquires ductility.
(vi) As metal ion is surrounded by electron cloud in all directions, the metallic bonding has no directional properties.
(vii) As the electrons are free to more around the positive ions, the metals exhibit high electrical and thermal conductivity.
(viii) The metallic lustre is due to the reflection of light by the electron cloud.
(ix) As the metallic bond is strong enough, the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.
(x) High thermal conductivity of metals is due to thermal excitation of many electrons from the valence bond to the conduction band.

There are three types of electron movement viz.,
1. lone pair becomes a bonding pair.
2. bonding pair becomes a lone pair
3. a bond breaks and becomes another bond
Type 1: A lone pair to a bonding pair

Type 2: A bonding pair to a lone pair

Type 3: A bonding pair to an another bonding pair

S_N2 Mechanism :
(i) The rate of S_N2 reaction depends upon the . concentration of both alkyl halide and the nucleophile.
(ii) Rate of reaction = is [alkylhalide] [nucleophile]. It follows second order kinetics and occurs in one step. 
(iii) This reaction involves the formation of a transition state in which both the reactant molecules are partially bonded to each other. The attack of nucleophile occurs from the back side (i.e opposite to the side in which the halogen is attacked).
(iv) The carbon at which substitution occurs has inverted configuration during the course of reaction just as an umbrella has tendency to invert in a wind stonri. is inversion of configuration is called Walden inversion; after paul walden who 1st discovered the inversion of configuration of a compound in S_N2reaction.
(v) S_N2reaction of an optically active haloalkane . is always accompanied by inversion of configuration at the asymmetric centre.
(vi) When 2 - Bromooctane is heated with sodium hydroxide, 2 - octanol is formed with invesion of configuration. (-) - 2 - Bromo octane is heated with sodium hydroxide (+) - 2 - Octanol is formed in which - OR group occupies a position opposite to what bromine had occupied,

(a) (-).2 - Bromo octane
(b) Transition State
(c) (+) 2 - Octanol (product)

A small piece of Na dried by pressing between the folds of a filter Paper is taken in a fusion tube and it is gently heated.
When it melts to a shining globule, put a pinch of the organic compound on it. Heat the tube till reaction ceases and becomes red hot. Plunge it in about 50 mL of distilled water taken in a china dish and break the bottom of the tube by striking against the dish. Boil the contents of the dish for about 10 mts and filter. This filtrate is known as lassaignes extract or sodium fusion extract and it used for detection of nitrogen, sulfur and halogens present in organic compounds.
If nitrogen is present it gets converted to sodium cyanide which reacts with freshly prepared furro,sulphate and feiric ion followed by conc. HCI and gives a Prussian blue color or green color precipitate. It confirms the presence of nitrogen. HCI is added to dissolve the lreenish precipitate of ferrous hydrbxide iroduced by the excess of NaOH on Feson which would otherwise markthe Prussian blue piecipitate. The following reaction takes part in the formation of Prussian blue.


from organic compounds
\(FeSo_{ 4 }+2NaOH\longrightarrow Fe(OH)_{ 2 }+Na_{ 2 }{ SO }_{ 4 }\)
from organic compounds
\(6FeCN+Fe(OH)_{ 2 }\longrightarrow Na_{ 4 }[Fe(CN)]_{ 6 }+2NaOH\)
Sod.ferrocyanide
\(3Na_{ 4 }[Fe(CN)_{ 6 }]+FeC1_{ 3 }\longrightarrow Fe_{ 4 }[Fe(CN)]_{ 3 }+12NaCI\)
ferric ferrocyanidePrussian blue or greenppt
Incase if both N & S are present, a blood red color is obtained due to the following reactions.
\(\mathrm{Na}+\mathrm{C}+\mathrm{N}+\mathrm{S} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{NaCNS}\)
sodium sulphocyanide
\(3 \mathrm{NaCNS}+\mathrm{FeCl}_3 \longrightarrow \mathrm{Fe}(\mathrm{CNS})_3+3 \mathrm{NaCl}\)
ferric sulphocyanide
(Blood red colour).

A ther mochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change (\(\Delta\)H).
Conventions adopted in thermochemical equations:
(i) The coefficients in a balanced thermochemical equation refer to number of moles of reactants and products involved in the reaction.
(ii) The enthalpy change of the reaction \(\Delta\)H has unit kJ.
(iii) When the chemical reaction is reversed, the value of H is reversed in sign with the same magnitude.
(iv) Physical states (gas, liquid, aqueous and solid) of all species is important and must be specified in a thermochemical reaction since H depends on the phases of reactants and products.
(v) If the thermochemical equation is multiplied throughout by a number, the enthalpy change is also be multiplied by the same number value.
(vi) The negative sign \(\Delta H_r^o\) of indicates the reaction to be an exothermic and the positive sign of \(\Delta H_r^o\) indicates an endothermic type of reaction.

Zn + NO₃^- ⟶ Zn + NH₄⁺² in basic medium
Zn ⟶ Zn⁺² + 2e (oxidation) NO₃^- ⟶ NH₄⁺ (Reduction)
Step-1: Balance all atoms other than hydrogen and oxygen.
Zn ⟶ Zn⁺²       NO₃^- ⟶ NH₄⁺
Step-2: For balancing oxygen atom in alkaline medium, add H20 molecules on the side deficient of oxygen atom.
Then balance hydrogen atom, add H20 molecule to the side deficient in hydrogen and equal number of OH- on the other side.
Zn ⟶ Zn⁺²       NO₃^- ⟶ NH₄^-+3H₂O
NO₃^- + 10H₂O ⟶ NH₄⁺ + 3H₂O
NO₃^- + 10H₂O ⟶ NH₄⁺ + 3H₂O + 10OH^-
Step-3: Add electrons to balance charge
Zn ⟶ Zn⁺² +2e
NO₃^- + 7H₂O + 8e ⟶ NH₄⁺ +10OH^-
       Zn ⟶ Zn+2 + 2e x 8
NO3- + 7H2O + 8e ⟶ NH4+ + 10OH-
___________________________________
8Zn + NO₃^- + 7H₂O ⟶ 8Zn⁺² + NH₄⁺ + 10OH^-
This is the balanced equation.

Characteristics of p-block elements:
(i) General electronic configuration is ns²np^1-6 (Group 13 to 18).
(ii) They are generally non-metals but some metals and metalloids are present.

• Non-metallic character increases from left to right along period.
• Metallic character increases on moving down a group.

(iii) They have high electron affinity and ionization enthalpy.

• Both the properties increase along the period.
• Both the properties decrease on moving down the group.

(iv) Oxidising character.

• Increases on moving from left to right along a period.
• Reducing character increases from top to bottom in a group.

(v) s-block and p-block elements are collectively called representative elements.

The electronic configuration of the element with 2K, 8L and 5M electrons will be
1s² 2s² 2p_x² 2p_y² 2p_z² 3s² 3p_x¹ 3p_y¹ 3p_z1
(i) Total number of electrons = 2 + 8 + 5 = 15
∴ Atomic number of the element = 15
(ii) Total number of s-electrons = 2 + 2 + 2 = 6
(iii) Total number of p-electrons = 6 + 3 = 9
(iv) Since, the atom is neutral
∴  Number of protons = Number of electrons = Atomic number = 15
(v) valency of the element = 3

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.

In \(\overset { .. }{ \underset { .. }{ O } } =C=\overset { .. }{ \underset { .. }{ O } } \)
Formal charge on carbon = \({ N }_{ v }-\left[ { N }_{ 1 }+\cfrac { { N }_{ b } }{ 2 } \right] =4-\left[ 0+\cfrac { 8 }{ 2 } \right] =4-4=0\)
Formal charge on oxygen = \({ N }_{ V }-\left[ { N }_{ 1 }+\cfrac { { N }_{ 2 } }{ 2 } \right] =6-\left[ 4+\cfrac { 4 }{ 2 } \right] =6-6=0\)

Particulate pollutants are small solid particles and liquid droplets suspended in air. Many of particulate pollutants are hazardous.
Examples : dust, pollen, smoke, soot and liquid droplets (aerosols) etc,.
They are blown into the atmosphere by volcanic eruption, blowing of dust, incomplete combustion of fossil fuels induces soot. Combustion of high ash fossil fuels creates fly ash and finishing of metals throws metallic particles into the atmosphere.
The non- viable particulates are small solid particles and liquid droplets suspended in air. They help in the transportation of viable particles. There are four types of non-viable particulates in the atmosphere. They are classified according to their nature and size as follows.
(i) Smoke : Smoke particulate consists of solid particles (or) mixture of solid and liquid particles formed by combustion of organic matter. 
For example, cigarette smoke, oil smoke, smokes from burning of fossil fuel, garbage and dry leaves.
(ii) Dust: Dust composed of fine solid particles produced during crushing and grinding of solid materials.
For example, sand from sand blasting, saw dust from wood works, cement dust from cement factories and fly ash from power generating units.
(iii) Mists : They are formed by particles of spray liquids and condensation of vapours in air.
For example, sulphuric acid mist, herbicides and insecticides sprays can form mists.
(iv) Fumes : Fumes are obtained by condensation of vapours released during sublimation, distillation, boiling and calcination and by several other chemical reactions.
For example : organic solvents, metals and metallic oxides form fume particles.

Carbon halogen bond is a polar bond as halogens are more electro negative than carbon. The carbon atom exhibits a partial positive charge (\(\delta \)+) and halogen atom a partial negative charge (\(\delta \)-)
The C -X bond is formed by overlap of sp³ orbital of carbon atom I with half filled p-orbitat of the halogen atom. The atomic size of halogen increases from fluorine to iodine, which increases the C - X bond length. Larger the size, greater is the bond length, and weaker is the bond formed. The bond strength of C - X decreases from C - F to C - I in CH₃X

No. of, moles of glucose ; n₂ = \(\frac{\mathrm{m}}{\mathrm{M}}=\frac{2.82}{180}=0.016 \mathrm{~mol} \)
No. of. moles of water ; n₁ = \(\frac{30}{18}=1.67 \mathrm{~mol} \)
\(\mathrm{X}_{1}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{1.67}{1.67+0.016}=0.99\)
\(\mathrm{X}_{1}+\mathrm{X}_{2}=1 \)
\(\therefore \mathrm{X}_{2}=1-\mathrm{X}_{1}=1-0.99=0.01 \).

A reaction that does occur under the given set of conditions is called a spontaneous reaction.
Example:
1. A waterfall runs downhill, but never up, spontaneously.
2. Heat flows from hotter object to a colder one.
3. Ageing process.

Empirical formula : It is the simplest formula of a compound which gives the ratio of the number of different atoms present in one molecule of the compound.

i) Both (A) and R are correct and (R) is the correct explanation of (A)