(i) When atoms combines to form molecules, their individual atomic orbitals lose their identity and forms new orbitals called molecular orbitals.
(ii) The shapes of molecular orbitals depend upon the shapes of combining atomic orbitals.
(iii) The number of molecular orbitals formed is the same as the number of combining atomic orbitals. Half the number of molecular orbitals formed will have lower energy than the corresponding atomic orbital, while the remaining molecular orbitals will have higher energy. The molecular orbital with lower energy is called bonding molecular orbital and the one with higher energy is called anti-bonding molecular orbital. The bonding molecular orbitals are represented as \(\sigma\) (Sigma), \(\pi\) (Pi), \(\delta\) (delta) and the corresponding antibonding orbitals are denoted as \(\sigma\)*, \(\pi\)* and \(\delta\)*.
(iv) The electrons in a molecule are accommodated in the newly formed molecular orbitals. The Filling of electrons in these orbitals follows Aufbau's principle, Pauli's exclusion principle and Hund's rule as in the case of filling of electrons in atomic orbitals.
(v) Bond order gives the number of covalent bonds between the two combining atoms. The bond order of a molecule can be calculated using the following equation.
Bond order\(={N_b-N_a\over 2}\)
Where, N_b = Total number of electrons present in the bonding molecular orbitals
N_a = Total number of electrons present in the antibonding molecular orbitals and A bond order of zero value indicates that the molecule doesn't exist.

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)

(ii) 2KMnO₄ + 10 HCI \(\rightarrow\) KCl + MnCl₂ + H₂O + Cl₂
(iii) Balance the equation atomically (except O and H).
2KMnO₄ + 10 HCI \(\rightarrow\) 2KCl + 2MnCl₂ + H₂O + Cl₂
(iv) Balance chlorine atoms by adding HCI and multiplying Cl₂ by 5.
2KMnO₄ + 16HCI \(\rightarrow\) 2KCl + 2MnCl₂ + H₂O + 5Cl₂
(v) To balance O and H, H₂O is multiplied by 8.
2KMnO₄ + 16HCI \(\rightarrow\) 2KCl + 2MnCl₂ + 8 H₂O + 5Cl₂

(i) Number of electrons in outermost shell: The number of electrons present in the outermost shell increases from 1 to 8 as we proceed in a period.
(ii) Number of shells: As we move from left to right in a period the shells remains the same. The number of shells present in the elements corresponds to the period number. For example all the elements of 2^nd period have on 2 shells (K, L)
(iii) Valency: The valency of the elements increases from left to right in a period. With respect to hydrogen, the valency of period elements increases from 1 to 4 and then falls to one. With respect to oxygen, the valency increases from 1 to 7.
(iv) Metallic character: The metallic character of the elements decreases across a period.
For example: 3rd period  

(i) T₁= 265 + 273 = 538 K
T₂ = 100 + 273 = 373 K
\(\eta =\left( \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \right) \times 100\)
\(=\left( \frac { 538-373 }{ 538 } \right) \times 100\)
\(=\frac { 165 }{ 538 } \times 100\)
\(\eta =42.75%\)%
(ii) T₁ = 265 + 273 = 538 K
T₂ = 35 + 273 = 308 K
\(\eta =\left( \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \right) \times 100\)
\( =\left( \frac { 538-308 }{ 538 } \right) \times 100\)
\(=\frac { 230 }{ 538 } \times 100\)
\(\eta =42.75%\)%.

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.

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.

Applying Bohr's postulates to a hydrogen like atom (one electron species such as H, He⁺ and Li²⁺ etc...)the radius of the n^th orbit and the energy of the electron revolving in the n^th orbit were derived. The results are as follows:
r_n = \(\frac { (0.529){ n }^{ 2 } }{ x } \mathring { A } \)  ...(1)
E_n = \(\frac { -13.6({ z }^{ 2 }) }{ { n }^{ 2 } } ev\quad { atom }^{ -1 }\)   or ...  (2) or
E_n = \(\frac { (-1312.8){ z }^{ 2 } }{ { n }^{ 2 } } kJ\quad { mol }^{ -1 }\) .....(3)

(i) CH₃ - OH : Does not exist as resonance hybrid due to absence of π-electrons.
(ii) R-CO NH₂ : Can exist as resonance hybrid due to the presence of non-bonding electrons on N and n-electrons on C = O bond.

(iii) CH₃-CH = CH-CH₂NH₂: Does not exist as resonance hybrid, because the lone pair on N-atom is not conjugated with n-electrons of the double bond

Standard characteristics of drinking water.

Addition of an inert gas to a reaction at equilibrium, at constant volume has no effect.

Gay Lussac's law:
At constant volume, the pressure of a fixed mass of a gas is directly proportional .to temperature.
\(P\propto T\) (or) \(P\over T\) = Constant
It can be graphically represented as shown here:

Lines in the pressure vs temperature graph are known as isochores ( constant volume) of a gas.

This can be removed by boiling the hard water followed by filtration. Upon boiling, these salts decompose into insoluble carbonate which leads to their precipitation. The magnesium carbonate thus formed further hydrolysed to give insoluble magnesium hydroxide.
\(Ca(HCO_3)_2\rightarrow CaCO_3+H_2O+CO_2\)
Mg(HCO₃)₂ \(\rightarrow\) MgCO₃ + H₂O + CO₂
MgCO₃ + H₂O \(\rightarrow\)Mg(OH)₂ + CO₂
The resulting precipitates can be removed by filtration

Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation in moles.

T = 298K ; ΔU= - 742.4 kJ mol^-1
ΔH=?
ΔH=ΔU+Δn_gRT
ΔH=ΔU+(n_p-n_r)RT
ΔH=-742.4 +\((2-\frac{3}{2})\)\(\times\)8.314 \(\times\) 10^-3 \(\times\) 298
=-742.4 + (0.5 \(\times\) 8.314 \(\times\)10^-3 \(\times\)298)
=-742.4 + 1.24
=-741.16 kJ mol^-1

(i) Lithium metal is used to make useful alloys. For example with lead it is used to make 'white metal' bearings for motor engines, with aluminium to make aircraft parts, and with magnesium to make armour plates. It is used in thermonuclear reactions.
(ii) Lithium is also used to make electrochemical cells.
(iii) Lithium carbonate is used in medicines
(iv) Sodium is used to make Na/Pb alloy.
(v) Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
(vi) Potassium chloride is used as a fertilizer. Potassium hydroxide is used in the manufacture of soft soap. It is also used as an excellent absorbent of carbon dioxide.
(vii) Caesium is used in devising photoelectric cells.

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 

Calcium Sulphate (plaster of paris), CaSO₄. 1/2H₂O :
It is a hemihydrate of calcium sulphate. It is obtained when gypsum, CaSO₄ .2H₂O,is heated to 393 K
2(CaSO₄.2H₂O) ⟶ 2CaSO₄,H₂O + 3H₂O
Above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, CaSO₄ is formed. This is a known 'dead burnt plaster'.
It has a remarkable property of setting with water. on mixing with an adequate quantity of water it forms a plastic mass that gets into hard solid in 5 to 15 minutes.

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)

The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is generally known as the metallic bond.

a) The product formed is ethylmagnesium iodide (Grignard reagent)
\({ C }_{ 2 }{ H }_{ 5 }+Mg\overset { dryether }{ \rightarrow } { { C }_{ 2 }H }_{ 5 }Mgl\)
                                 Ethyl magnesium iodide
b) The Grignard carbon is highly basic and reacts with acidic protons of polar solvents like water to form an alkani so all reagents should be pure and dry.

This is one method to purify and separate liquids present in the mixture having their boiling point close to each other. In the fractional distillation, a fractionating column is fitted with distillation ask and a condenser. A thermometer is fitted in the fractionating column near the mouth of the condenser. This will enable to record the temperature of vapour passing over the condenser. The process of separation of the components in a liquid mixture I at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. The process of fractional distillation is repeated. This method finds remarkable application in distillation of petroleum, coal-tar and crude oil.

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 = \(\frac{30}{18}=1.67 \mathrm{~mol} \)
\(\mathrm{X}_{1}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{1.67}{1.67+0.016}=0.99\)
\(\mathrm{X}_{1}+\mathrm{X}_{2}=1 \)
\(\therefore \mathrm{X}_{2}=1-\mathrm{X}_{1}=1-0.99=0.01 \).

Molar mass of ¹²C =  12.00 g mol^-1.
Mass of 6.023\(\times\)10²³ carbon atom =  12.0 g
\(\therefore\) Mass of 1 carbon atom \(={12\over6.023\times{10}^{23}}\)
= 1.992\(\times\)10^-23 g.

\(\Delta { G }^{ \ominus }\) = -RT ln K = -2.303 RT log K.
R = 8.314 JK^-1 mol^-1;T = 300 K; K=10
\(\Delta { G }^{ \ominus }=-2.303 \times 8.314 \)  JK^-1 mol^-1 x (300 K) x log 10
= -5527 J mol^-1= -5.527 kJ mol^-1

\(\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}\)
r_A and r_B rates of diffusion of gases A and B and the M_A and M_B are their molecular weighs.

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