(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

\(P_A^o\) = 10 torr, P_solution = 9 torr
W_A = 20 g  W_B = 1 g
M_A = 200 g mol^-1  M_B = ?
\({\Delta P\over P_A^o}={W_B\times M_A\over M_B\times W_A}\)
\({10-9\over 10}={1\times 200\over M_B\times 20}\)
\(M_B={200\over 20}\times 10=100\ g\ mol^{-1}\)

(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.

PV = Boyle's constant.
But \(PV=nRT=\frac{w}{M}\times RT\)
=\(\frac{8}{10} mol \times0.0821 L atm K^{-1} mol^{-1}\times 300 K \)
= 12.315 L atm.
Hence Boyle's constant is 12.315 L atm.

T=298K
\(\Delta { G }_{ r }^{ 0 }\)=- 13.6 kJ mol^-1
=-13600 J mol^-1
ΔG⁰=- 2.303 RT log K_eq
log K_eq=\(\frac { 13.6kJmol^{ -1 } }{ 2.303\times 8.314\times { 10 }^{ -3 }JK^{ -1 }mol^{ -1 }\times 298K } \)
log K_eq = 2.38
K_eq = antilog (2.38)
K_eq = 239.88.

4(1) + 2x + 7(-2) = 0
4 + 2x - 14 = 0
2x - 10 = 0
2x = 10
x = 5
Oxidation number of P in H₄P₂O₇ is +5.

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.

De Broglie wavelength of the tennis ball equal to 5400 \(\overset { 0 }{ A } \).
m = 54 g 
V = ?
\(\lambda=\frac{h}{mV}\)
\(V=\frac{h}{m\lambda}\)
\(\mathrm{v}=\frac{6.626 \times 10^{-34} \mathrm{JS}}{54 \times 10^{-3} \mathrm{~kg} \times 5400 \times 10^{-10} \mathrm{~m}}=2.27 \times 10^{-26} \mathrm{~ms}^{-1}\)

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) 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).

(i) An organic compound ®of molecular formula C₂H₆0 is Ethanol CH₃-CH₂OH.
(ii) Ethanol reacts with thionyl chloride in the presence of pyridine to give CH₃-CH₂Cl Ⓑ as product

(iii) Ethyl chloride on treatment with alcoholic KOH, undergoes dehydrohalogenation to give C₂H₄, ethylene © as the product.

(iv) Ethylene on reaction with Cl₂ yield ethylene dichloride as Ⓓ as the product.

(i) Mutual sharing of one or more pair of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond
(ii) If two atoms share just one pair of electron, a single covalent bond is formed as in the case of hydrogen molecule (H₂).
(iii) If two or three electron pairs are shared between the two combining atoms, then the covalent bond is called double bond and triple bond respectively, as in the case of O₂ and N₂ molecules respectively.

\(K_c={[NH_3]^4[O_2]^5\over [NO]^4[H_2O]^6}\)
\(4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \rightleftharpoons 4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})}\)

(ii) Balance the changes in O, N, by multiplying the oxidant and reductant by suitable numbers.
4MnO₂ + 2O₂ + KOH\(\rightarrow\)K₂MnO₄ + H₂O
(iii) Balance the equation atomically (except O and H).
4MnO₂ + 2O₂ + KOH\(\rightarrow\) 4K₂MnO₄ + H₂O
(iv) Balance oxygen and hydrogen atoms by multiplying H₂O by 4.
4MnO₂ + 2O₂ + 8KOH \(\rightarrow\) 4K₂MnO₄ + 4H₂O

(a) Atomic radius is the distance between the centre of its nucleus and the outermost shell containing the electron.
(b) Difficulties in determining atomic radius
(i) The size of an atom is very small (~1.2\(\mathring A\) i.e 1.2 x 10 ^-10)
(ii) The atom is not a rigid sphere; it is more like a spherical cotton ball rather than like a cricket ball.
(iii) It is not possible to isolate an atom and measure its radius.
(iv) The size of an atom depends upon the type of atoms in its neighborhood and also the nature of bonding between them.

Louis de Broglie proposed that all forms of matter showed dual character. To quantify this relation, he Jerived an equation for the wavelength of a matter wave. He combined the following two equations of energy of which one represents wave character (hu) and the other represents the particle nature (mc²).
Planck's quantum hypothesis:
E = hv      ....(1)
Einsteins mass-energy relationship:
E = mc²     .... (2)
From (1) and (2)
hv = mc²
hc/\(\lambda\) = mc²
\(\therefore \lambda ={h\over mc}\)        ...(3)
The equation (3) represents the wavelength of photons whose momentum is given by mc. (Photons have zero rest mass).
For a particle of matter with mass m and moving with a velocity v, the equation (3) can be written as
 \(\lambda ={h\over mv}\)       ....(4)
This is valid only when the particle travels at speed much less than the speed of Light.

(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)

\(CO+\frac { 1 }{ 2 } { O }_{ 2 }\longrightarrow { CO }_{ 2 }\)
\({ \triangle }G_{ (reaction) }^{ 0 }={ \sum { G } }_{ f(products) }^{ 0 }-{ \sum { G } }_{ f(reactants) }^{ 0 }\)
\({ \triangle G }_{ (reaction) }^{ 0 }=\left[ { G }_{ { CO }_{ 2 } }^{ 0 } \right] -\left[ { G }_{ CO }^{ 0 }+\frac { 1 }{ 2 } { G }_{ { O }_{ 2 } }^{ 0 } \right] \)
\({ \triangle G }_{ (reaction) }^{ 0 }\) =-394.4+[137.2+0]
\({ \triangle G }_{ (reaction) }^{ 0 }\) =-257.2 kJ mol^-1
\({ \triangle G }_{ (reaction) }^{ 0 }\) of a reaction at a given temperature is negative hence the reaction is spontaneous.

The element which occupies group number (16) and period number (2) is oxygen. (B) is D₂O which is used as a moderator in nuclear reactions.
So (A) must be deuterium, which is an isotope of hydrogen
    \(2\underset { (A) }{ { D }_{ 2 } } +{ O }_{ 2 }\rightarrow 2\underset { (B) }{ { D }_{ 2 }O } \)
So (B) is D₂O
(A) adds to (C) as follows :
\(3 \mathrm{D}_{2}+\mathrm{C}_{3} \mathrm{H}_{6} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{2}\)
So (D) is 1,2 - dideutero propane. 

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.

(i) Low ionisation enthalpy of metal atoms.
(ii) High electron gain enthalpy of non-metal atoms.
(iii) High lattice enthalpy of compound formed.

i) Acetic acid: Solid carbon dioxide reacts with Grignard reagent to form addition product which on hydrolysis yields carboxylic acids.

ii) Acetone :
\(\underset { Iodomethane }{ { CH }_{ 3 }-I+Mg } \overset { dryether }{ \longrightarrow } \underset { Methyl \ magnesium \ iodide }{ { CH }_{ 3 }MgI } \)
iii) Ethyl acetate: Ethy1chloroformate reacts with Grignard reagent to form esters.

iv) Iso propyl alcohol : Aldehydes other than formaldehyde, react with Grignard reagent to give addition product which on hydrolysis yields secondary alcohol.

v) Methyl cyanide : Grignard reagent reacts with cyanogen chloride to from alkyl cyanide

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.

Molar heat of sublimation is defined as the change in enthalpy when one mole of a solid is directly converted into the gaseous state at its sublimation temperature.

(i) A hot air inside the balloon rises because of its decreased density and causes the balloon to float.
(ii) If you take a helium balloon outside on a chilly day, the balloon will crumble. Once you get back into warm area, the balloon will return to its original shape. This is because, in accordance with Charles' law, a gas like helium takes up more space when it is warm.

NO₃^- (N in NO₃^-)
Oxidation number of nitrogen = x
Oxidation number of oxygen = - 2
x + 3 (- 2) = - 1
x - 6 = -1
x = + 5
Hence, the oxidation number of nitrogen in NO₃^- is + 5.

iv) (A) is correct but( R) is not correct