(i) C-Br is more polar than C-H (due to electronegativity )
(ii) C-O is more polar than C-N
(iii) C-O is more polar than C-S

(i) (a) Carbon Monoxide:
Carbon monoxide is a poisonous gas produced as a result of incomplete combustion of coal are firewood. It is released into the air mainly by automobile exhaust. It binds with haemoglobin and form carboxy haemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced. This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring of eye sight and cardiac arrest.
(b) Carbon dioxide:
Carbon dioxide is released into the atmosphere mainly by the process of respiration, burning of fossil fuels, forest fire, decomposition of limestone in cement industry etc.Green plants can convert CO₂ gas in the atmosphere into carbohydrate and oxygen through a process called photosynthesis. The increased CO₂ level in the atmosphere is responsible for global warming. It causes headache and nausea.
ii) a) Carbon Monoxide :
It binds with haemoglobin and form carbory haemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced. This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring of eye sight and cardiac arrest.
b) Carbon dioxide :
It causes headache and nausea.

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

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.

Boyle's law states that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
\(V\propto\frac{1}{P}\) (T and n are fixed)
If the pressure of the gas increases, volume will decrease and if the pressure of the gas decreases, the volume will increase. So PV = Constant.

Molecular mass of H₂0 = 18
gram molecular mass of H₂0 = 18g
1 mol of H₂0 = 18g of H₂0 = 6.023 x 10²³molecules of water
1.8g of H₂0 = \({6.023\times10^{23}\over18}\times1.8 \)
\(=6.023\times10^{22} molecules\)

Hydrogen bonding is because of dipole-dipole interactions between the molecule in which hydrogen atom is covalently bonded to a highly electronegative atom.

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

6Li + N₂ ⟶ 2Li₃ N

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)

Freons (CFC): The chloro uoro derivatives of methane and ethane are called freons.
Uses:
(i) Freons are a used as refrigerants in refrigerators and air conditioners. 
(ii) It is used as a propellant for aerosols and foams
(iii) It is used as propellant for foams to spray out deodorants, shaving creams, and insecticides

The scope of thermodynamics:
(i) To derive feasibility of a given process.
(ii) It also helps in predicting how far a physical (or) chemical change can proceed, until the equilibrium conditions are established.

\(Equivalent \ mass ={sum \ of \ the \ atomic \ masses \ of \ atoms \ present \ in \ the \ ion \over charge \ on \ the \ ion}\)
For CO₃ ^-2 ion
\(Equivalent \ mass ={1 x atomic \ mass \ of C + 3 x atomic \ mass \ of 'O'\over 2}\)
\(={12+3\times 16\over 2}={60\over2}=30\ g\ eq^{-1}\)

It is defined as the number of molecules of a gas that get diffused in unit time.
rate of diffusion = \(\frac{Volume\ of \ the\ gas\ diffused}{time\ taken}\)

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

Efflorescence is the spontaneous loss of water by a hydrated salt, which occurs when the aqueous vapor pressure of the hydrate is greater than the partial pressure of the water vapour in the air. This is the property of salts
Ex: Glauber's salt- Na₂SO₄·10H₂O
Epsom salt - MgSO₄·7H₂O

(i) The aromatic hydrocarbon Ⓐ is Benzene, C₆H₆.
(ii) Benzene reacts. with Cl₂ to give chlorobenzene (C₆H₅CI), as Ⓑ as the product.

(iii) Chlorobenzene Ⓑ acts with sodium hydroxide to give (C₆HsOH) phenol, © as the product.

(iv) Chlorobenzene reacts with ammonia to give Aniline, C₆H₅NH₂Ⓓ as the product.

(i) When atoms combine to form molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.
(ii) The shape 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 and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.
(iv) The bonding molecular orbitals are represented as σ" (sigma), π (Pi), ઠ(delta) and the corresponding anti-bonding orbitals are called σ*π"*, 7t* and ઠ*.
(v) The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Aufbau's Principle, Pauli's exclusion principle and Hund's rule as in the case of filling of electrons in the atomic orbitals.
(vi) 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:
\(Bond\quad order=\cfrac { { N }_{ b }-{ N }_{ a } }{ 2 } \)
N_b = Number of electrons in bonding molecular orbitals.
N_a =; Number of electrons in anti-bonding molecular orbitals.
(vii) A bond order of zero value indicates that the molecule does not exist.

Step-1: To find atoms undergoing change in O.N.
\(\overset { +2-2 }{ CuO } +\overset { -3+1 }{ { NH }_{ 3 } } \rightarrow \overset { 0 }{ Cu } +\overset { 0 }{ { N }_{ 2 } } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step-2: To find total increase and decrease in O.N.
CuO ➝ Cu (decrease of 2 unit per atom)
NH3 ➝ N2 (increase of3 unit per atom)
Total decrease = 2 x 3 = 6
Total increase = 3 x 2 = 6
Step-3: To balance the total decrease and increase in O.N, multiply two by 3 and NH3 by 2.
3CuO + 2NH₃ ⟶ Cu + N₂ + H₂O
Step-4: To balance all atoms other than 'H' and 'O'
3CuO + 2NH₃ ⟶ 3Cu + N₂ + H₂O
Step-5: To balance 'O' atoms
3CuO + 2NH₃ ⟶ 3Cu + N₂ + 3H₂O
hydrogen atoms balance by themselves.
Hence the final equation is 
3CuO + 2NH₃ ⟶ 3Cu + N₂ + 3H₂O

Energy acquired by the electron (as kinetic energy) after being accelerated by a potential difference of 1 KV
i.e., 1000 volts = 1000 eV
= 1000 x 1.602 x 10^-19 J
= 1.602 x 10^-16 J (1 eV = 1.602 x 10^-19 J)
Energy in Joules = charge on the electron in coulombs x potential difference in volts
i.e., Kinetic Energy (KE) = \(\frac{1}{2}\) mv²
= 1.602 x 10^-16 J or \(\frac{1}{2}\) x 9.1 x 10^-31 v²
v²  = 3.521 x 10¹⁴ or v = 1.88 x 10^-1 ms^-1
\(\therefore \lambda=\frac{h}{mv}=\frac{6.626\times 10^{-34}kgm^2 s{-1}}{9.1\times 10^{-31}kg\times 1.88\times 10^{-1}ms^{-1}}\)
= 3.87 x 10^-11 m.

\({ H }_{ 2(g) }+\frac { 1 }{ 2 } { O }_{ 2 }\longrightarrow { H }_{ 2 }{ O }_{ (1) }\quad \quad { \Delta H }^{ o }=-285.8\quad KJ\quad (1)\)
\({ C }_{ 9graphite) }+{ O }_{ 2 }\longrightarrow { CO }_{ 2 }\quad ​​​​{ \Delta H }^{ o }=-393.5\quad KJ\quad (2)\)
\({ CH }_{ 4(g) }+{ 2O }_{ 2 }\longrightarrow { CO }_{ 2(g) }+{ 2H }_{ 2 }{ O }_{ (1) }\quad { \Delta H }^{ o }=-890\quad KJ\quad (3)\)
equation (1) X₂ + (2) - (3)
\({ C }_{ (graphite) }+{ 2H }_{ 2(g) }\longrightarrow { CH }_{ 4(g) }\quad { \Delta H }_{ f }^{ o }=-74.7\quad KJ\)
(2 x -285.8) + (-393.5) - (-890.4)
= -571.6 - 393.5 + 890.4
= -965.1 + 890.4
= -74.7 KJ

(i) ₄Be - 1s² 2s² ; ₅B - 1s² 2s² 2p¹

• The I.E. of Be is more than that of B though the nuclear charge of boron atom (Z = 5) is greater than that of beryllium atom (Z = 4). This can be explained as follows.
• Boron atom (Z = 5; 1s² 2s² 2P_x¹ 2p_y⁰ 2p_z⁰) is having one unpaired electron in the 2p-subshell. Be-atom (Z = 4; 1s² 2s²) is having paired electrons in the 2s-subshell.
• As the fully filled 2s-subshell in Be-atom is more stable than B-atom due to symmetry, more energy would be needed to remove an electron from Be-atom. Hence, Be has high I. E. Hence I.E of Be > B

(ii) ₇N - 1s² ,2s² ,2p_x¹, 2p_y¹, 2p_z¹
₈O -1s² ,2s² ,2p_x², 2p_y¹, 2p_z¹
∴ O has lower I.E. than N.

The group metal - 1 which is present in common salt is sodium.
So (A) is sodium
Sodium reacts with hydrogen (B) to give, sodium hydride (C). In sodium hydride the hydrogen is present in -1 oxidation state.
\(2\underset { (A) }{ Na } +\underset { (B) }{ { H }_{ 2 } } \rightarrow 2\underset { (C) }{ NaH } \)
So (B) hydrogen and (C) is sodium hydride
H₂ reacts with oxygen gas (D) to give an universal solvent, water (E) follows:
\(2\underset { (B) }{ { H }_{ 2 } } +{ O }_{ 2 }\rightarrow 2\underset { (D) }{ { H }_{ 2 }O } \)
So (E) is water. Water is the universal solvent
Water (E)  reacts with sodium  (A) follow to give (F), which is a strong base.
\(2\underset { (E) }{ { H }_{ 2 }O } +\underset { (A) }{ 2Na } \rightarrow 2\underset { (F) }{ NaOH } +{ { H }_{ 2 } } \)
So (F) is sodium hydroxide.

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