Calculation of K_P

Total no. of moles in the equilibrium mixture = 1 - α + α + α = (1 + a) mol.
Let the total pressure of equilibrium mixture =p atm
Partial pressure of PCI_5,\({ PCl }_{ 3 }=\cfrac { \alpha }{ 1+\alpha } \times p\quad atm\) 
Partial pressure of \({ PCL }_{ 3 }=\cfrac { \alpha }{ 1+\alpha } \times p\quad atm\) 
Partial pressure of Cl₂=\({ PCl }_{ 2 }=\cfrac { \alpha }{ 1+\alpha } \)
\({ K }_{ P }=\cfrac { P_{ PCl_{ 3 } }\times { P }_{ cl_{ 2 } } }{ { P }_{ PC{ l }_{ 5 } } } =\cfrac { \left( \frac { \alpha }{ 1+\alpha } p\quad atm \right) \times \left( \cfrac { \alpha }{ 1+\alpha } p\quad atm \right) }{ \frac { 1-\alpha }{ 1+\alpha } p\quad atm } =\cfrac { { a }^{ 2 }p }{ 1-{ a }^{ 2 } } \)
P=4 atm and α.= 10%= \(\cfrac { 10 }{ 100 } =0.1\)
\({ K }_{ P }=\cfrac { \left( 0.1 \right) \times \left( 0.1 \right) \times \left( 4atm \right) }{ 1-\left( 0.1 \right) ^{ 2 } } =\cfrac { 0.04 }{ 0.99 } =0.004atm\)
= 0.96 atm

(i) According to molecular orbital theory the atomic orbitals of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band gap.
(ii) The bonding molecular orbitals are completely filled with an electron pair in each and the anti-bonding molecular orbitals are empty.
(iii) Absence of band gap accounts for high electrical conductivity of metals.
(iv) High thermal conductivity is due to thermal excitation of many electrons from the valence band to the conduction band.
(v) With an increase in temperature, the electrical conductivity decreases due to vigorous thermal motion of lattice ions that disrupts.

(i) The vapour pressure of a solution containing a nonvolatile, non-electrolyte solute is always lower than the vapour pressure of the pure solvent. Consider a closed system in which a pure solvent is in equilibrium with its vapour. At equilibrium the molar Gibb's free energies of solvent in the liquid and gaseous phase are equal (\(\triangle\)G = 0). When a solute is added to this solvent, the dissolution takes place and its free energy (G) decreases due to increase in entropy. In order to maintain the equilibrium, the free energy of the vapour phase must also decrease. At a given temperature, the only way to lower the free energy of the vapour is to reduce its pressure. Thus the vapour pressure of the solution must decrease to maintain the equilibrium.

We know that from the Raoult's law the relative lowering of the vapour pressure is equal to the mole fraction of the solute From the above equation,it is clear that the relative lowering of vapour pressure depends only on the mole fraction of the solute (x_B)and is independent of its nature. Therefore, relative lowering of vapour pressure is a colligative property.
(ii) \(\frac { \triangle P }{ { P }_{ A }^{ 0 } } =\frac { { W }_{ B }\times { M }_{ A } }{ { M }_{ B }\times { W }_{ A } } \)
\(\frac { \triangle P }{ { P }_{ A }^{ 0 } } \) - relative lowering of vapour pressure
W_A and W_B - weights of solvent and solute.
M_A and M_B  - Molar masses of solvent and solute

(i) Solvay process - in this process ammonia is converted to ammonium carbonate, which is then converted to ammonium bicarbonate by passing excess carbon dioxide in sodium chloride solution saturated with ammonia.
(ii) The ammonium bicarbonate formed reacts with soctium chloride to give sodium bicarbonate. As sodium bicarbonate has poor solubility, it gets precipitated.
(iii) The sodium bicarbonate is isolated and is heated to give sodium carbonate.
(iv) The equations involved in this process is as below:
i) \(2NH_3 + H_2O + CO_2 ⟶\underset{Ammonium\ carbonate}{ (NH_4)2CO_3}\)
ii) \((NH_4)2CO_3 + H_2O + CO_2 ⟶ \underset{Ammonium\ bicarbonate}{2NH_4HCO_3}\)
iii) \(NH_4HCO_3 + NaCl⟶\underset{Ammonium\ chloride}{NH_4Cl + NaHCO_3}\)
iv) \(2NaHCO_3\xrightarrowΔ\underset{Sodium\ carbonate}{Na_2CO_3 }+ CO_2 + H_2O\)

The variation of the probability of locating the electron on a sphere with nucleus at its centre depends on the azimuthal quantum number of the orbital in which the electron is present.

For 1s orbital, l = 0, m = 0,\(f(\theta)=\frac{1}{\sqrt 2}\)and \(g(\varphi)=\frac{1}{\sqrt 2\pi}\) Therefore, the angular distribution function is equal to \(\frac{1}{2\sqrt \pi }\) 
i.e. it is independent of the angle \(\theta\) and \(\varphi\). Hence, the probability of finding the electron is independent of the direction from the nucleus. So, the shape of the s orbital is spherical.

For p orbitals, l = 1 and the corresponding m values are -1, 0 and +1. The angular distribution functions are quite complex and are not discussed here, The shape of the p orbital is shown in Figure (b). The three different m values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as P_x, P_y and P_z and the angular distribution for these orbitals shows that the lobes are along the x, y arid z axis respectively. As seen in the Figure the 2p orbitals have one nodal plane

For 'd' orbital l = 2 and the corresponding m values are -2, -1, 0, +1, +2. The shape of the d orbital looks like a 'clover leaf'.
The five m values give rise to five d orbitals namely d_xy, d_yz, d_zx, d_x²-y² and d_z². The 3d orbitals contain two nodal planes.

For 'f' orbital, l = 3 and the m values are -3, -2,-1,0, +1, +2, +3 corresponding to seven f orbitals. f_z³, f_xz²,f_yz²,f_xyz,f_{z(x² - y²)},f_{x(x² - 3y²)},f_{y(3x² - y²)} which are shown in Figure. There are 3 nodal planes in the f-orbitals.

Step - 1 : To find out atoms undergoing change in O.N.
\(\overset { +1+7-2 }{ KMnO_{ 4 } } +\overset { +1-2+1 }{ KOH } +\overset { +1-1 }{ { H }_{ 2 }{ O }_{ 2 } } \rightarrow +\overset { +1+6-2 }{ K_{ 2 }MnO_{ 4 } } +\overset { 0 }{ { O }_{ 2 } } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step - 2 : To find out total increase and decrease in O.N .
\(\overset { +7 }{ KMnO_{ 4 } } \rightarrow \overset { +6 }{ K_{ 2 }MnO_{ 4 } } \) (decrease of 1 unit per atom)
\(\overset { -1 }{ { H }_{ 2 }{ O }_{ 2 } } \rightarrow \overset { 0 }{ { O }_{ 2 } } \) (increase of 1 unit per atom or 2 units per atom)
Total increase = 2
Total decrease = 2
Step - 3 : To balance the total increase and decrease in O.N, multiply KMnO₄ by 2.
2KMnO₄ + KOH + H₂O₂ ⟶ K2MnO₄ + O₂ + H₂O
Step - 4 : To balance all atoms other than 'H' and 'O'
2KMnO₄ + 2KOH + H₂O₂ ⟶ 2K₂MnO₄ + O₂ + H₂O
number of oxygen atoms on LHS = 12
number of oxygen atoms on RHS = 11
Hence, multiply H₂O in RHS by 2. The equation becomes,
2KMnO₄ + 2KOH + H₂O₂ ⟶ 2K₂MnO₄ + O₂ + H₂O
Hydrogen atoms balance by themselves. The balanced equation is
2KMnO₄ + 2KOH + H₂O₂ ⟶ 2K₂MnO₄ + O₂ + 2H₂O

Heavy water is chemically similar to ordinary water (H₂0). However D₂0 reacts more slowly than H₂0 in chemical reactions.
(i) When compounds containing hydrogen are treated with D₂0, hydrogen undergoes an exchange for deuterium
2NaOH + D₂0 \(\rightarrow\) 2NaOD + HOD
HCI + D₂0\(\rightarrow\) DCI + HOD
NH₄CI + 4D₂0 \(\rightarrow\) ND₄CI + 4HOD .
These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound.
For example, when D₂0 is treated with of hypo-phosphorus acid only one hydrogen atom is exchanged with deuterium. It indicates that, it is a monobasic acid.
H₃P0₂ + D₂0 \(\rightarrow\) H₂DP0₂ + HDO
(ii) It is also used to prepare some deuterium compounds:
Al₄C₃ + 12D₂0 \(\rightarrow\) 4AI(OD)₃ + 3CD₄
CaC₂ + 2 D₂0\(\rightarrow\) Ca(OD)₂ + C2D2
Mg₃N₂ + 6D₂0 \(\rightarrow\) 3Mg(OD)₂ + 2 ND₃
Ca₃P 2+ 6D₂0\(\rightarrow\) 3Ca(OD)₂ +2PD₃

% Efficiency = \([\frac{T_1-T_2}{T_1}]\times100\)
T₁=110+273=383 K.
T₂=25+273=298 K.
% Efficiency = \([\frac{383-298}{383}]\times100\)
=22%.