The temperature at which the liquid and vapour phases are at equilibrium is called the boiling point and condensation point of the liquid.

The Lewis structure for BeH₂ molecule is as follows : H : Be : H
There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, BeH₂ is of the type AB₂. It has a linear structure.

Dipole moments of each Be-H bond are equal and opposite in direction. Therefore, they nullify each other. Hence, BeH₂ has a net zero dipole moment.

(i) The vapour pressure of the solution increases with increase in temperature. The variation of vapour pressure with respect to temperature of pure water is given by the curve-A.
(ii) At 100^oC, the vapour pressure of water is equal to 1 atm. Hence, the boiling point of water is 100^o-C (T^o_b)'
(iii) When a solute is added to water, the vapour pressure of the resultant solution is lowered. The variation of vapour pressure with respect to temperature for the solution is given by curve-B. 
(iv) From the graph, it is evident that the vapour pressure of the solution is equal to 1 atm. pressure at the temperature T_b which is greater than T^o_b' The difference between these two temperatures (T_b - T^o_b) gives the elevation of boiling point.
\(\Delta { T }_{ b }={ T }_{ b }-{ { T }_{ b }^{ o } }\)

This method is to purify liquids from non-volatile impurities, and used for separating the constituents of a liquid mixture which differ in their boiling points.
There are various methods of distillation depending upon the difference in the boiling points of the constituents. The methods are
(i) simple distillation
(ii) fractional distillation and
(iii) steam distillation.
The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver. is method is called simple distillation. Liquids with large difference in boiling point (about 40K) and do not decompose under ordinary pressure can be purified by simply distillation Eg. The mixture of C₆H₅NO₂ (b.p 484K) & C₆H₆(354K) and mixture of diethyl ether (b.p 308K) and ethyl alcohol (b.p 351K)

Temporary hardness of water is due to the presence of bicarbonates of calcium and magnesium in water i.e., Ca(HCO₃)₂ and Mg(HCO₃) in water. Permanent hardness of water is due to the presence of soluble chlorides and sulphates of calcium and magnesium i.e., CaCI₂, CaSO₄, MgCl₂ and MgSO₄.

(i) Size of the atom: The new electron which was added experiences stronger attraction to its nucleus if the atoms are smaller in size.
\(Atomic\ size \ \infty{1\over Electron \ gain \ enthalpy}\)
(ii) Nuclear charge: The new electron which was added experiences stronger attraction to its nucleus if the atom possess greater nuclear charge.
Nuclear charge \(\infty\) Electron gain enthalpy
(iii) Electronic configuration: An atom with stable electronic configuration has no tendency to gain an electron. Such atoms have zero or almost zero electron gain enthalpy.

Sodium hydroxide is prepared commercially by the electrolysis of brine solution in Castner-Kellner cell using a mercury cathode and a carbon anode. Sodium metal is discharged at the cathode and combines with mercury to form sodium amalgam. Chlorine gas is evolved at the anode. The sodium amalgam thus obtained is treated with water to give sodium hydroxide.
At cathode: Na⁺ + e^- ⟶ Na(amalgam)
At anode: Cl^- ⟶ 1/2Cl₂ ↑ + e^-
2Na(amalgam) + 2H₂O ⟶ 2NaOH + 2Hg + H₂↑

Heat of Transition: The heat of transition is defined as "The change in enthalpy when one mole of an element changes from one allotropic form to another. For example, the transition of diamond into graphite may be represented as
C_(diamond) ⟶ C_(graphite) ΔH_trans = + 13.81 KJ
S_(monoclinic) ⟶ S_(rhombic) ΔH_trans = -0.067 KJ

Equivalent weight of H₃PO₄
= \(\frac { Molecular\ mass }{ No.of\ replaceable\ { H }^{ + } } =\frac { 98 }{ 1 } =98\)
Equivalent weight of Ca(OH)₂
= \(\frac { Molecular\ mass }{ No.of\ replaceable\ {O H }^{ - } } =\frac { 74 }{ 1 } =74\)

1. Heavy atoms like gold, platinum have heavy nucleus. Heavy nucleus contains large amount of positive charge.
2. When a beam of \(\alpha\)-particles is shot at a thin gold foil, most of them pass through without much effect.
3. Some are deflected back due to enormous repulsive force of heavy nucleus.
4. If thin foil is made of light aluminium, then the number of \(\alpha\) -particles deflected back will be negligible.