W₁ = 1 kg
W₂ = 68.4g
M₂ = 342 g mol^-1
Tb = 373.1K
Kb = 0.52K.kg mol^-1
T-TO = Kbm
M = 0.2 
T - T^O = K_b x m
T - T^O = 0.52 x 0.2 
T - T^O = 0.104
T - 373 .1 = 0.104
T = 0.104 + 373.1
T = 373 .204

(ii) Bond order \(\alpha\) Stability
Greater the bond order, more stable the molecule is
Eg: N₂(B.O = 3) is much stable than O₂(B.O = 2)
Bond order \(\alpha {1\over Bond \ length}\)
Greater the bond order shorter is the bond length.
Eg: Nitrogen: B.O. = 3; Bond length =110pm Oxygen: B.O. = 2; Bond length = 121pm.

According to Le Chatelier's principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance.
Let us consider the reaction
\({ H }_{ 2 }\left( g \right) +{ I }_{ 2 }\left( g \right) \rightleftharpoons 2HI\left( g \right) \)
The addition of H₂ or I₂to the equilibrium mixture, disturbs the equilibrium. In order to minimize the stress, the system shifts the reaction in a direction where H₂ and I₂ are consumed i.e., the formation of additional HI would balance the effect of added reactant. Hence, the equilibrium this to the right (forward direction) i.e. the forward reaction takes place until the equilibrium is re established. Similarly, removal of HI (product) also favours the forward reaction.

(a) Molarity = \(\frac { Number\ of\ moles\ of\ solute }{ Volume\ of\ the\ solution\ in\ L } \)
Given Molarity(m) = 0.250M
No. of moles = 0.110 mol
Volume of solution = ?
Volume of the Solution in L = \(\frac { 0.110 }{ 0.250 } \) = 0.44 L
(b) Molarity = \(\frac { Number\ of\ moles\ of\ solute }{ Volume\ of\ the\ solution\ in\ L } \)
Molarity (M) = 0.20 M
No. of moles(m) = 0.150m
Volume of the solution = ?
Voilume of solution in Litre = \(\frac { 0.150 }{ 0.20 } \)
= 0.75 L
= 750 ml

Solid in liquids:
1.When you add sugar to water at a particular temperature, it dissolves to form sugar solution. If you continue to which the added sugar remains as solid and the resulting solution is called a saturated solution. Here, as in the previous cases a dynamic equilibrium is established between the solute molecules in the solid phase and in the solution phase.
Sugar (Solid) In this process \(\rightleftharpoons \) Sugar (Solution)
Rate of dissolution of solute =Rate of crystallisation of solute
Gas in liquids:
1.When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecules in the gaseous state and those dissolved in the liquid.
2. In carbonated beverages the following equilibrium exist
\({ CO }_{ 2 }\left( g \right) \rightleftharpoons { CO }_{ 2 }\left( s \right) \)
3. Henry's law is used to explain such gas-solution equilibrium processes

(i) The B-in BF₃ undergoes sp² hybridisation and has a triangular planar geometry. The N-in NH3 undergoes sp³ hybridisation and has pyramidal shape with one lone pair on N-atom.
(ii) Both S in SiF₄ and CI in ClO; undergo sp³ hybridisation.
(iii) Since the electronegativity of s is less than 0, H₂S possess a lower bond angle then H₂0.

(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

(i) Ripening of fruits and vegetables in few days.
(ii) Tarnishing of silver in few months.
(iii) Rusting of iron slowly.

Among anthracene and cyclopentadiene - anthracene is aromatic in nature.

Anthracene		Cyclopentadiene
1.	Planar structure	Planar structure
2.	Contains 14 delocalised \(\pi\) electrons (4n + 2 = 14\(\pi\)e-)	Contains 4 non-delocalised \(\pi\) electrons
3.	Hence aromatic	Hence non-aromatic

Mole fraction of solute = \(\frac { Moles\ of\ the\ component }{ Total\ number\ of\ moles\ of\ all } \)
No.of moles = \(\frac { Mass }{ Molecular\ mass } \)
No. of moles of ethanol = \(\frac { 56 }{ 46 } \) = 1.28 moles
No. of moles of water = \(\frac { 36 }{ 18 } \) = 2 moles
No. of moles fraction of ethyl/alchohol = \(\frac { 1.28 }{ \left( 1.28+2 \right) \quad } \)
= \(\frac { 1.28 }{ 3.28 } \) = 0.39

Isomerism:
Presence of double bond in alkene provides the possibility of both structural and geometrical isomerism.
Structural Isomerism:
The first two member's ethene C₂H₄ and propene C₃H₆ do not have isomers because the carbon atoms in the molecules can be arranged only one distinct way.
However from the third member of alkene family butene C₄H₁₀ structural isomerism exists.
(i) CH₃-CH = CH-CH₃  1-Butene
(ii) CH₂=CH-CH₂-CH₃ 2-Butene
(iii) \({ CH }_{ 2 }=\overset { \underset { | }{ { CH }_{ 3 } } }{ C } -{ CH }_{ 3 }\) 2-Methyl-1-propene
structures (i) & (ii) are position isomers. structures (i) & (iii), (ii) & (iii) are chain isomers. 
Geometrical isomerism:
It is a type of stereoisomerism and it is also called cis-trans isomerism. Such type of isomerism results due to the restricted rotation of doubly bounded carbon atoms.
if the similar groups lie on the same side, then the geometrical isomers are called C is-isomers. When the similar groups lie on the opposite side, it is called a Trans isomer.
for example: the geometrical isomers of 2-Butane is expressed as follow

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)

Restricted rotation around C=N (oximes) gives rise· to geometrical isomerism in oximes. Here 'syn' and 'anti' are used instead of cis and trans respectively. In the syn isomer the H atom of a doubly bonded carbon and -OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond. For Eg:

Homologous series: A series of organic compounds each containing a characteristic functional group and the successive members differ from each other in molecular formula by a CH₂ group is called homologous series.
Alkanes: Methane (CH₄), Ethane (C₂H₆), Propane (C₃H₈) etc.
Alcohols: Methanol (CH₃OH), Ethanol (C₂H₅OH) Propanol (C₃H₇OH) etc. Compounds of the homologous series are represented by a general formula.
Alkanes C_nH_2n+2, Alkenes C_nH_2n, Alkynes C_nH_2n-2 and can be prepared by general methods. They show regular gradation in physical properties but have almost similar chemical property.

(i) The arrangement of atoms must be identical or almost same in every resonance structure.
(ii) The energy content of all the canomical forms must be almost same.
(iii) Each canocial form must have the same number of unpaired electrons.

The type of mutual sharing of one or more pairs of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond. If two atoms share just one pair of electron a single covalent bond is formed as in the case of hydrogen molecule. If two or three electron pairs are shared between the two combining atoms, then the covalent bond is called a double bond or a triple bond, respectively

True.
Some haloalkanes yield a mixture of olefins in different amounts. It is explained by Saytzeff's Rule, which states that 'In a dehydrohalogenation reaction, the preferred product is that alkene which has more number of alkyl groups attached to the doubly bonded carbon (more substituted double bond is formed).
\({ CH }_{ 3 }-\underset { \overset { | }{ Br } }{ CH } -{ CH }_{ 2 }-{ CH }_{ 3 }\)
2-Bromopropane

(i) Silver Oxide(moist):

(ii) Sodium hydrogen sulphide:
\(\underset { Bromo \ ethane }{ { CH }_{ 3 }{ CH }_{ 2 }Br+NaSH } \overset { alcohol/{ H }_{ 2 }O }{ \underset { \triangle }{ \longrightarrow } } \underset { Ethan \ ethiol }{ { CH }_{ 3 }{ CH }_{ 2 }SH+NaBr } \)
(iii) Potassium cyanide :
\(\underset { Bromo \ ethane }{ { CH }_{ 3 }{ CH }_{ 2 }-Br+KCN } \longrightarrow \underset { Ethyl \ cryanide }{ { CH }_{ 3 }{ CH }_{ 2 }-CN+KBr } \)

\(\overset { \underset { | }{ Cl } }{ \underset { \overset { | }{ Cl } }{ CH } } -{ CH }_{ 3 }\quad (A)\) 1, 1,-dichloro ethane
\(\underset { \overset { | }{ Cl } }{ { CH }_{ 2 } } -\underset { \overset { | }{ Cl } }{ { CH }_{ 2 } } \) (B) 1, 2, dichloro ethane
\(\overset { \underset { | }{ Cl } }{ \underset { \overset { | }{ Cl } }{ CH } } -{ CH }_{ 3 }\underset { aq\quad KOH }{ \longrightarrow } \left[ \overset { \underset { | }{ OH } }{ \underset { \overset { | }{ OH } }{ CH } } -{ CH }_{ 3 } \right] \underset { -{ H }_{ 2 }) }{ \longrightarrow } { CH }_{ 3 }CHO\)
(A)                                                                    (C)
1, 1, dichloro ethane                                      Acetaldhyde
\(A \ \ \overset { \underset { | }{ Cl } }{ \underset { \overset { | }{ Cl } }{ CH } } -{ CH }_{ 3 }\)  -1,1 dichloro ethane
B  \(\underset { \overset { | }{ Cl } }{ { CH }_{ 2 } } -\underset { \overset { | }{ Cl } }{ { CH }_{ 2 } } \)-1, 2 dichloro ethane
C CH₃CHO -Acetaldehyde 
D \(\underset { \overset { | }{ OH } }{ { CH }_{ 2 } } -\underset { \overset { | }{ OH } }{ { CH }_{ 2 } } \)-Ethyleneglycol

(w) = 0.16 g
(x) = 0.35 g
\(\% S=\frac{32}{233} \times \frac{x}{\mathrm{w}} \times 100=\frac{32}{233} \times \frac{0.35}{0.16} \times 100=30.04 \%\)

Weight of solute (W₂) = 20 mass percent of solution means 20 g of ethylene glycol
Weight of solvent (water) W₁ = 100 - 20 = 80 g
ΔT_f = K_f m
\(={K_f\times W_2\times 1000\over M_2\times W_1}\)
\(={1.86\times 20\times 1000\over 62\times 80}\)
= 7.5 K
The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e 7.5 K lower than the normal freezing point of water (273 - 7.5K) = 265.5 K

H₂(g) + I₂(g) \(\rightleftharpoons \) 2HI(g)
At a particular instant, the concentration of H₂, I₂and HI are found to be 0.2 mol L^-1, 0.2 mol L^-1 and 0.6 mol L^-1 respectively. From the above information we can predict the direction of reaction as follows.
\(Q={[HI]^2\over[H_2][I_2]}={0.6\times 0.6\over 0.2\times 0.2}=9\)
Since Q < K_c, the reaction will proceed in the forward direction.

The balanced equation for the formation of HCI is,
H₂(g) + CI₂(g) \(\rightarrow\) 2 HCI (g)
As per the stoichiometric equation, under given conditions,
To produce 2 moles of HCI, 1 mole of chlorine gas is required.
To produce 44.8 litres of HCI, 22.4 litres of chlorine gas are required.
\(\therefore\) To produce 11.2 litres of HCI,

= 5.6 litres of chlorine are required.

\(\therefore\) The empirical formula is Fe₂O₃.

(i) To produce colours, fireworks experts bum the metal and chlorine together in a vapour, where the two elements are gases instead of solid
(ii) The burning excites the electron pushing them into a higher than normal energy level. As the electrons returns to their normal level, they release their extra energy as a colourful burst of light.
(iii) True blue fireworks are the hardest to make since the compound copper chloride breaks down in a hot flame.

Alkali metal oxides M₂O, M₂O₂ and MO₂ are easily hydrolyzed by water to form the hydroxides according to the following reactions:
M₂O + H₂O ⟶ 2M⁺ + 2OH^-
M₂O₂ + 2H₂O ⟶ 2M⁺ + 2OH^- + H₂O₂
2MO₂ + 2H₂O ⟶ 2M⁺ + 2OH^- + H₂O₂ + O₂

Equivalent mass of a salt:
It is defined as the number of parts by mass of the salt that is produced by the neutralization of one equivalent of an acid by a base. Therefore the equivalent mass of the salt is equal to its molar mass.

(i) Metallic hydrides are obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Hence, they are called interstitial hydrides.
(ii) The hydrides show properties similar to parent metals and hence they are also known as metallic hydrides.
(iii) They are mostly non-stoichiometric with variable composition (TiH_1.5-1.8 and PdH_0.6-0.8)
(iv) Some are relatively light, inexpensive and thermally unstable which makes them useful for hydrogen storage applications. Example, TiH₂, ZrH₂, ZnH₂.

(i) Variation of volume of the gas sample with temperature at constant
(ii) Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P₁ to P₅.
(iii) i.e. P₁ < P₂ < P₃ < P₄ < P₅ When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.
(iv) All gases are becoming liquids if they are cooled to sufficiently low temperatures.
(v) In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.

Charles' law states that for a fixed mass of a gas at constant pressure, the volume is directly proportional to temperature (K).
\(V\propto T\) (or)  = \(\frac{V}{T}\) Constant

Volume vs Temperature:
If a balloon is moved from an ice water bath to a boiling water bath, the gas molecules inside move faster due to increased temperature and hence the volume increases.

(i) Metals comprise more than 78% of all known elements.They are present on the left side of the periodic table.
(ii) They are usually solids at room temperature. [Mercury is an exception (Hg-liquid), gallium (303K) and cesium (302K) also have very low melting points].
(iii) Metals usually have high melting and boiling points.
(iv) They are good conductors of heat and electricity.
(v) They are malleable and ductile, and also can be flattened into thin sheets by hammering and drawn into thin wires.

(i) The distribution of electrons into orbitals, s, p, d and f of an atom is called its electronic configuration. The electronic configuration of an atom is characterized by a set of four quantum numbers, n, 1, m and s. Of these the principal quantum number (n) defines the main energy level known as shells.
(ii) The position of an element in the periodic table is related to the configuration of that element and thus reflects the quantum numbers of the last orbital filled.
(iii) The electronic configuration of elements in the periodic table can be studied along the periods and groups separately for the best classification of elements.
(iv) Elements placed in a horizontal row of a periodic table is called a period. There are seven periods.
(v) A vertical column of the periodic table is called a group. A group consists of a series of elements having similar configuration of the outermost shell. There are 18 groups in periodic table.

It occurs naturally as a result of nuclear reactions induced by cosmic rays in the upper atmosphere.
\(_{ 7 }^{ 14 }{ N }+_{ 0 }^{ 1 }{ n }\rightarrow _{ 6 }^{ 12 }{ C }+_{ 1 }^{ 3 }{ H }\)
\(_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 2 }{ H }\rightarrow _{ 1 }^{ 3 }{ H }+_{ 1 }^{ 1 }{ H }\)
\(_{ 3 }^{ 6 }{ Li }+_{ 0 }^{ 1 }{ n }\rightarrow _{ 2 }^{ 4 }{ He }+_{ 1 }^{ 3 }{ H }\) 

n = 4; f orbital l = 3 \(\Rightarrow\) m₁= - 3, -2,-1, 0, +1, +2, +3
out of two electrons, one electron occupies 4f orbital with m₁ = -3 and another electron occupies 4f orbital with m₁ = -2.
All the four quantum numbers for the two electrons are

Molar mass of A g₂CO₃ = \(\{2\times atomic\ mass\ of \ Ag+atomic\ mass\ of\ C +3\times atomic\ mass \ of\ 'O'\)
= 2 x 108 + 12 + 3 x 16
= 276 u
Molar mass of AgO = atomic mass of Ag+ atomic mass of '0'
= 108 + 16 = 124g mol^-1
When silver carbonate is heated, carbon dioxide leaves as gas and silver oxide remains as a residue.
According to the equation,
One mol of Ag₂C0₃leaves = 2 mol of AgO as residue
276 g of Ag₂0 = 124g of AgO
\(11.04 \ of \ Ag_2O={124\over276}\times 11.04g \ of \ AgO\)
= 4.96g

The given reaction is N₂(g) + 3H₂(g)  \(\rightarrow\) 2NH₃(g); \(\Delta\)H° = -92.4 kJ mol^-1
\(\Delta { H }^{ ° }=2\Delta { H }_{ f }^{ o }({ NH }_{ 3 })-[\Delta { H }_{ f }^{ o }({ N }_{ 2 })+3\Delta { H }_{ f }^{ o }({ H }_{ 2 })]\)
By definition, the standard enthalpy of formation of elements is equal to zero.
-92.4 = 2\(\Delta \)\({ H }_{ f }^{ o }\)(NH₃)-(0+3\(\times\)0)
i.e., for 2 mol of NH₃ 2\(\Delta\)\({ H }_{ f }^{ o }\)(NH₃)-90+3\(\times\)0)
i.e., for 2 mol of NH₃ 2 \(\Delta\)\({ H }_{ f }^{ o }\)(NH₃) = -92.4
for 1 mol of NH₃ \(\Delta\)\({ H }_{ f }^{ o }\) (NH₃) = \(\frac { -92.4 }{ 2 } =46.2\) kJ mol^-1

\(Average \ atomic \ mass \ of \ carbon ={12\times \%abundance \ of \ C^{12}+13 \times \% \ abundance \ of \ C^{13}+14\% abundance of c^{14} \over 100} \)
given % abundances Of C¹² = 98.892
C¹³ = 1.108
C^I4 = 2 x 10^-10
Since 2 x 10^-10is so small compound to 100, it can be neglected .
\(Average \ atomic \ mass \ of \ carbon ={12\times 98.892 +13 \times 1.108 \over 100} \)
= 12.01 a.m.u

(i) The coefficients in a balanced thermochemical equation refer to number of moles of reactants and products involved in the reaction.
(ii) The enthalpy change of the reaction \(\Delta\)H_r has to be specified with appropriate sign and unit.
(iii) When the chemical reaction is reversed, the value of \(\Delta\)H is reversed in sign with the same magnitude.
(iv) The physical states (gas, liquid, aqueous, solid in brackets) of all species are important and must be specified in a thermochemical reaction, since \(\Delta\)H depends on the physical state of reactants and products.
(v) If the thermochemical equation is multiplied throughout by a number, the enthalpy change is also multiplied by the same number.
(vi) The negative sign of \(\Delta\)H_r indicates that the reaction is exothermic and the positive sign of \(\Delta\)H_r indicates an endothermic reaction.

\(\lambda =\frac{h}{\sqrt{2KEm}}\ \lambda=\frac{h}{mv}or \frac{h}{p}\)
E = Kinetic energy
m = mass of particle               
\(\lambda\) = 4.8 pm = 4.8 x 10^-12 m     
h = 6.63 x 10^-34 J.S.               
m = 9.11 x 10^-31 kg             
 4.8 x 10^-12 = \(\frac{6.63\times 10^{-34}}{\sqrt{12\times E\times 9.11\times 10^{-31}}}\)
E = 1.047 x 10^-14 J.

Compressibility factor (Z) = \(\frac{PV}{nRT}\)
Given: n = 1; P = 50 atm; V = 350 x 10^-3 L = 0.35 L R = 0.0821 L atm K^-1 mol^-1 ;
T = 27 + 273 = 300 K
Substituting these values in the above equation
\(Z=\frac{(50 atm) (0.35 L)}{(1 mol (0.821 Latm K^{-1} mol^{-1})(300K)}\)
= 0.711
Since, Z < 1, the gas shows negative deviation from ideal behaviour.

(i) Fluorine decomposes cold water.
\(2F_2+2H_2O_2 \rightarrow2H_2F_2+O_2 \\ 3F_2+3H_2O_2 \rightarrow 2H_2F_2+O_3(Ozonised \ oxygen)\)
(ii) Chlorine decomposes cold water forming HCI and HCIO (hypo chlorous acid
\(Cl_2+H_2O\rightarrow2HCl+HClO\)

The heat absorbed, when one mole of a solid melts at its melting point reversibly, is called molar heat of fusion. The entropy change is given by
\({ \triangle S }_{ f }=\frac { { \triangle H }_{ f } }{ { T }_{ f } } \).
Where ΔH_f is molar .heat of fusion. T_f is melting point.

(i) The second law of thermodynamics helps us to predict whether the reaction is feasible or not
(ii) It tells about the direction of the flow of heat.
(iii) It also tells that energy cannot be completely converted into equivalent work.

P+ HNO₃\(\rightarrow\) H₃PO₄ + NO₂ + H₂O

Equalise the increase I decrease in O N by multiplying P species by +1 and N species by +5

(l) BeO is almost insoluble in water because Be²⁺ is a small cation with high polarising power and O^2- is small anion.
∴ the lattice energy very high.
When BeO is dissolved in water, the hydration energy of its ions is not sufficient to overcome high lattice energy.
(ii) On the other hand, Be²⁺ is a small cation and is a large anion. Hence Be²⁺ can easily polarize, \(SO_4^{2-}\) ions, making BeSO₄ unstable the lattice energy of BeSO₄ is not very high and so it is soluble in water.

Equivalent mass of Ba(OH)₂
Molar mass of Ba(OH)₂ = 171.34 g/mol
Acidity of the Ba(OH)₂ = 2
Equivalent mass of the Ba(OH)₂ = \(\frac { Molar \ mass \ of \ the \ base }{Acidity \ of \ the \ base}\)
\(=\frac{171.34}{2}=85.5\)

Atomic mass of potassium = 39
No. of moles = \(\frac { Mass }{ Molar\ mass } \)
No.of moles = \(\frac { 19.5 }{ 39 } \) = 0.5 moles

6x + 12(1) + 6(-2) = 0
6x + 12 -12 = 0
6x = 0
x = 0
Oxidation number of C in C₆H₁₂O₆ is zero.

(i) Although IE₁ values of alkaline earth metals are higher than that of alkali metals, the IE₂ values of alkaline earth metals are much smaller than those of alkali metals.
(ii) This occurs because in alkali metals the second electron is to be removed from a cation, which has already acquired a noble gas configuration.
(iii) In the case of alkaline earth metals the second electron is to be removed from a monovalent cation, which still has one electron in the outermost shell.
(Iv) Thus, the second electron can be removed more easily in the case of group 2 elements than in group 1 elements

48 parts by mass of oxygen are made available from 294 parts by mass of K₂Cr₂O₇
∴ 8 parts by mass of oxygen will be furnished by =\(\frac{294\times8}{48}=49\)
Equivalent mass of K₂Cr₂O₇ = 49 g equiv^-1

ΔH_f (NaCl)=30.4 kJ = 30400 J mol^-1
ΔS_f (NaCl)=28.4 JK^-1mol^-1
T_f=?
ΔS_f =\(\frac { \Delta { H }_{ f } }{ { T }_{ f } } \)
T_f=\(\frac { \Delta { H }_{ f } }{ \Delta { S }_{ f } } \)
T_f=\(\frac { 30400J{ mol }^{ -1 } }{ 28.4J{ K }^{ -1 }mol^{ -1 } } \)
T_f=1070.4K

Water can easily dissolve all ionic compounds. Due to H-bonding with polar molecules, even covalent molecules like alcohols and carbohydrates dissolve in water.
Water can hydrolyse many metallic and nonmetallic oxides, hydrides, phosphides and other salts.
P₄O_10(s) + 6H₂O ➝ 4H₃PO_4(aq)

K₂CrO₄
2(1) + x + 4(-2) = 0
2 + x - 8 = 0
x - 6 = 0
x = +6
Oxidation number of Cr in K₂CrO₄ is +6

If a substance which gives a line emission spectrum is placed in an external electric field, its lines get split into a number of closely spaced lines. This phenomenon is known as Stark effect.

Electronic configuration of the element is 3d¹ 4S².
The element belongs to fourth period and group 3.
The element is scandium (Z = 21).

In a group, the density increases with increase in atomic number. But the density of ₁₉K is less than ₁₁Na.
Reason:
Atomic volume ofK is nearly twice of Na, but its mass (39) is not exactly double of Na (23).
Since density is inversely related to volume, the greater volume K has lower density than Na

(i) Lothar Meyer plotted the physical properties such as atomic volume, melting point and boiling point against atomic weight and obtained a periodically repeated pattern.
(ii) Lothar Meyer observed a change in length of that repeating pattern.
(iii) In 1868, Lothar Meyer had developed a table of the elements that closely resembles the modern periodic table.

p-orbitals :
(a) For p orbitals l = 1 and the corresponding m values are -1,0 and + 1. The angular distribution functions are quite complex
(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 and z axis respectively.
(c) The 2p orbitals have one nodal plane

Metal Hydride (Hydrogen Sponge):
The best studied binary hydrides are the palladium - hydrogen system. Hydrogen interacts with palladium in a unique way, and forms a limiting monohydride, PdH. Upon heating, H atoms diffuse through the metal to the surface and recombine to form molecular hydrogen. Since no other gas behaves this way with palladium, this process has been used to separate hydrogen gas from other gases.

\(2 \mathrm{Pd}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{PdH}(\mathrm{s})\)
The hydrogen molecule readily adsorbs on the palladium surface, where it dissociates into atomic hydrogen. The dissociated atoms dissolve into the interstices or voids (octahedral/tetrahedral) of the crystal lattice.
Technically, the formation of metal hydride is by chemical reaction but it behaves like a physical storage method, i.e., it is absorbed and released like a water sponge. Such a reversible uptake of hydrogen in metals and alloys is also attractive for hydrogen storage and for rechargeable metal hydride battery applications.

At a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
\(V\alpha \frac { 1 }{ P }\) at constant T& n.
Mathematical form: P₁V₁ = P₂ V₂ = K

(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).

The word Aufbau in German means 'building up'. In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them.
Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals. The order of filling of various orbitals as per the Aufbau principle which is in accordance with (n + l) rule.

C₁ V₁ = C₂ V₂
6M(V₁) = 0.25 M x 500 ml
\(\mathrm{V}_{1}=\frac{0.25 \times 500}{6}\)
V₁ = 20.83 ml

Ozone is formed by a series of reactions that occur from the sun shines

ΔH_f⁰ CO₂ = -393.5 kJ mol^-1
ΔH_f⁰ CO = -111.31 kJ mol^-1
ΔH_f⁰ (H₂O) = - 242 kJ mol^-1
CO₂(g) + H₂(g) - CO(g) + H₂O(g)
ΔH_f⁰ = ?
ΔH_f⁰ = Σ(ΔH_f⁰)_products-Σ(ΔH_f⁰)_reactions
ΔH_f⁰ = [ΔH_f⁰(CO) + ΔH_f⁰(H₂O)] - [ΔH_f⁰(CO₂) + ΔH_f⁰(H₂)]
ΔH_f⁰ = [-111.31 + (-242)] - [-393.5 + (0)]
ΔH_f⁰ = [-353.31] + 393.5
ΔH_f⁰ = 40.19
ΔH_f⁰ = +40.19 kJ mol^-1.

Work is defined as the force (F) multiplied by the displacement(x).
-w = F.x.

2NH₃ + H₂O + CO₂ ⟶ (NH₄)₂ CO₃
(NH₄)₂ CO₃ + H₂O + CO₂ ⟶ 2NH₄ HCO₃
2NH₄HCO₃ + NaCl ⟶ NH₄Cl + NaHCO₃
2NaH CO₃ ⟶ Na₂ CO₃ + CO₂ + H₂O.