\(q_{rev}= (-\Delta_f H^{\ominus})=-286 kJ\ mol^{-1}=286000\ J\ mol^{-1}\)
\(\Delta{S}{_{(Surroundmgs)}}=\frac{q_{rev}}{T}=\frac{286000\ J\ mol^{-1}}{298\ K}\) = 959 J K^-1 mol^-1.

\(\Delta { G }^{ \ominus }\) = -RT ln K = -2.303 RT log K.
R = 8.314 JK^-1 mol^-1;T = 300 K; K=10
\(\Delta { G }^{ \ominus }=-2.303 \times 8.314 \)  JK^-1 mol^-1 x (300 K) x log 10
= -5527 J mol^-1= -5.527 kJ mol^-1

We will use the ideal gas equation for this calculation as below:
\({P=nRT\over V}={{2mol\times 0.0821Latm.K^{-1}.mo^l{-1}\times (70+273K)}\over{6dm^3}}\)
= 9.39 atm.

= (12 x 12) + (22 x 1) + (11 x 16)
= 342 g /mol

= (39 x 2) + (2 x 52) + (7 x 16)
= 294 g mol.

= (1 x 39) + (1 x 55) + (4 x 16)
= 158 g / mol

Molecular mass = Density x Molar volume
Molar volume of CO₂ = 2.24 x 10^-2m³
Density of CO₂ 1.977 kg m^-3
\(\therefore \text { Molecular mass of } \mathrm{CO}_{2}=1.977 \times 10^{3} \mathrm{~g} \mathrm{~m}^{\not -3} \times 2.24 \times 10^{-2} \mathrm{~m}^{\not -3}\)
= 1.977 X 10¹ x 2.24
= 44g

Cl(g) + e^- ⟶Cl^-(g)    AH = 348 kJ mol^-1
For one mole (35.5g) 348 kJ is released.
\(\therefore \text { For } 17.5 \mathrm{~g} \text { chlorine, } \frac{348 \mathrm{~kJ}}{35.5 \mathrm{~\not g}} \times 17.75 \mathrm{~\not g} \text { energy leased. }\)
∴ The amount of energy released =\(\frac { 348 }{ 2 } \) =174 kJ

Noble gases: Ioniation energy ranging from 2372 KJmol^-1 to 1037 kJ mol^-1.
For element X, the IE₁ value is in the range of noble gas, moreover for this element both IE₁ and IE₂ are higher and hence X is the noble gas.
For Y, the first ionisation energy is low and second ionisation energy is very high and hence Y is most reactive metal. For Z, both IE₁ and IE₂ are higher and hence it is least reactive.

Electronic Configuration: (n - 1)d² ns²
for n = 5, the electronic configuration is,
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d² 5s²
Atomic number: 40
4^th group 5^th period (d block element) = Zirconium

Atomic number: 120
IUPAC temporary symbol: Unbinilium
IUPAC temporary symbol: Ubn
Possible electronic configuration :[Og] 8s²

The main basic difference between Mendeleev's periodic table and modem periodic table is that first one is constructed on the basis of atomic weight and the later is constructed on the basis of atomic number.

(iv) \({ K }\overset { +7 }{ \underset { \underset { 2\times { 3e }^{ - } }{ \downarrow } }{ M } } n{ O }_{ 4 }+{ H }_{ 2 }\overset { -1 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ C_{ 2 } } } } { O }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+\overset { +2 }{ M } n{ SO }_{ 4 }+\overset { +4 }{ C } { O }_{ 2 }+{ H }_{ 2 }O\)
2KMnO₄ + 5H₂C2O₄ + H₂S0₄ \(\longrightarrow \) Mn0₂ + Na₂S0₄ + KOH
2KMnO₄+ 5H₂C₂O₄ + H₂S0₄ \(\longrightarrow \) K₂SO₄ + 2MnSO₄ + 10CO₂ + H₂O
2KMnO₄ + 5H₂C₂O₄ + 3H₂S0_{4 \(\longrightarrow \)}  K₂S0₄ + 2MnS0₄ + lOC0₂ + 8H₂0

(iii) \(\overset { 0 }{ \underset { \underset { 2e^{ - } }{ \downarrow } }{ Cu } } { O }_{ 7 }+H\overset { +5 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ N } } } { O }_{ 3 }\longrightarrow \overset { +2 }{ Cu } \left( { No }_{ 3 } \right) _{ 2 }+\overset { +4 }{ N } { O }_{ 2 }+{ H }_{ 2 }O\)
Cu +2HNO₃ \(\longrightarrow \)  Cu(NO₃)₂ + NO₂ + H₂O
Cu + 2HNO₃ + 2HNO₃ \(\longrightarrow \) Cu(NO₃)₂ + 2NO₂ + 2H₂O
Cu + 4HNO₃ \(\longrightarrow \) Cu (NO₃)₂ + 2No₂ + 2H₂O

 (ii) \({ K }\overset { +7 }{ \underset { \underset { 3e^{ - } }{ \uparrow } }{ M } } n{ O }_{ 4 }+{ Na }_{ 2 }\overset { +4 }{ \underset { { 2e }^{ - } }{ \underset { \downarrow }{ S } } } { O }_{ 3 }\longrightarrow \overset { +4 }{ M } { nO }_{ 2 }+{ Na }_{ 2 }\overset { +6 }{ s } { O }_{ 4 }+KOH\)
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \)   MnO₂ + Na₂SO₄ + KOH
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \)  2MnO₂ + 3Na₂SO₄ + KOH
\(\Rightarrow\) 2KMNO₄ + 3NaSO₃ + H₂O \(\longrightarrow \) 2MnO₂ + ₃Na₂SO₄ +2KOH

State Function: Enthalpy, entropy, temperature, free energy
Path Function: Heat, work.

(i) The third law of thermodynamics states that the entropy of pure crystalline substance at absolute zero is zero.
(ii) It can also be stated as it is impossible to lower the temperature of an object to absolute zero in a finite number of steps
(iii) Mathematically, \(\lim _{ T\rightarrow 0 }{ S=0 } \) for a perfectly ordered crystalline state.

It is impossible to construct a machine that absorbs heat from a hot source and converts it completely into work by a cyclic process without transferring a part of heat to a cold sink

Lattice energy is defined as "The amount of energy required to completely remove the constituent ions from its crystal lattice to an infinite distance." It is also referred as lattice enthalpy

The heat of neutralisation is defined as "The change in enthalpy when one gram equivalent of an acid is completely neutralised by one gram equivalent of a base or vice versa in dilute solution"

The calorific value is defined as "The amount of heat produced in calories (or joules) when one gram of the substance is completely burnt." The SI unit of calorific value is J kg^-1. It is usually expressed in cal g ^-1.

Molar heat capacity is defined as "The amount of heat absorbed by one 'mole of the substance to raise its temperature by 1 kelvin". The SI unit of molar heat capacity is JK^-1 mol^-1

The heat of combustion of a substance is defined as "The change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen". It is denoted by ΔH_C

Gibbs free energy is defined as G = H - TS

(i) non-spontaneous
(ii) non-spontaneous
(iii) spontaneous

(i) Entropy is a measure of the molecular disorderliness (randomness) of a system. dS = dq_rev/T
(ii) The entropy (S) is equal to heat energy exchanged (q) divided by the temperature (T) at which the exchange takes place. Therefore, The SI unit of entropy is JK^-1

The property that is independent of the mass or the size of the system is called an intensive property.
Examples: Refractive index, Surface tension, density, temperature, Boiling point, Freezing point, molar volume, etc.,

The enthalpy change of a reaction either at constant volume or constant pressure is the same whether it takes place in a single or multiple steps provided the initial and final states are same.

The first law of thermodynamics, also known as the law of conservation of energy, states that "The total energy of an isolated system remains constant though it may change from one form to another."
The mathematical statement of the First Law is: ΔU=q+w
Where q - the amount of heat supplied to the system; w - work done on the system

(i) ground state :

This type of electronic configuration have ten possible arrangement (i.e. Half filled configuration is More).

a) Principal quantum number defines energy and size of an orbital.
b) Azimuthal quantum number defines shape of an orbital
c) Magnetic quantum number defines spatial orientation (direction) of an orbital.

2Ca + O₂ ⟶ 2CaO

Mg (OH)₂ ; Magnesium hydroxide.

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

Energy electrons symmetry
This is due to the symmetrical distribution and exchange energy of given d- electrons. Symmetry leads to stability.
Exchange energy:
If two or more electrons with the same spin are present in degenerate orbitals, there is a possibility for exchanging their positions. During exchange process, the energy is released and the released energy is called exchange energy. If more number of exchanges are possible, more exchange energy in released. More number of exchanges are possible only in case of half filled and fully filled configurations.
For example, in chromium the electronic configuration is [Ar]3d⁵ 4s¹. The 3d orbital is half filled and there are ten possible exchanges as shown in figure. On the other hand only six exchanges are possible for [Ar]3d⁴ 4s² configuration. Hence, exchange energy for the half filled configuration is more. This increases the stability of half filled 3d orbitals.

The exchange energy is the basis for Hund's rule, which allows maximum multiplicity, that is electron pairing is possible only when all the degenerate orbitals contain one electron each.

Average atomic mass
= \(\frac { (78.99\times 23.99)+(10\times 24.99)+(11.01\times 25.98) }{ 100 } \)
= \(\frac { 2430.9 }{ 100 } \)
= 24.31 u

Mol.mass = 2(H) + 1(S) + 4(0)
= 2(1) + 1(32) + 4(16)
= 2 + 32 + 64 = 98

Mol.mass = 3(H) + 1(B) + 3(0)
= 3(1) + 1(11) + 3(16)
= 3 + 11 + 48 = 62

Mol.mass = 3(C) + 6(H) + 1(0)
= 3(12) + 6(1) + 1(16)
= 36 + 6 + 16 = 58

i) urea [CO(NH₂)₂]
Mol.mass = 1 (C) + 2(N) + 4(H) + 1(0)
= 1(12) + 2(14) + 4(1) + 1(16)
 = 12 + 28 + 4 + 16 = 60  
ii) Acetone [CH₃ COCH₃]
Mol.mass = 3(C) + 6(H) + 1(0)
= 3(12) + 6(1) + 1(16)
= 36 + 6 + 16 = 58
iii) Boric Acid [H₃ BO₃]
Mol.mass = 3(H) + 1(B) + 3(0)
= 3(1) + 1(11) + 3(16)
= 3 + 11 + 48 = 62
iv) Sulphuric  Acid [H₂ SO₄]
Mol.mass = 2(H) + 1(S) + 4(0)
= 2(1) + 1(32) + 4(16)
= 2 + 32 + 64 = 98

Lanthanides: [₅₄Xe] 4f^1-14 5d¹ 6s²
Actinides : [₈₆Rn] 5f^0-146d^0-2 7s²

It is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to set of rules.

Gram equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.

One mole is the amount of substance of a system, which contains as many elementary particles as there are atoms in 12 g of carbon -12 isotope.

a) Electron-precise hydrides:
These have required number of electron to represent their conventional d lewis structure. All the elements of carbon group (14) form such hydrides.
Eg CH₄, C₂H₆, SiH₄, GeH₄
b) Electron - deficient hydrides:
There act as Lewis acids (ie) electron acceptors. The elements from group (13) form such hydrides.
Eg: B₂H₆
c) Electron - rich hydrides:
There have excess electron, which are present as lone pairs. Elements of group 15-17 form such hydrides. They behave as Lewis bases (ie) electron donors.
Eg: NH₃ ,H₂O, HF

NOTE :
At any condition ice cube does not sink in water.
At 0°C, ice and liquid water will be in equilibrium and will coexist. Hence no freezing or melting will occur.

NOTE:
At any condition ice cube does not sink in water.
At 0⁰C, ice and liquid water will be in equilibrium and will coexist. Hence no freezing or melting will occur.

The total number of electrons are less in the cation than the neutral atom while the nuclear charge remains the same. Therefore the effective nuclear charge of the cation is higher than the corresponding neutral atom. Thus the successive ionisation energies, always increase in the following order
IE₁ < IE₂ < IE₃ < .....

It is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Mg + I.E₁ ⟶ Mg⁺ +1e^-     .....(i)
Mg⁺ + I.E₂ ⟶ Mg²⁺ + 1e^-  ....(ii)
Mg²⁺ + I.E₃ ⟶ Mg³⁺ + 1e^-  ....(iii)
(i) The step (iii) which involves the formation of Mg³⁺ requires higher ionisation energy.
(ii) Mg²⁺ consist of 10 electrons (2, 8) attaining the stable noble gas configuration of argon (Z = 10).
(iii) Since the valence orbital is completely filled, more energy will be required to remove electrons.

No the above definition is incorrect. The correct definition is Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge.
Z_eff=Z-S
Where Z is the atomic number and 'S' is the screening constant.

Ions of different elements having the same number of electrons are called isoelectronic ions.

The modem periodic law states that, "the physical and chemical properties of the elements are periodic functions of their atomic numbers.

Given
T₁ = 127°C = 127 + 273 = 400 K
T = 47°C = 47 + 273 = 320 K
% efficiency η = ?
η = \(\left[ \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \right] \) x 100
η =\(\left[ \frac { 400-320 }{ 400 } \right] \) x 100
η = \(\left[ \frac { 80 }{ 400 } \right] \) x 100
η = 20%

\(\mathrm{E}_{n}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}\)
Second excited state
\(\therefore E_{3}=\frac{-13.6}{9} \mathrm{eV}\)
n = 3
E₃ = -1.51 eV

(i) spontaneous reaction: A reaction that occurs under the given set of conditions without any external driving force is called a spontaneous reaction.
(ii) Criteria for spontaneity of a process: The spontaneity of any process depends on three different factors.
ΔH = -ve, ΔS = +ve, ΔG = -ve.

\({ H }_{ 2 }\overset { -1 }{ \underset { \overset { \uparrow }{ (ie\times 2) } }{ { O }_{ 2 } } } +{ \underset { \overset { \downarrow }{ Ie^{ - } } }{ Fe } }^{ 2 }+\rightarrow { Fe }^{ 3+ }+\overset { -2 }{ {H}_{2} O} \)
\({ H }_{ 2 }{ O }_{ 2 }+{ 2Fe }^{ 2+ }+\rightarrow Fe^{ 3+ }+{ H }_{ 2 }O\)
\(\Rightarrow \) \({ H }_{ 2 }{ O }_{ 2 }+{ 2Fe }^{ 2+ }+{ 2 }H^{ + }\rightarrow 2Fe^{ 3+ }+{ 2H }_{ 2 }O\)

The balanced equation for the combustion of ethane
C₂H₆ + \(\frac { 7 }{ 2 } \)O₂ \(\longrightarrow \) 2CO₂ +3H₂O
2C₂H₆ + 7O₂ \(\longrightarrow \) 4CO₂ + 6H₂O
To produce 4 moles of CO₂, 2 moles of ethane is required
To produce 1 moles (44 g) of CO₂ required
Number of moles of ethane
\(=\frac{2 \mathrm{~mol} \text { ethane }}{4 \not \mathrm{molCO}_{2}} \times 1 \not \mathrm{molCO}_{2}\)
= \(\frac { 1 }{ 2 } mole\quad of\quad ethane\)
= 0.5 mole of ethane

Rate of diffusion \(\text { (r) } \alpha \frac{1}{\sqrt{M}}\)
\(\mathrm{M}_{\mathrm{NH}_{3}}=17 ; \mathrm{M}_{\mathrm{HCl}}=36.5 \)
\(\therefore \mathrm{r}_{\mathrm{NH}_{3}}>\mathrm{r}_{\mathrm{HCl}} \)
Hence white fumes are first formed near HCL.

1) The unit of molecular mass is atomic mass unit [amu]. The unit of molar mass is gram per mole.
2) Molecular mass is the mass of one molecule while molar mass is the mass of one mole of molecules (6.022 x 10²³)
(i) Molecular mass of CO₂ = 1(C) + 2(0) = 12 + 32 = 44 amu
or 7.304 x 10^-23 g
(ii) Molar mass of CO₂ = 44 g mol^-1.

i) ₂₅Mn - 1s², 2S^2, 2p⁶, 3s², 3p⁶, 4s², 3d⁵
₂₃Mn²⁺ - 1s², 2S^2, 2p⁶, 3s², 3p⁶, 3d⁵
ii) ₂₄Cr - 1s^2, 2S²,2p⁶, 3s², 3p⁶, 4s¹, 3d³
₂₁Cr³⁺ - 1s^2, 2S²,2p⁶, 3s², 3p⁶, 3d³

On heating incineration might take place due to the increase of pressure

The larger the size of the molecule, the greater will be Van der Waals' attraction. Therefore greater the deviation from ideal behaviour. So bromine will deviate more from ideal behaviour because it has bigger atoms.

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

ii) 1 mol of x + 1 mol of y + 3 mol of z₂

ii) 50 atoms of x + 25 atoms of y + 50 molecules of z₂

1. Gypsum is used in making drywalls or plaster boards.
2. Another important use of gypsum is the production of plaster of Paris. Gypsum is heated to about 300 degree Fahrenheit to produce plaster of paris, which is also known as gypsum plaster. It is mainly used as a sculpting material.
3. Gypsum is used in making surgical and orthopedic casts, such as surgical splints and casting moulds.
4. Gypsum plays an important role in agriculture as a soil additive, conditioner, and fertilizer. It helps loosen up compact or clay soil, and provides calcium and sulphur, which are essential for the healthy growth of a plant.
5. Gypsum is used in toothpastes, shampoos, and hair products.
6. Gypsum is a component of portland cement, where it acts as a hardening retarder to control the speed at which concrete sets.

Calcium Sulphate (plaster of paris), CaSO₄. 1/2H₂O :
It is a hemihydrate of calcium sulphate. It is obtained when gypsum, CaSO₄ .2H₂O,is heated to 393 K
2(CaSO₄.2H₂O) ⟶ 2CaSO₄,H₂O + 3H₂O
Above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, CaSO₄ is formed. This is a known 'dead burnt plaster'.
It has a remarkable property of setting with water. on mixing with an adequate quantity of water it forms a plastic mass that gets into hard solid in 5 to 15 minutes.

This is due to the smaller size and relatively high charge as Be²⁺ ion. Moreover Beryllium is a non - metal. But Magnesium is a metal. In the case of beryllium the ionisation energy and electronegativity of the halides are almost similar; but there is a vast difference in IE and electronegativity between Mg and halides.

The largest use of plaster of paris is in the building industry as well as plasters. It is used for immobilising the affected part of organ where there is a bone fracture or sprain. It is also employed in dentistry, in ornamental work and for making casts of statues and busts.

The lattice energy of LiF is higher due to the smaller size of Li⁺ and F^-. So LiF has lower solubility.

6Li + N₂ ⟶ 2Li₃ N

If there is a tiny sticky area on the walls of the container of a gas, then the collision of the molecules will not be elastic. So the gas will behave as a real gas. So the pressure will be less than the calculated value.

Molar mass = density x Molar volume
= 1.965 x 2.24 x 10^-2
= 4.4016 x 10^-2 kg/mol
= 4.4016 x 10^-2 x 10³ g/mol
= 44.016 g/mol

Interstitial hydrides have metallic bonding. These are non - stoichiometric; their composition varies with temperature and pressure. The crystal lattice expands due to the inclusion of dihydrogen and there is distortion of crystal lattice. So they have lower density than the parent metal. Recent studies have shown that except the hydrides of Ni, Pd, Ce & Ac, other hydride are different from the parent metals.

(i) WO₃ + 3H₂ ➝ W + 3H₂O
(ii) H₂ + Cl₂ ➝ 2HCl.

a) HCl is a gas. It is an electron rich covalent hydride. It is a covalent molecule with strong covalent bond. However the forces between the molecules are not as strong.
b) NaH is a solid. It is an ionic hydride of group I sodium metal. It is an insoluble hydride

Hydrogen has electronic configuration 1s¹. So it resembles ns¹ general valence shell configuration of alkali metals.
Its similarity with alkali metals are:
1. Form uni positive (H⁺) ions like Na⁺, K⁺, Cs⁺
2. Form halides, (HX), oxides (H₂O) peroxides (H₂O₂) and sulphides (H₂S) like (Nax, Na2O, Na₂O₂, Na₂S)
3. Acts as reducing agent. 
4. Has +1 oxidation state.
5. If it is placed in the halogen family it should be the most electro negative element which is not true.

(i) no. of electrons: 35 (given)
no. of protons : 35
(ii) Electronic configuration
1s² 2S² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵
(iii) Last electron:

4P_x        4P_y     4p_z
last electron present in 4P_y orbital y
n = 4, l = 1 m₁ = either + 1 or -1 and s = -1/2

(i) \({ K }_{ 2 }\overset { +6 }{ \underset { \underset { 2\times { 3e }^{ - } }{ \uparrow } }{ Cr_{ 2 } } } { O }_{ 7 }+K\overset { -1 }{ \underset { { 1e }^{ - } }{ \underset { \downarrow }{ I } } } +{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ \overset { +3 }{ Cr } }_{ 2 }({ SO }_{ 4 })_{ 3 }+\overset { 0 }{ I } _{ 2 }+{ H }_{ 2 }O\)
K₂Cr₂O₇ + 6KI + H₂SO₄ \(\longrightarrow \)  K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
K₂Cr₂O₇ + 6KI + H₂SO₄ \(\longrightarrow \)  K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + H₂O
K₂Cr₂O₇ + 6KI + 7H₂SO₄  \(\longrightarrow \)  4k₂SO₄ + Cr₂(SO₄)₃ + ₃I₂ + 7H₂
ii) \({ K }Mno_{ 4 }+{ Na }_{ 2 }{ So }_{ 3 }\longrightarrow { MnO }_{ 2 }+{ Na }_{ 2 }{ So }_{ 4 }+KOH\)
 \({ K }\overset { +7 }{ \underset { \underset { 3e^{ - } }{ \uparrow } }{ M } } n{ O }_{ 4 }+{ Na }_{ 2 }\overset { +4 }{ \underset { { 2e }^{ - } }{ \underset { \downarrow }{ S } } } { O }_{ 3 }\longrightarrow \overset { +4 }{ M } { nO }_{ 2 }+{ Na }_{ 2 }\overset { +6 }{ s } { O }_{ 4 }+KOH\)
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \)   MnO₂ + Na₂ SO₄ + KOH
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \)  2MnO₂ + 3Na₂SO₄ + KOH
\(\Rightarrow\) 2KMNO₄ + 3NaSO₃ + H₂O \(\longrightarrow \) 2MnO₂ + ₃Na₂ SO₄ + 2KOH
iii) \(Cu+{ HNO }_{ 3 }\longrightarrow Cu\left( { No }_{ 3 } \right) _{ 2 }+{ No }_{ 2 }+{ H }_{ 2 }O\)
\(\overset { 0 }{ \underset { \underset { 2e^{ - } }{ \downarrow } }{ Cu } } { O }_{ 7 }+H\overset { +5 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ N } } } { O }_{ 3 }\longrightarrow \overset { +2 }{ Cu } \left( { No }_{ 3 } \right) _{ 2 }+\overset { +4 }{ N } { O }_{ 2 }+{ H }_{ 2 }O\)
Cu +2HNO₃ \(\longrightarrow \)  Cu(NO₃)₂ + NO₂ + H₂O
Cu + 2HNO₃ + 2HNO₃ \(\longrightarrow \) Cu(NO₃)₂ + 2NO₂ + 2H₂O
Cu + 4HNO₃ \(\longrightarrow \) Cu (NO₃)₂ + 2No₂ + 2H₂O
iv) \({ KMn }O_{ 4 }+{ H }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ MnSO }_{ 4 }+{ CO }_{ 2 }+{ H }_{ 2 }O\)
\({ K }\overset { +7 }{ \underset { \underset { 2\times { 3e }^{ - } }{ \downarrow } }{ M } } n{ O }_{ 4 }+{ H }_{ 2 }\overset { -1 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ C_{ 2 } } } } { O }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+\overset { +2 }{ M } n{ SO }_{ 4 }+\overset { +4 }{ C } { O }_{ 2 }+{ H }_{ 2 }O\)
2KMnO₄ + 5 H₂C2O₄ + H₂S0₄ \(\longrightarrow \) Mn0₂ + Na₂S0₄ + KOH
2KMnO₄+ 5 H₂C₂O₄ + H₂S0₄ \(\longrightarrow \) K₂SO₄ + 2MnSO₄ + 10CO₂ + H₂O
2KMnO₄ + 5 H₂C₂O₄ + 3H₂S0_{4 \(\longrightarrow \)}  K₂S0₄ + 2MnS0₄ + 10C0₂ + 8 H₂0

\(\overset { o }{ Z } n\longrightarrow { Z }n^{ 2+ }\)
\(\overset { +5 }{ N } { O }_{ 3 }^{ - }\longrightarrow \overset { 2+ }{ NO } \)
(1) \(\Rightarrow \) Zn \(\rightarrow\) Zn²⁺ +2e^-                                  .....(3)
(2) \(\Rightarrow \) \({ NO }_{ 3 }^{ - }+{ 3e }^{ - }+{ 4H }^{ + }\longrightarrow NO+{ 2H }_{ 2 }O\)......(4)

half reaction \(\Rightarrow \) \({ S }_{ 2 }{ O }_{ 3 }^{ 2- }\longrightarrow { S }_{ 4 }{ O }_{ 6 }^{ 2- }\)
\({ I }_{ 2 }\longrightarrow { I }^{ - }\)
 

\(\overset { +3 }{ C } _{ 2 }{ O }_{ 4 }^{ 2- }\longrightarrow \overset { +4 }{ C } { O }_{ 2 }\)
\(\overset { +6 }{ Cr } _{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { Cr }^{ 3+ }\)
(1)  \(\Rightarrow \) \({ C }_{ 2 }{ O }_{ 4 }^{ 2- }\longrightarrow { 2CO }_{ 2 }+{ 2e }^{ - }\)
\({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { 2Cr }^{ 3+ }+{ 7H }^{ 2 }O\)

Half reactions are:
\(\overset { +7 }{ M } { nO }_{ 4 }^{ - }\longrightarrow { Mn }^{ 2+ }\)
and \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }\)
(1) \(\Rightarrow \) \({ MnO }_{ 4 }^{ - }+{ 8H }^{ - }+5e^{ - }\longrightarrow { Mn }^{ 2+ }+{ 4H }_{ 2 }O\)
(2) \(\Rightarrow \) \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+{ 2e }^{ - }\)

ii) 

iii) 

iv) 

Empirical formula = Na₂ SH₂₀ O₁₄ 
\(\left[ \begin{matrix} { Na }_{ 2 }{ SH }_{ 20 }{ O }_{ 14 } \\ =(2\times 23)+(1\times 32)+(20\times 1)+14(16) \\ =46+32+20+234 \\ =322 \end{matrix} \right] \)
n = \(\frac { molar\quad mass }{ caluclated\quad empirical\quad formula\quad mass } =\frac { 322 }{ 322 } =1\)
Molecular formula = Na₂ SH₂₀O₁₄
Since all the hydrogen in the compound are present as water
\(\therefore \) The molecular formula is Na₂ SO₄ 10H₂O.

Empirical Formula = C₆ H₆O
n = \(\frac { molar\ mass }{ calculated\ eprirical\ formula\ mass } \)
= \(\frac { 2\times \ vapour\ density }{ 94 } \frac { 2\times 47 }{ 94 } =1,\) 
Molecular formula (C₆H₆O)  x 1 = C₆H₆O.

2Al + Fe₂O₃ \(\longrightarrow \)  Al₂O₃ + 2Fe

Molar mass of Al₂O₃ format = 6 mol x 102 g mol^-1 = 612 g
[  Al₂O₃ : (2 x 27) + 3(16) = 54 + 48 = 102] = 612 g
Excess reagent = Fe₂O₃
Amount of excess reagent left at the end of the reaction = 1 mol x 160 g mol^-1
= 160g [ Fe₂O₃ : (2 x 56) + (3 x 16) = 112 + 48 = 160] = 160 g

Astronauts must wear space suits filled with air, whenever they leave a space craft and are exposed to the environment of space.
Dangers experienced on moon's space :
1. In space there is no air to breathe and no air pressure.
2. Space is extremely cold and filled with dangerous radiations.
3. If they do not wear space suits, there may be bleeding due to high body pressure.
4. The space suits prevent astronauts from impacts of small bits of space dust.
5. There is no atmosphere on the moon and there are dangers from micro-meteorite impacts.
6. The astronauts are protected from these dangers on wearing space suits.

mass of one atom  = 6.645  x 10^-23 g
\(\therefore\) mass of 1 mole of an atom = 6.645  x 10^-23 g x 6.022 x 10²³ = 40 g
\(\therefore\) number of moles of element in 0.320 kg = \(\frac { 1\quad mole }{ 40g } \times 0.320kg\)
= \(\frac { 1mol\times 320g }{ 40g } \)
= 8 mol 

The solution to Schrodinger equation gives the permitted total energy values called eigen values and the corresponding wave function represent atomic orbitals.

(i) Due to small size of Nitrogen atom.
(ii) Due to polar nature of N-H bond.
(iii) Due to inter molecular H- bonding which are stronger than London forces present in other hydrides. Other hydrides lack H - bonding