(i) Albert Einstein proposed that light has dual nature. i.e. like photons behave both like a particle and as a wave.
(ii) Louis de Broglie extended this concept and proposed that all forms of matter showed dual character.
(iii) He combined the following two equations of energy of which one represents wave character (hv) and the other represents the particle nature (mc²).

Given  \(\Delta\)x = 0.1  or 0.1 x 10^-10 m or 10^-11 m.
h = 6.626 x 10^-34 kg m² s^-1, m = 9.11 x 10^-31 kg
According to uncertainty principle,
\(\Delta x,m(\Delta v)=\frac{h}{4\pi}\)
i.e., \(\Delta v=\frac{h}{4\pi\times m\times \Delta x}\)
= \(\frac{6.626\times 10^{-34}kg m^2 s^{-1}}{4\times 3.14\times 9.11\times 10^{-31}kg \times 10^{-11}m}\)
= 5.79 x 10⁶ ms^-1.

KE = \(\frac{1}{2}\)mv² = 5.85 x 10^-25 J
By de Broglie equation \(\lambda=\frac{h}{mv}\)
But \(\lambda=\frac{\nu}{V}=\frac{h}{mv}\)
v = \(\frac{mv^2}{h}=\frac{2\times 5.85\times 10^{-25}J}{6.026\times 10^{-34}J}\)
= 1.77 x 10⁹ s^-1.

Given m = 9.1 x 10^-31 kg; v = 10³ ms^-1
h = 6.6 x 10^-34 kg m² sec^-1
\(\lambda=\frac{h}{mv}=\frac{6.6\times 10^{-34}kgm^2 sec^{-1}}{9.1\times 10^{-31}kg\times 10^3 ms^{-1}}\)
= 7.25 x 10^-7 m.

Similarities:
(i) Both have similar shape.
(ii) Both have same angular momentum = \(\sqrt{l(l+1)} \frac{h}{2\pi}\)
Dissimilarities:
(i) Is orbital has no node while 2s orbital has one node.
(ii) Energy of 2s orbital is greater than Is orbital.
(iii) The size of the 2s orbital is larger than Is orbital.

This is because, according to this principle, if one electron of an atom has same particular values, for the four quantum numbers, then all the other electrons in that atom are excluded from having the same set of values.

According to J. C. Maxwell, whenever an electron is subjected to acceleration, it emits radiation and loses energy. As a result of this, its orbit should become smaller and smaller and finally it should drop into the nucleus by following a spiral path. This means that atom would collapse and thus Rutherford's model failed to explain stability of atoms. Another drawback of the Rutherford's model is that it gives no information about the electronic structure of an atom.

According to this theory, atom was assumed to consist of a sphere of uniform distribution of about 10^-10 m positive charge with electrons embedded in it such that the number of electrons equal to the number of positive charges and the atom as a whole is electrically neutral.

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

Empirical formula = C₂H₃O₃
Empirical formula mass = 24 + 3 + 48 = 75
Molecular mass = 2 x Vapour density
= 2 x 75 = 150
n = \({Molecular \ mass \over Empirical \ formula \ mass }={150\over 75}=2\)
Molecular formula = (Empirical formula)_n
\(\therefore\) Molecular formula = C₂H₃O₃ x 2
Molecular formula = C₄H₆O₆

Factors influencing ionization enthalpy:
(i) Size of the atom:
If the size of an atom is larger, the outermost electron shell from the nucleus is also larger and hence the outermost electrons experience lesser force of attraction. Hence it would be more easy to remove an electron from the outermost shell. Thus, ionization energy decreases with increasing atomic sizes.
Ionization enthalpy \(\infty{1\over Atomic \ size}\)
(ii) Magnitude of nuclear charge:
As the nuclear charge increases, the force of attraction between the nucleus and valence electrons also increases. So, more energy is required to remove a valence electron. Hence I.E increases with increase in nuclear charge.
Ionization enthalpy \(\alpha \ nuclear \ charge\)
(iii) Screening or shielding effect of the inner electrons:
The electrons of inner shells form a cloud of negative charge and this shields the outer electron from the nucleus. This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. If screening effect increases, ionization energy decreases.
Ionization enthalpy  \(\infty{1\over Screening\ effects}\)
(iv) Penetrating power of subshells s, p, d, and f:
The s-orbital penetrate more closely to the nucleus as compared to p-orbitals. Thus, electrons in s-orbitals are more tightly held by the nucleus than electrons in p-orbitals. Due to this, more energy is required to remove a electron from an s-orbital as compared to a p-orbital. For the same value of 'n', the penetration power decreases in a given shell in the order.
s>p>d>f.
(v) Electronic configuration:
If the atoms of elements have either completely filled or exactly half filled electronic configuration, then the ionization energy increases.

Step - 1 : To find atoms undergoing change in O.N.
\(\overset { +1-1 }{ KBr } +\overset { +4-2 }{ MnO_{ 2 } } +\overset { +1+6-2 }{ { H }_{ 2 }SO_{ 4 } } \rightarrow +\overset { +1+1+6-2 }{ KHSO_{ 4 } } +\overset { +2+6-2 }{ MnSO_{ 4 } } +\overset { -1-2 }{ { H }_{ 2 }O } +\overset { 0 }{ { Br }_{ 3 } } \)
Step - 2 : To find the total increase and decrease
\(\overset { +4 }{ MnO_{ 4 } } \rightarrow \overset { +2 }{ MnSO_{ 4 } } \) (Decrease in O.N. of 2 units per atom)
KBr^-1Br₂ (increase in O.N. of I unit per atom)
Total decrease = 2 x 1 = 2
Total increase = 1 x 2 = 2
Step - 3 : To balance the total increase and decrease in O.N, multiply KBr by 2.
2KBr + MnO₂ + H₂SO₄ ⟶ KHSO₄ + MnSO₄ + H₂O + Br₂
Step - 4 : To balance all atoms other than hydrogen and oxygen atoms
2KBr + MnO₂ + H₂SO₄ ⟶ 2KHSO₄ + MnSO₄+ H₂O + Br₂
Since SO₄^-2 (sulphate) radical does not undergo any change in O.N.
balance them RHS = 3 (SO₄²) radical; RHS = I (SO₄^-2)
Hence multiply H₂SO₄ in LHS by 3. The equation now becomes
2KBr + MnO₂ + 3H₂SO₄ ⟶ 2KHSO₄ + MnSO₄ + H₂O + Br₂
To balance 'O' atoms, multiply H₂O in RHS by 2.
2KBr + MnO₂ + 3H2SO4  ⟶ 2KHSO₄ + MnSO₄ + 2H₂0 + Br₂
This is the balanced equation.

\(\frac{1}{\lambda}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]\)
n₁ = 3,  n₂ = 5
\(\frac{1}{12818}=R[\frac{1}{9}-\frac{1}{25}]=\frac{16R}{9\times 25}\)
or 12418 = \(\frac{9\times 25}{16\times R}\)...(1)
When n₁ = 2,
 \(\frac{1}{\lambda}=R[\frac{1}{4}-\frac{1}{9}]=\frac{5}{36}R\)
\(\lambda=\frac{36}{5R}\)......(2)
Dividing equation (2) by equation (1)
\(\frac{\lambda}{12818}=\frac{36}{5R}\times \frac{16R}{9\times 25}=\frac{64}{125}\)
\(\lambda=\frac{64}{125}\) x 12818 = 6562.8 \(\overset { o }{ A } \)         

(i) In metal displacement reactions, we learnt that zinc replaces copper from copper sulphate solution. Let us examine whether .the reverse reaction takes place or not. As discussed earlier, place a metallic copper strip in zinc sulphate solution. If copper replaces zinc from zinc sulphate solution, Cu²⁺ ions would be released into the solution and the colour of the solution would change to blue. But no such change is observed. Therefore, we conclude that among zinc and copper, zinc has more tendency to release electrons and copper to accept the electrons.

(ii) Let us extend the reaction to copper metal and silver nitrate solution. Place a strip of metallic copper in silver nitrate solution taken in a beaker. After some time, the solution slowly turns blue. This is due to the formation of' Cu²⁺ ions, i.e. copper replaces silver from silver nitrate. The reaction is
(iii) It indicates that between copper and silver, copper has the tendency to release electrons and silver to accept electrons.

(iv) From the above experimental observations, we can conclude that among the three metals, namely, zinc, copper and silver, the electron releasing tendency is in the following order
Zinc> Copper> Silver.

Combination reaction: Redox reactions in which two substances combine to form a single compound are called combination reaction.

Molar mass = \(\frac { weight\ of\ the\ substance\ \times \ Molar\ volume }{ Volume\ of\ the\ substance\ at\ STP } \)
Molar volume at STP = 2.24 x 10^-2 m³
= 22.4 L (or) 22400 cc
Molar mass of the gas at STP = \(\frac { 23.2\times 22.4 }{ 20 } \)
= 25.984 g

Molar mass of HCl = 36.5 g
The molar volume of 36.5 g (1 mole) of He 1 = 2.24 x 10^-2m³.
\(\therefore\) The volume of 146 g (4 moles) of HCI  = \(\frac { 2.24\times { 10 }^{ -2 } }{ 36.5 } \times 146\)
= 8.96 x 10^-2m³
No. of molecules in 146 g of HCI = 4 N
=  4 x Avogadro Number
=  4 x 6.023 x 10²³
=  24.092 x 10²³
=  2.4092 x 10²⁴ molecules. 

H₂C₂O₄+KMnO₄ + H₂SO₄ \(\rightarrow\) H₂SO₄ + MnSO₄ + CO₂ + H₂O

Equalise the increase I decrease in O N by multiplying Cu species by 5 and Mn species by 1
5H₂C₂O₄+KMnO₄ + H₂SO₄ \(\rightarrow\) K₂SO₄ + MnSO₄ + 5CO₂ + H₂O
Balance all other atoms except H and O atoms
5H₂C₂O₄+2KMnO₄ + 3H₂SO₄ \(\rightarrow\) K₂SO₄ + 2MnSO₄ + 10CO₂ + H₂O
Balance O atom by adding H₂O on the side falling short of oxygen atoms.
5H₂C₂O₄+2KMnO₄ + 3H₂SO₄ \(\rightarrow\) K₂SO₄ + 2MnSO₄ + 10CO₂ + H₂O+ 7H₂O
The balanced equation is
5H₂C₂O₄+2KMnO₄ + 3H₂SO₄ \(\rightarrow\) K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂O​​​​​

Apple turns brown when cut since the surface is exposed to air and undergoes oxidation. It can be prevented by dipping sliced apples in lemon juice. Lemon juice is an antioxidant which takes in all the available oxygen and prevents it from reaching the apple's tissues

Mass of oxygen = 1.5 - 1.05
= 0.45 g
0.45g of oxygen combines with 1.05 g of metal.
ஃ 8g of oxygen combines with \(\frac{8\times1.05}{0.45}\) g of metaI
= 18.66 g of metal
ஃequivalent mass of metal = 18.66g equ^-1

(i) Equivalent Mass of Acids:
 E = \(\frac { Molar\ mass\ of\ the\ acid }{ basicity\ of\ the\ acid } \) 
(ii) Equivalent Mass of Bases
E = \(\frac { Molar\ mass\ of\ the\ base }{ Acidity\ of\ the\ base } \)
(iii) Equivalent Mass of Oxidising agent:
E =  \(\frac { Molar\ mass\ of\ the\ oxidising\ agent }{ no.\ of\ moles\ of\ electrons\ gained\ by\ one\ mole\ of\ the\ oxidsing\ agent } \)

Newlands law of Octaves:
When elements are arranged in increasing order of their atomic weights every eighth element resembles its properties with the first one just like the eight note of a musical table.

Limitations: Failed for heavier elements beyond calcium.

(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.