Second electron is to be removed from cation in which same number of protons attract less number of electrons more srongly, therefore, more energy is needed to remove second electron.

Y(39) : 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹
The element is Ytterium(39) It belongs to 5th period and 3rd group due to 3 valence electrons.

Δv = Uncertainty in speed = 100 - 99.99 = 0.01 %
\(\Delta V=40 \mathrm{~m} \mathrm{~s}^{-1} \times \frac{0.01}{100}=40 \times 10^{-4}\)
= 4 x 10^-3 m s^-1
According to Heisen berg's Principle
\(\Delta x . \Delta V=\frac{h}{4 \pi m}\)
\(\Delta x=\frac{h}{4 \pi m \Delta v}\)
where Δx uncertainty in position.
'h' is planck's constant,
'm' is mass of electron,
π = 3.142
\(\Delta x=\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} \mathrm{~s}^{-1}}{4 \times 4 \times 10^{-3} \mathrm{~ms}^{-1} \times 9.11 \times 10^{-31} \mathrm{~kg} \times 3.142}\)
Δx = 1.447 x 10^-2 m

The relative stability of various classes of carbonium ions may be explained by the number of no-bond resonance structures that can be written for them. Such structures are obtained by shifting the bonding electrons from an adjacent C-H bond to the electron deficient carbon so the positive charge originally on carbon is dispersed to the hydrogen. This manner of electron release by assuming no bond character in the adjacent C-H bond is called Hyperconjugation. Greater the hyperconjugation, greater will be the stability of alkenes.

The decomposition of a compound by heat is called pyrolysis. This process when applied to alkanes is known as cracking.

Mass percent of an element 
= \( \frac{\text{ Mass of that element in the compund} \times 100 }{\text{ Molar mass of } Na_2SO_4}\)
= (2\(\times\)22.99) + 32.06 + (4\(\times\)16.00) = 142.04 g
Mass percent of Sodium
= \(\frac{45.98 \times 100}{142.04}\)
= 32.37
Mass percent of Sulphur  
= \(\frac{32.06\times100}{142.04}\)
= 22.57
Mass percent of  oxygen  
= \(\frac{64\times100}{142.04} \)
= 45.06

C < O < F is increasing order of electron gain enthalpy due to decrease in atomic size.

Given results are tabulated as follows:

The empirical formula of the given compound is Na₂CO₃.
Now, Empirical formula mass of the compound = \(=(2 \times 23 \mathrm{u})+(1 \times 12 \mathrm{u})+(3 \times 16 \mathrm{u})=46 \mathrm{u}+12 \mathrm{u}+48 \mathrm{u}=106 \mathrm{u}\)
Then, \(n=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{106 \mathrm{u}}{106 \mathrm{u}}=1\)
Therefore, molecular formula of the given compound is given by, Molecular formula = 1 x Empirical formula = \(1 \times \mathrm{Na}_{2} \mathrm{CO}_{3}=\mathrm{Na}_{2} \mathrm{CO}_{3}\)

The main points of difference between inductive and resonance effects are given below.

Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down a group. However, adding an electron to the 2p-orbital leads to greater repulsion than adding an electron to the larger 3p-orbital. Hence the element with most negative electron gain enthalpy is chlorine; the one with the least negative electron gain enthalpy is phosphorus.

(i) Wurtz reaction: Alkanes are produced by heating an alkyl halide with sodium metal in dry ether solution.
CH₃Br + 2Na + BrCH₃ \(\overset { dry\ ether }{ \longrightarrow } \) CH₃CH₃ + 2NaBr
                                                    Ethane
(ii) Hydrogenation: Alkenes react with hydrogen in presence of Ni or Pd catalyst to form saturated compounds.
CH₂ = CH₂ + H₂ \(\overset { Ni }{ \underset { 373K }{ \longrightarrow } } \) CH₃ - CH₃
                                       Ethane

6.022 x 1022 molecules of oxygen
6.022 x 1023 molecules of oxygen = 1 mol
6.022 x 1022 molecules of oxygen = 1 mol x \(\frac { 6.022\times { 10 }^{ 22 } }{ 6.022\times { 10 }^{ 23 } } \) = 0.1 mol

392 g of sulphuric acid
Molar mass of H₂SO₄ = 2 x 1 + 32 + 4 x 16 = 98 g
98 g of sulphuric acid = 1 mol
392 g of sulphuric acid = 1 mol x\(\frac { 392 \ g }{ (98 \ g) } \) = 4 mol

ionization\(({ _{ 7 }N }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 3 }),\) enthalpy of nitrogen is greater than oxygen \(({ _{ 8 }O }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 4 }),\)due to extra stable half-filled 2p-orbitals. Similarly, ionization enthalpy of phosphorus is \(({ { _{ 15 }{ P } } }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2 }p^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 3 })\) greater than sulfur on\(({ { _{ 16 }{ S } } }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2 }p^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 4 })\) moving down the group, ionization enthalpy decreases with increasing atomic size. so, the increasing order of first ionization enthalpy is S.

The uncertainty in the speed is 2%, i.e.,
\(45 \frac { 2 }{ 100 } =0.9m{ s }^{ -1 }\)
Using the equation Heisenberg's principle,
\(\triangle x=\frac { h }{ 4\pi m.\triangle v } =\frac { 6.6\times{ 10 }^{ -34 } Js }{ 4\times3.14\times40g \times{ 10 }^{ -3 } \ kg \ g^{-1}\times 0.9 m s^{-1} }\)
= 1.46×10^–33 m
This is nearly ~ 10¹⁸ times smaller than the diameter of a typical atomic nucleus. As mentioned earlier for large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.

\(\begin{array}{l} \text { (a) } \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\\ \quad \quad \quad \quad \text { Hex }-1 \text { - ene } \end{array}\)
\(\begin{array}{l} \text { (b) } \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\\ \quad \quad \quad \quad \text { Hex }-2 \text { - ene } \end{array}\)
\(\begin{array}{l} \text { (c) } \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{3}\\ \quad \quad \quad \quad \text { Hex - } 3 \text { - ene } \end{array}\)

The aromatic compounds apparently contain alternate double and single bonds in a cyclic structure, and resemble benzene in chemical behaviour. They undergo substitution reactions rather addition reactions. This characteristic behaviour is called Aromatic character or Aromaticity.
Conditions for Aromaticity:
(i) An aromatic compound is cyclic and planar.
(ii) Each atom in an aromatic ring has a p-orbital. These p-orbitals must be parallel so that a continuous overlap is possible around a ring.
(iii) The cyclic \(\pi\)-molecular orbital formed by the overlap of p-orbitals must contain (4n + 2) \(\pi\) electrons, where (n = 0, 1, 2, 3, 4 etc.)

Limitations of Bohr's model of an atom:
(i) It could not explain spectrum of multi-electron atoms.
(ii) It could not explain Zeeman and Stark effects.
(iii) It could not explain shape of molecules.
(iv) It was not in accordance with Heisenberg's uncertainty principle.
Quantum Mechanical Model: It was developed on the basis of Heisenberg's uncertainty principle and dual behaviour of matter. Main features of this model are given below :
(i) The energy of electrons in an atom is quantized i.e. can only have certain values.
(ii) The existence of quantized electronic energy levels is a direct result of the wave-like properties of electrons.
(iii) Both, the exact position and velocity of an electron in an atom cannot be determined simultaneously.
(iv) The orbitals are filled in increasing order of energy. All the information about the electron in an atom is stored in orbital wave function.\(\Psi \)
(v) From the value of\(\Psi \)²  at different points within atom, it is possible to predict the region around the nucleus where electron most probably will be found.

Atomic Number (Z): The atomic number of an element is equal to the number of protons present inside the nucleus of its atoms. Since, an isolated atom has no net charge on it, in neutral atoms, the total number of electrons is equal to its atomic number. 
Atomic number (Z) = Number of protons in the nucleus of an atom
= Number of electrons in the neutral atoms
Mass Number (A): The sum of the number of neutrons and protons in the nucleus of an atom is called its mass number. Mass number is denoted by A. Thus, for an atom, Mass number (A) = Number of protons (P) + Number of neutrons (n) A = P + n
Neutron: It is neutral particle. It is present in the nucleus of an atom. Expect hydrogen (which contains only one electron and one proton but no neutron), the atoms of all other elements including isotopes of hydrogen contain all the three fundamental particles called neutron, proton and electron.
The relation between mass number, Atomic no. and no. of neutrons is given by the equation:
A = Z+n I
Where A = Mass number
Z = Atomic number
n = Number of neutrons in the nucleus.

Factors affecting Ionization enthalpy.
(i) Atomic size. With the increase in atomic size, the number of electron shells increases and thus the force of attraction between the electrons and the nucleus decreases. Therefore the ionization enthalpy decreases.
(ii) Nuclear charge. As the nuclear charge increases the attraction for the electron also increases that's why ionization enthalpy increases.
(iii) Screening or shielding effect. In a multi-electron atom, the electron present in the inner shells shield the electrons in the valence shell as a result these electrons experience less attraction from the nucleus. This leads to lesser ionization enthalpy.
Variation along a period. On moving from left to right in a period the nuclear charge increases and the atomic size decreases as a result ionization enthalpies are expected to increase.
Variation within a group. On moving down the group as the atomic size of the elements increases that's why ionization enthalpy decreases down the group

(a) Group 1 because they have largest atomic size.
(b) Group 18 because they have stable electronic configuration.
(c) Be(1s² 2s²). It is because 'Be' has completely filled s-orbital which is more stable therefore, needs higher energy to remove electrons than B(1s² 2s² 2p¹).
(d) It is because 'N' has half filled orbital which is more stable and more energy is needed to remove electron in 'N' than O.