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

\(\%\ of\ P=\frac{62}{222}\times\frac{weight\ of\ Mg_2P_2O_7\ formed\times100}{weight\ of\ organic\ compound}\)
\(=\frac{62}{222}\times\frac{0.22}{0.12}\times100\)
= 51.20%

According to uncertainty principle,
\(\Delta \)x. m\(\Delta \)v = \(\frac { h }{ 4\pi } \) or \(\Delta \)v =\(\frac { h }{ 4\pi m\Delta x } \); h = 6.626 x 10^-34 kg m^-2s^-1; m = 10 g = 10^-2 .kg
\(\Delta \)x, = 10^-5 m ; \(\Delta \)v = \(\frac { (6.626\times 10^{ -34 }kg\quad m^{ 2 }s^{ -1 } }{ 4\times 3.143\times (10^{ 2 }kg)\times (10^{ -5 }m) } \) = 5.27 x 10^-28 mv

(i) V and VI or VI and VII form a pair of metamers since they differ in the number of carbon atoms on the either side of the functional group, i.e. O-atom.
(ii) I and V, I and VI , I and VII; II and V, II and VI, II and VII ; III and V, II and VI; III and VII;IV and V ; IV and VI and IV and VII are all functional group isomers.
(iii) I and II, III and IV and, VI and VII represent position isomerism.
(iv) I and III, I and IV, II and III and II and IV represent chain isomerism.
(a) CH₃COH₃, (b) H - CH = CH₂

First member of each group of representative elements ( i.e. s and p-block elements ) shows anomalous behaviour due to
(i) small size
(ii) high ionisation enthalpy
(iii) high electronegativity and
(iv) absence of d- orbitals.
For example in s-block elements, lithium shows anoimalous behaviour from rest of the alkali metals.
(i) Compounds of lithium have significant covalent character. While compounds of other alkali metals are predominantly ionic.
(ii) Lithium reacts with nitrogen to form lithium nitride while other alkali metals do not form nitrides.
In p-block elements, first member of each group has four orbitals, one 2s- and three 2p-orbitals in their valence shell. So, these elements show a maximum covalency of four ehile other members of the same group or different group show a maximum covalency beyond four due to availability of vacant d-orbitals.

Molar mass of methane ( CH₄ ) = 12 + 4\(\times\)1 = 16 g
16 g of methane contains = 6.022\(\times\)10²³ moilecules
16 g of methane will contain 
= \(\frac{7.022 \times 10^{23}}{(16 g)}\)
= 6.022 \(\times\)10²² molecules
(ii) Number of electrons in  6.022 \(\times\)10²² molecules of methane. 1 molecule of methane contains = 6 + 4 = 10 electrons  6.022 \(\times\)10²² moleculesof methane contain electrons =  6.022 \(\times\)10²² 2 molecules\(\times\)10
= 6.022\(\times\)10²³ molecules

(i) Molar mass of H₂O = 2 x atomic mass of hydrogen + 1 x atomic mass of oxygen
= 2 x 1.0079 u + 1 x 16.00 u
= 18.0158
(ii) Molar mass of CO₂ = 1 x atomic mass of Carbon +2 x atomic mass of oxygen
= 1 x 12.01 u +2 x 16.00 u 
= 44.01 u
(iii) Molar mass of CH4 =1 x atomic mass of Carbon +4 x atomic mass of Hydrogen
= 1 x 12.01 u +4 x 1.0079 u
= 16.0416

Step 1. Conversion of mass per cent to grams : Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present.
Step 2. Convert into number moles of each element : Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound
Moles of hydrogen = \(\frac{4.07 \mathrm{~g}}{1.008 \mathrm{~g}}=4.04\)
Moles of carbon = \(\frac{24.27 \mathrm{~g}}{12.01 \mathrm{~g}}=2.021\)
Moles of chlorine = \(\frac{71.65 \mathrm{~g}}{35.453 \mathrm{~g}}=2.021\)
Step 3. Divide each of the mole values obtained above by the smallest number amongst them : Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl. In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.
Step 4. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements : CH₂Cl is, thus, the empirical formula of the above compound.
Step 5. Writing molecular formula : (a) Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula.
For CH₂Cl, empirical formula mass is
12.01 + (2 x 1.008) + 35.453
= 49.48 g
(b) Divide Molar mass by empirical formula mass
\(\frac{\text { Molar mass }}{\text { Empirical formula mass }}=\frac{98.96 \mathrm{~g}}{49.48 \mathrm{~g}}\)
= 2 = (n)
(c) Multiply empirical formula by n obtained above to get the molecular formula
Empirical formula = CH₂Cl, n = 2. Hence molecular formula is C₂H₄Cl₂.

Hyperconjugation: 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 arrived by shifting the bonding electrons from an adjacent C-H bond to the electron-deficient carbon. In this way, 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 or No-Bond Resonance. The greater the hyperconjugation, the greater will be 'the stability of the compound. The increasing order of stability can be shown as.

Inductive Effect: The inductive effect refers to the polarity produced in a molecule as a result of higher electronegativity of one atom compared to another. Atoms or groups which lose electron towards a carbon atom are said to have + I Effect.
Those atoms or groups which draw electron away from a carbon atom are said to have -I Effect.
Common examples of -I effect are:
NO₂, F, Cl, Br, I, OH etc.
Examples of +1 effect are (Electron releasing)
(CH₃)₂C-, (CH₃)₂CH-, CH₃CH₂-, CH₃- etc.
Electromeric effect: The electromeric effect refers to the polarity produced in a multiple bonded compound as it is approached by a reagent.

The atom A has lost its share in the electron pair and B has gained this share. As a result A acquires a positive charge and B a negative charge. It is a temporary effect and takes place only in the presence of a reagent.
(a) -I-effect as shown below:
As the number of halogen atoms decreases, the overall -1- effect decreases and the acid strength decreases accordingly.

(b) +I-effect as shown below:
As the number of alkyl groups increases, the +I-effect increases and the acid strength decreases accordingly.

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

Electronegativity:
1. The tendency of an atom in a molecule to attract the shared pair of electrons towards itself is known as electronegativity.
2. There is no specific unit for electronegativity. In the modern periodic table:
3. In the modern periodic table: as we move left to right, across a period the nuclear charge increases, atomic size decreases, and electronegativity value increases.
4. As we move down the group there is an increase in the atomic number also nuclear charge and a decrease in the electronegativity value.
5. In general, metal shows a lower electronegativity value compared to on-metals.

A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.
(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of B will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.
(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.

The balanced equation for the combustion of methane is :
\({ CH }_{ 4 }(g)+2{ O }_{ 2 }(g)\rightarrow { CO }_{ 2 }(g)+{ 2H }_{ 2 }O(g)\)
(i) 16 g of CH₄ corresponds to one mole.
(ii) From the above equation, 1 mol of CH₄ (g) gives 2 mol of H₂O (g).
2 mol of water (H₂O) = 2 × (2 + 16) = 2 × 18 = 36 g
1 mol H₂O = 18 g H₂O ⇒ \(\frac{18g H_2O}{1 mol H_2O} = 1\)
Hence, 2 mol H₂O × \(\frac{18g H_2O}{1 mol H_2O} \)
= 2 × 18 g H₂O = 36 g H₂O

electrons in the atom of any element is called its atomic radius. It refers to both covalent or metallic radius depending on whether the element is a non-metal or a metal. Ionic radius. The Ionic radii can be estimated by measuring the distances between cations and anions in ionic crystals.

The digits in a properly recorded measurement are known as significant figures. It is also defined as follows. The total numbers of figures in a number including the last digit whose value is uncertain is called number of significant figures.

Cations : Na⁺, Mg²⁺ Al³⁺
Anions : N^3- , O^2-,F^-

General configuration for 17th group elements is ns² np⁵ . In the third period, the principal quantum number for valence shell is three, so the electronic configuration of valence shell for the given element is 3s², 3p⁵. Third period starts from atomic number, Z = 11 and end at Z = 18. Hence, the atomic number of the given element is 10 + 7 = 17.

Mass number is a whole number because it is the sum of the number of protons and number of neutrons whereas atomic mass is fractional because it is the average relative mass.

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

Statement (b) is incorrect.
The d-block has 10 columns because a maximum of 10 electrons can occupy all the orbitals in a d subshell.

(a) Free radicals are atoms or group of atoms having one electron in excited state
(b) 3° > 2° > 1°.
(c) 400 g mol^-1, because it will contain at least one atom 'S', i.e., 32 g of sulphur.
(d) CH₄.
(e) It is estimated by taking mass of BaSO₄ formed.
(f) It is done to convert NaCN to HCN and N₂S to H₂S if nitrogen and sulphur are present because they will interfere with the AgNO₃ test for halogens, H₂S and HCN are removed as gases.

(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.