(i) ‘hexane’ indicates the presence of 6 carbon atoms in the chain. The functional group chloro is present at carbon 2. Hence, the structure of the compound is CH₃CH2CH₂CH₂CH(Cl)CH₃.
(ii) ‘pent’ indicates that parent hydrocarbon contains 5 carbon atoms in the chain. ‘en’ and ‘ol’ correspond to the functional groups C=C and -OH at carbon atoms 4 and 2 respectively. Thus, the structure is CH₂ = CHCH₂CH (OH)CH₃
(iii) Six membered ring containing a carbon-carbon double bond is implied by cyclohexene, which is numbered as shown in (I). The prefix 3-nitro means that a nitro group is present on C-3. Thus, complete structural formula of the compound is (II). Double bond is suffixed functional group whereas NO₂ is prefixed functional group therefore double bond gets preference over –NO₂ group:

(iv) ‘1-ol’ means that a -OH group is present at C-1. OH is suffixed functional group and gets preference over C=C bond. Thus the structure is as shown in (II):

(v) ‘heptanal’ indicates the compound to be an aldehyde containing 7 carbon atoms in the parent chain. The ‘6-hydroxy’ indicates that -OH group is present at carbon 6. Thus, the structural formula of the compound is:CH₃CH(OH)CH₂CH₂CH₂CH₂CHO. Carbon atom of –CHO group is included while numbering the carbon chain.

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 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.

Significant figures are meaningful digits which are known with certainty plus one which is estimated or uncertain. The uncertainty is indicated by writing the certain digits and the last uncertain digit.
There are certain rules for determining the number of significant figures. These are stated below:
1. All non-zero digits in a number are significant
2. The zeros between two non-zero digits are always significant
3. The zeros written to the left of the first non-zero digit in a number are non-significant. They simply indicate the position of the decimal point.
4. All zeros placed to the right of a decimal point in a number are significant.
(i) 0.0025 - There are 2 significant figures.
(ii) 208 - There are 3 significant figures.
(iii) 5005 - There are 4 significant figures.
(iv) 126,000 - There are 3 significant figures.
(v) 500.0 - There are 4 significant figures.
(vi) 2.0034 - There are 5 significant figures.

Homolytic fission is breaking of a bond in such a manner that each atom takes one electron each to form free radicals.
\(A-B\longrightarrow A.+B\)
(b) Organic ions which contain a negatively charged carbon atom are called carbanions. e.g., \(\overset { \ominus }{ { CH }_{ 3 } } \) is carbanion.

(i) a) ₁H = 1s¹, H^- = 1s²
b) ₁₁Na = 1s²2s², 2p⁶, 3s¹, Na⁺ = 1s²2s²2p⁶
c) ₈O = 1s²2s², 2p², 2p⁴, O^2- = 1s²2s²2p⁶
d) ₉F = 1s²2s²,2p⁵, F^- = 1s²2s²2p⁶.
(ii) To obtain an atomic number of an element fill the orbitals in order of their increasing energies up to the given orbital configuration.
a) 1s²2s², 2p⁶, 3s² (Z = 12)
b) 1s²2s², 2p³ (Z = 7)
c) 1s²2s², 2p⁶, 3s²,3p⁵ (Z = 17).
(iii) a) [He]2s¹ represents ₃Li (lithium)
b) [Ne] 3s²3p³ represents ₁₅P (phosphorus)
c) [Ar]4s²3d¹ represent ₂₁Sc (scandium)

When n = 5, l = 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s and 5p increases is 5s < 4d < 5p. The total number of orbitals available are 9. The maximum number of electrons that can be accommodated is 18; and therefore 18 elements are there in the 5^th period.

To obtain an atomic number of an element fill the orbitals in order of their increasing energies up to the given orbital configuration.
a) 1s²2s², 2p⁶, 3s² (Z = 12)
b) 1s²2s², 2p³ (Z = 7)
c) 1s²2s², 2p⁶, 3s²,3p⁵ (Z = 17).

Isoelectronic species have the same number of electrons but different atomic numbers. Number of positive charge shows the number of electrons lost and number of negative charges shows the number of electrons gained by an atom. Calculation of number of electrons have been shown below.
\(_{ 11 }Na^{ + }=11-1=10{ e }^{ - }, \ _{ 19 }K^{ + }=19-1=18{ e }^{ - },\)
\(_{ 12 }Na^{ 2+ }=12-2=10{ e }^{ - }, \ _{ 20 }Na^{ 2+ }=20-2=18{ e }^{ - },\)
\(_{ 16 }S^{ 2- }=16+2=18{ e }^{ - }, \ _{ 18 }Ar=18{ e }^{ - }\)
Hence, isoelectronic species are
\(Na^{ + }and \ Mg^{ 2+ } \ { k }^{ + },{ Ca }^{ 2+ },{ s }^{ 2- } \ and \ Ar\)

Since 69.9% and 30.1% of iron and oxygen are present in 100 grams of iron oxide.
Step I - Calculation of number of moles
Number of moles of iron present in 100 gram of iron oxide\( =\frac{\text{ weight in grams (iron)}}{\text{molecular weight of iron}} =\frac{69.9}{55.8}\)=1.25mole
Number of moles of oxygen present in 100 gram of iron oxide \(= \frac{\text{weight in grams (oxygen) }}{ \text{molecular weight of oxygen}}=\frac{30.1}{16}\)=1.88moles
Step II - Calculation of ratio of number of moles of iron to number of moles of oxygen
Ratio of iron to oxygen \(=\frac{ 1.25}{1.88} = \frac{1}{1.5}\)
On multiplying the given ration with 2 we get 2/3
Hence the empirical formula of iron oxide is Fe₂O₃
Step III - Calculation of n
Emphirical formula = Fe₂O₃ = (2 x 55.85) + (3 x 16.00) = 159.7 g mol^-1
Molar mass of Fe₂O₃=159.69
n = \(\frac{molar mass}{ emphirical formula mass } \) = \(\frac{159.8}{159.7}\) = 1

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.

(a) 3-Ethyl-4,7-dimethyl nonane
(b) 1-Ethyl-2, 5-dimethyl cyclohexane

Modem Periodic Law states that physical and chemical properties of the elements are a periodic function of their atomic numbers.

(a) σ_{C – C}: 4; σ_C–H : 6; π_C=C :1; π C≡C:2
(b) σ_{C – C}: 3; σ_C–H: 6; π_C=C: 2.

Isoelectronic species are those which have same number of electrons.
(i) Na has 10 electrons.Therefore, the species
N^3-, O^2-, F^-, Mg²⁺, Al³⁺ etc., each of which has also 10 electrons and hence, isoelectronic with it
(ii) Cl^- has 18 electrons. Therefore, the species P^3-, S^2-, Ar, K⁺ and Ca²⁺, each one of which contains 18 electrons and hence , isoelectronic with it.
(iii) Ca²⁺ has 18 electrons. Therefore, the species P^3-, S^2-, Ar and K⁺, each of which also contains 18 electrons and hence, isoelectronic with it.
(iv) Rb⁺ has 36 electrons. Therefore, the species Br^-, Kr or Sr²⁺ each of which also has 36 electrons and hence, isoelectronic with it.

Mendeleev's periodic law : It states that the properties of the elements are a periodic function of their atomic weights
Modern periodic law : It states that the properties of the elements are a periodic function of their atomic numbers. Thus, change in the base of classification of elements from atomic weight to atomic number is the basic difference between Mendeleev's periodic law and the modern periodic law.

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.

In a period, the non-metallic character increases from left to right. Thus, among B, C, Nand F, non-metallic character decreases in the order: F > N> C> B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F> N> C> B > Si, i.e., option (c) is correct.

(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.

(a) 1 mole of N₂ needs 3 moles of H₂ ,
Therefore, 2 moles of N₂ needs 6 moles of H₂ .
But we have only 2 moles of H₂ , so H₂ is limiting reagent.
3 moles of H₂ gives 2 moles of NH₂.
2 moles of H₂ gives \(\frac{2}{3} \times 2=1.33\) moles of NH₃ .
(b) 18 g of water (1 mole) containing 10 x 6.022 x 10²³ electrons = 6.022 x 10²⁴ electrons.
[ \(\because\)1 molecule of H₂O = 2 + 8 = 10 e)
(c) 'M is molarity
'd is density of solution
'm' is molality
\(m=\frac{M \times 1000}{1000 \times d-M \times \text { Molar mass }}\)
\(=\frac{1 \times 1000}{1000 \times 1.10-1 \times 58.5}\)
\(=\frac{1000}{1100-58.5}=\frac{1000}{1041.5}=m=0.96 \mathrm{~mol} / \mathrm{kg} .\)
(d) It is ratio of number of moles of solute (component) to the total number of moles of solute and solvent (all components).
(e) X_glucose + x_H2O = 1
\(\Rightarrow\) X_H2O = 1 - 0.1 = 0.9