(i) The basic units for length is meter (m), for time is second(s) and for mass is kilogram (kg).
\(28.7pm\times \frac { { 10 }^{ -12 }m }{ 1pm } =28.7\times { 10 }^{ -11 }m\) 
(ii) The basic units for length is meter (m), for time is second(s) and for mass is kilogram (kg)
\(15.15\mu s\frac { { 10 }^{ -12 }s }{ 1\mu s } =1.515\times { 10 }^{ -5 }s\)
(iii) The basic units for length is meter (m), for time is second(s) and for mass is kilogram (kg). 
\(25365mg\times \frac { 1g }{ 1000mg } \times \frac { 1kg }{ 1000g } =2.5365\times { 10 }^{ -2 }kg\)

The compound is \(\mathrm{C}_2 \mathrm{H}_4\).
The atomic mass of C is 12 grams and the atomic mass of hydrogen is 1 gram.
Hence the molecular mass of ethene is \(2 \times 12+1 \times 4=28\) grams.
The amount of C is \(12 \times 2=24\) grams.
Hence mass percentage of carbon in ethene can be given as follows-
\( \frac{24}{28} \times 100  =85.71 \%\)

Same group elements have similar valence shell electronic configuration. Therefore, have similar physical and chemical properties.

(a) Due to highly negative electron gain enthalpy they act as good oxidising agents as they can gain electrons easily.
(b) Electronic configuration of noble gases is such that all subshells are completely filled. Hence, their electron gain enthalpy is almost zero.

sodium extract is boiled with nitric acid to decompose NaCN and Na₂S if present.
\(\mathrm{NaCN}+\mathrm{HNO}_{3} \longrightarrow \mathrm{NaNO}_{3}+\mathrm{HCN} \uparrow\)
\(\mathrm{Na}_{2} \mathrm{~S}+2 \mathrm{HNO}_{3} \longrightarrow 2 \mathrm{NaNO}_{3}+\mathrm{H}_{2} \mathrm{~S} \uparrow\)
If cyanide and sulphide are not removed, they will react with AgNO₃ and hence, will interfere with the silver nitrate test for halogens
\(\mathrm{NaCN}+\mathrm{AgNO}_{3} \longrightarrow \underset{\text { White ppt }}{\mathrm{AgCN}}+\mathrm{NaNO}_{3}\)
\(\mathrm{Na}_{2} \mathrm{~S}+2 \mathrm{AgNO}_{3} \longrightarrow \underset{\text { Black } \mathrm{ppt}}{\mathrm{Ag}_{2} \mathrm{~S}}+2 \mathrm{NaNO}_{3}\)

Al₂O₃ (alumina) is most suitable adsorbent for column chromatography.

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.

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

(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.

(ii) Those species which are positively charged or electron deficient are called electrophiles.
\(\text { e.g. } \mathrm{H}^{+}, \mathrm{AlCl}_{3}, \mathrm{Cl}^{+}\)

(iii) Fuse the organic compound with sodium metal. Sodium reacts with 'C' and 'N' present in organic compound to form NaCN.
\(\mathrm{Na}+\mathrm{C}+\mathrm{N} \longrightarrow \mathrm{NaCN}\)
Add FeSO₄ to L.E. (Lassaigne's extract)
\(6 \mathrm{NaCN}+\mathrm{FeSO}_{4} \longrightarrow \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+\mathrm{Na}_{2} \mathrm{SO}_{4}\)
Dilute H₂SO₄ is added to convert Fe²⁺ to Fe³⁺ and blue colour is formed due to formation of ferric ferrocyanide.
\(4 \mathrm{Fe}^{3+}+3 \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\begin{aligned} &\longrightarrow \mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}+12 \mathrm{Na}^{+}\\ &\text { Blue colouration } \end{aligned}\) 

(i) 1 mole of C₂H₆ contains 2 moles of carbon atoms.
Number of moles of carbon atoms in 3 moles of C₂H₆
= 2 × 3 = 6
(ii) 1 mole of C₂H₆ contains 6 moles of hydrogen atoms.
Number of moles of carbon atoms in 3 moles of C₂H₆
= 3 × 6 = 18
(iii) 1 mole of C₂H₆ contains 6.023 × 10²³ molecules of ethane.
Number of molecules in 3 moles of C₂H₆
= 3 × 6.023 × 10²³ = 18.069 × 10²³

\((C{ H }_{ 3 })_{ 3 }{ C }^{ + }>{ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }^{ + }>C{ H }_{ 2 }=C{ H }{ CH }_{ 2 }^{ + }>C{ H }_{ 3 }{ CH }_{ 2 }^{ + }\)

The Ist ionisation enthalpy of magnesium is higher than that of Na due to higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of first electron, Na + formed has the electronic configuration of neon (2, 8). The higher stability of the completely filled noble gas configuration leads to very high second ionisation enthalpy for sodium. On the other hand, Mg⁺ formed after losing first electron still has one more electron in its outermost (3s) orbital. As a result, the second ionisation enthalpy of magnesium is much smaller than that of sodium.

Within a period, the oxidising character increases from left to right. Therefore, among F, 0 and N, oxidising power decreases in the order: F > 0 > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Further because 0 is more electronegative than Cl, therefore, 0 is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is: F > 0 > Cl > N, i.e., option (b) is correct.