Hyperconjugation interaction in \({ \left( { CH }_{ 3 } \right) }_{ 3 }\overset { + }{ C } \) is greater than in \({ CH }_{ 3 }\overset { + }{ C } { H }_{ 2 }\) as the \({ \left( { CH }_{ 3 } \right) }_{ 3 }\overset { + }{ C } \)  has nine C-H bonds.  In \(\overset { + }{ C } { H }_{ 3 }\), vacant p orbital is perpendicular to the plane in which C-H bonds lie; hence cannot overlap with it. Thus, \({ CH }_{ 3 }^{ + }\) lacks hyperconjugative stability.

(a) When organic compound is fused with sodium metal and then extracted by water, it is called Lassaigne's extract. Yes.
(b) Blood red colour.
(c) Lassaigne's extract is prepared in distilled water since tap water contains Cl^- ions. No, oxygen cannot be detected by Lassaigne's test.

(a) This method is based upon the fact that nitrogenous compound is heated with copper oxide in an atmosphere of carbon dioxide yield free nitrogen.

(b) P₁ = 715 -15 = 700 mm Hg, P₂ = 760 mm Hg
\(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)
T₁ = 300 K, T₂ = 273 K
V₁ = 50 crrr', V₂ = ?
\(V_2=\frac{700\times50\times273}{300\times760}\)
= 41.9 cm³
\(\%\ of\ N=\frac{28}{22400}\times41.9\times\frac{100}{W}\)
= 17.46%​​​​​​​

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.

Step I. Calculation of volume of unused acid
Volume of NaOH solution required = 60 cm³
Normality of NaOH solution = 1/2 N
Normality of H₂SO₄ solution = l/N
Volume of unused acid can be calculated by applying normality equation
\(\underbrace { N_{ 1 }{ V }_{ 1 } }_{ Acid } =\underbrace { { N }_{ 1 }{ V }_{ 1 } }_{ Base } \)
\(1\times V=\frac{1}{2}\times 60=30cm^3\)
Step II. Calculation of volume of acid used
Volume of acid added = 50 cm³
Volume of unused acid = 30 cm³
Volume of acid used = (50 - 30) = 20 cm³
Step III. Calculation of percentage of nitrogen
Mass of compound = 0.50 g
Volume of acid used = 20 cm³
Normality of acid used = 1N
Percentage of N =\(\frac{1.4\times Volume\ of\ acid\ used\times Normality\ of\ acid\ used}{Mass\ of\ the\ compound}\)
\(=\frac{1.4\times20\times1}{0.50}=56\%\)

Step I. Calculation of mass of CO₂ produced
Mass of compound 0.20 g
Percentage of carbon = 69%
Percentage of carbon \(=\frac{12}{44}\times\frac{Mass\ of\ varbon\ dioxide\ formed}{Mass\ of\ compound}\)
\(69=\frac{12}{44}\times\frac{Mass\ of\ carbon\ dioxide\ formed}{(0.20g)}\)
\(\therefore \) Mass of CO₂ formed \(=\frac{69\times44\times(0.20g)}{12\times100}=\)0.506 g
Step II. Calculation of mass of H₂O produced
Mass of compound = 0.20 g
Percentage of hydrogen = 4.8 %
Percentage of hydrogen \(=\frac{2}{18}\times\frac{Mass\ of\ water\ formed}{Mass\ of\ compound}\times100\)
\(4.8=\frac{2}{18}\times\frac{Mass\ of\ formed}{(0.20 g)}\times 100\)
\(\therefore\) Mass of H₂O formed \(=\frac{4.8\times18\times(0.20g)}{2\times100}=\) 0.0864 g

Estimation of halogens: It involves oxidising the organic substance with fuming nitric acid in the presence of silver nitrate. The halogen of the substance is thus converted to silver halide which is separated and weighed:
Weight of organic compound = W gm
weight of silver halide = x g.
% of halogen = \(\frac{\text{At.wt.of halogen}\times100x}{\text{Mol.wt of silver halide}\times w}\)
Estimation of sulphur: The organic substance is heated with fuming nitric acid but no silver nitrate is added. The sulphur of the substance is oxidised to sulphuric acid which is then precipitated as barium sulphate by adding excess of barium chloride solution. From the weight of BaS04 so obtained the percentage of sulphur can be calculated.
% of sulphur = \(\frac{32(At.weight\ of\ S)}{233(mol\ weight\ of\ BaSO_4)}\times\frac{weight\ of\times100\ BaSO_4}{weight\ of\ organic\ compound}\)
Estimation of phosphorous: The organic substance is heated with fuming nitric acid whereupon phosphorous is oxidised to phosphoric acid. The phoshoric acid is
precipitated as ammonium phosphomolybdate, (NH₄)₃ PO₄·12MoO₃, by the addition of ammonia and ammonium molybdate solution which is then separated, dried and weighed.
% of P = \(\frac{31\times w_1\times 100}{1877\times w}\)
where Molar mass of (NH₄)₃ PO₄.12MoO₃ = 1877 g
If phosphorous is estimated as Mg₂P₂O₇
% of P = \(\frac{62\times w_1\times100}{222\times w}\)

(a) Ethyl alcohol, \({ C }_{ 2 }{ H }_{ 5 }OH\)
(b) Ethanol, CH₃CH₂OH sp³sp³
(c) Alcohol acts as a depressant on the central nervous system. It has a complex mode of action and affects multiple systems in the brain. Its long-term consumption may cause irreversible damage to the liver.
(d) Values possessed by Sohan are the loving and caring attitude towards his father. Values possessed by the father is love towards this child.