\({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
For a second order reaction
\(\mathrm{t}_{1 / 2}=\frac{2^{\mathrm{n}-1}-1}{(\mathrm{n}-1) \mathrm{k} \cdot\left[\mathrm{A}_{0}\right]^{\mathrm{n}-1}}\)
n = 2
\(\mathrm{t}_{1 / 2}=\frac{2^{\mathrm{2}-1}-1}{(\mathrm{n}-1) \mathrm{k} \cdot\left[\mathrm{A}_{0}\right]^{\mathrm{2}-1}}\)
\(\mathrm{t}_{1 / 2} =\frac{1}{ \mathrm{k}\left[\mathrm{A}_{0}\right]}\)

sc ⇒ 2r = a
⇒ r = a/2
bcc ⇒ 4r = \( \sqrt{3a} \) ⇒ r = \(\frac { \sqrt{3a}} {4}\)
fcc ⇒ 4r ⇒  \( \sqrt{2a} \) ⇒ r = \(\frac { \sqrt{2a}} {4} = \frac { a} {2\sqrt{2a}}\)
\(\left( \frac { a }{ 2 } :\frac { \sqrt { 3 } }{ 4 } a:\frac { a }{ 2\sqrt { 2 } } \right) \)

As we move from La to Lu, their metallic behaviour because almost similar to that of aluminium.

Mn²⁺ ⇒ 3d⁵ contains 5 unpaired electrons
n = 5,
\( \sqrt{n(n+ 2)} \) BM
\(= \sqrt{5(5+ 2)} = \sqrt{35} = 5.92 BM\)

(i) White precipitate formed is 2,4,6 tribromoaniline.

(ii) 

The agonist drug, morphine, is used as a pain killer, binds to the opioid receptors and activates them. This supress the neuro transmitters that causes pain.

Pheneol IS used as an antiseptic as well as a disifectant 0.2% of phenol is an antiseptic whereas 1% solution is a disinfectant.

Vitamin C.

Sugar differing in configuration at an asymmetric centre is known as epimers.

(i) D isomer
(ii) Lisomer

(i) This equation is purely empirical and valid over a limited pressure range.
(ii) The values of constants k and n also found vary with temperature.
(iiii) No theoretical explanations were given.

The metal is treated with strong oxidising agents such as concentrated HNO₃. As a result, protective oxide layer is formed on the surface of metal

(ii) 2,4-dichlorobenzaldehyde.

(iii) ethanal

Cathodic protection:
In this technique, unlike galvanising the entire surface of the metal to be protected need not be covered with a protecting metal. Instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded. This known as sacrificial protection. (or) Cathodic protection.

(i) First order reaction
(ii) First order reaction
(iii) \(\frac{1}{2}+2=2 \frac{1}{2}\); Pseudo first order reaction

Oxidation number of F = -1
Oxidation number of H = +1
Oxidation number of O in HOF =x
(+1) + x + (-1) = 0
x = 0
Oxidation number of O in HOF = 0

(i) Icosogens → B, Al, Ga, In, Tl
(ii) Tetragens → C, Si, Ge, Sn, Pb
(iii) Pnictogens → N, P, As, Sb, Bi
(iv) Chalcogens → O, S, Se, Te, Po

(i) The extraction of metals from their oxides can be carried out by using different reducing agents.
(ii) Consider the following reaction
\(\frac{2}{\mathrm{y}} \mathrm{M}_{\mathrm{x}} \mathrm{O}_{\mathrm{y}(\mathrm{s})} \rightarrow \frac{2 \mathrm{x}}{\mathrm{y}} \mathrm{M}_{(s)}+\mathrm{O}_{ 2(\mathrm{~g})}\)            (1)
(iii) The above reduction may be carried out with carbon. In this case the reducing agent carbon may be oxidized to either CO or CO₂
\(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}  \)                  (2)
\(2 \mathrm{C}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{(\mathrm{g})} \)                (3)
(iv) If CO is used as a reducing agent
\(2 \mathrm{CO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}\)           (4)
(v) A suitable reducing agent is selected based on the thermodynamics considerations.
(vi) We know that for a spontaneous reaction, the change in free energy (\(\triangle\)G) should be negative.
(vii) Therefore, thermodynamically, the reduction of metal oxide with a given reducing agent can occur if the free energy change for the coupled reaction is negative.
(viii) Hence, the reducing agent is selected in such a way that it provides a large negative \(\triangle\)G value for the coupled reaction.

(i) Perkin's Reactions:
When an aromatic aldehyde is heated with an aliphatic acid anhydride in the presence of the sodium salt of the acid corresponding to the anhydride, condensation takes place and an α, β unsaturated acid is obtained
This reaction is known as Perkin's reaction.

(ii) Knoevengal Reaction:
(a) Benzaldehyde condenses with malonic acid in the presence of pyridine forming cinnamic acid
(b) Catalyst - Pyridine
(c) Carbanion formed from malonic acid.

Fat Soluble Fat Soluble	Water Soluble Vitamins
Do not dissolve in water	Readily soluble in water
They are stored in fatty tissues and liver	These can't be stored
Eg: Vitamin A, D, E and K	Eg: Vitamin Band

(i) Thermodynamic principles can be employed to derive a relation between electrical energy and the maximum amount of work (W_max),
(ii) The maximum amount of work obtainable from the cell is the product of charge flowing per mole and maximum potential difference E, through which the charge is transferred.
W_max = -nFE ..... (1)
where, n = number of moles of electrons transferred
F = Faraday (96495 coulombs)
E = emf of the cell
According to thermodynamics,
W_max = ΔG ...(2)
From (1) and (2)
ΔG = - nFE
when E is positive, ΔG is positive ΔG will be negative and the cell reaction is spontaneous.

(a) Alcohols are soluble in water because they form intermolecular hydrogen bonding with water. Lower members are completely miscible with water and the higher members are not. This is because of the increase in size of hydrophobic (water repelling) alkyl group in the alcohol.
(b) Strong basic substances like organo metallic compounds (RMgX-Grignard reagent) are decomposed by alcohol.
R - OH + CH₃MgBr ⟶ R - O - Mg - Br + CH₄
Hence, alcohols cannot be used as a solvent for Grignard reagent.

(a) If we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell.
(b) So, AC current is used for this measurement to prevent electrolysis.

(i) Its small size
(ii) Its high electronegativity
(iii) Its high ionisation energy
(iv) Non-availability of d-orbital in the valence shell.
(v) Rather inert
(vi) High bond energy

Aldol condensation can also takes place between two different aldehydes or ketones or between one aldehyde and one ketone such an aldol condensation is called crossed or mixed aldol condensation. This reaction is not very useful as the product is usually a mixture of all possible condensation products and cannot be separated easily.
\(\underset { formldehyde }{ HCHO } +\underset { acataldehyde }{ { CH }_{ 3 }CHO } \overset { dil.NaOH }{ \longrightarrow } \underset { 3-hydroxy \ propanol }{ HO-{ CH }_{ 2 }-{ CH }_{ 2 }-CHO } \) 
\(\underset { formaldehyde }{ { HCHO } } +\underset { acetone }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 } } \overset { dil.NaOH }{ \longrightarrow } \underset { 4-hydroxy \ butan-2-one }{ HO-{ CH }_{ 2 }-{ CH }_{ 2 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 } } \)

Given data:
k₁ = 2.45 x 10^-5 sec^-1; T₁ = 273 K
k₂ = 16.2 x 10^-4 sec^-1; T₂ = 303 K
R = 8.314 JK^-1 mol^-1
Formula: \(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] \)
Solution:
\(\log { \frac { 16.2\times { 10 }^{ -4 } }{ 2.45\times { 10 }^{ -5 } } } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \left( \frac { 303-273 }{ 273\times 303 } \right) \)
\(1.8203=\frac { { E }_{ a } }{ 2.303\times 8.314 } \times \left[ \frac { 30 }{ 273\times 303 } \right] \)
∴ E_a = 52802.3 x 1.8203
= 96116 J / mol
E_a = 96.116 kJ mol^-1.

(i) The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-1)d orbital and ns orbital as the energy difference between them is very small.
(ii) At the beginning of the series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.
(iii) The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases.
(iv) Hence, the first and last elements show less number of oxidation states and the middle elements with more number of oxidation states.
(v) For example, the first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.
(vi) The relative stability of different oxidation - states of 3d metals is correlated with the extra stability of half filled and fully filled electronic configurations. Example: Mn²⁺(3d⁵) is more stable than Mn⁴⁺(3d³).

(i) Iron is one of the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.
(ii) Cast iron is used to make pipes, valves and pumps stoves etc.
(iii) Magnets can be made from iron and its alloys and compounds.
(iv) An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery.
(v) Nickel steel is used for making cables, automobiles and aeroplane parts.
(vi) Chrome steels are used for manufacturing cutting tools and crushing machines.

(i) Order of the reaction =1; \(\mathrm{t}_{1 / 2}=60 \mathrm{~s} ; \mathrm{k}=?\)
\(\mathrm{k}=\frac{0.6932}{\mathrm{t}_{\frac{1}{2}}} \)
\(=\frac{0.6932}{60}  \)
\(k =1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
(ii) \(\left[\mathrm{A}_{0}\right]=100 \% ; \mathrm{t}=180 \mathrm{~s} ;[\mathrm{A}]=? ; \mathrm{k}=1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
For first order reaction
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]} \)
\(1.155 \times 10^{-2} =\frac{2.303}{180} \log \left(\frac{100}{[A]}\right)  \)
\(\frac{0.01155 \times 180}{2.303} =\log \left(\frac{100}{[A]}\right)  \)
\(0.9027 =\log 100-\log [\mathrm{A}]  \)
\(\log [\mathrm{A}] =\log 100-0.9027 \)
\(\log [A]=2-0.9027 \)
\(\log [A]=1.0972 \)
[A] = antilog  of (1.0972)
[A] =12.51 %

a) if both assertion and reason are true and reason is the correct explanation of assertion.