phenol

\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A₀] = 0.1: [A] = 0.05
\(k= \left[ \frac { 1 }{ t _{1/2}} \right] ln\left[ \frac { { 0.1 } }{ 0.05 }\right] \)
\(k= \left[ \frac { 1 }{ t _{1/2}} \right] ln (2)\)
t_1/2 = In(2)/K

A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

Number of A ions = N_c/8 = 8/8 = 1
Number of B ions = N_f/2 = 6/2 = 3
Simplest formula = AB₃

a) Thorpe nitrile condensation
(i) Self condensation of two molecules of alkyl nitrile (containing $\alpha-\mathrm{H}$ atom) in the presence of sodium to form iminonitrile.

b) Levine and Hauser acetylation
(i) The nitriles containing α - hydrogen also undergo condensation with esters in the presence of sodamide in ether to form ketonitriles. This reaction is known as "Levine and Hauser" acetylation.
(ii) This reaction involves replacement of ethoxy (OC₂H₅) group by methylnitrile (-CH₂CN) group and is called as cyanomethylation reaction.

When Chlorobenzene is hydrolyzed with 6-8 % NaOH at 300 bar and 633 K, in a closed vessel. Sodium Phenoxide is formed which on treatment with dilute HCl gives phenol.

CCl₃ - NO₂ (trichloronitro. methane) is Chloropicrin.

During intramolecular dehydration, if there is a possibility to form a carbon-carbon double bond at different locations, the preferred location is the one that gives the more (highly) substituted alkene i.e., the stable alkene.

(i) 
                                             n-Nitro benzoate (Major Product)

(ii) 

The conversion of ore into metal oxide (oxidation) is called calcination. It is the process in which the ore is subjected to the action of heat at high temperature in the absence of air below its melting point. Example
\({ CaCO }_{ 3 }(limestone)\longrightarrow CaO+{ CO }_{ 2 }\uparrow \)
\({ MgCO }_{ 3 }\left( Magnesite \right) \longrightarrow { MgO }+{ CO }_{ 2 }\uparrow \)

Arrhenius equation is,
\(k=Ae^\left ({ \frac { -Ea }{ RT } } \right )\)
Here,
A \(\rightarrow\) Frequency factor
E_a \(\rightarrow\) Activation energy of the reaction
R \(\rightarrow\) Gas constant
T \(\rightarrow\) Absolute temperature (in K)

(i) sp³d³ hybridisation
(ii) Pentagonal bipyramidal structure.

(i) A basic repeating structural unit of a crystalline solid is called a unit cell.
(ii) A crystal is consisted of large number of unit cells.

Catenation is an ability of an element to form chain of atoms.
The conditions for catenation.
(a) The valency of element is greater than or equal to two.
(b) Element should have an ability to bond with itself
(c) The self bond must be as strong as Its bond with other elements
(d) Kinetic inertness of catenated compound towards other molecules.
(e) Carbon possesses all the above properties and forms a wide range of compounds with itself and with other elements such as H, O, N, S and halogens.

In this extraction, a basic flux, quick lime (CaO) is used, since the silica gangue present in the ore is acidic in nature. The quick lime combines with it to form calcium silicate (slag).
CaO_(s) + Sio_2(s) ⟶ CaSio_3(s)
Flux       Gangue    Slag

Anisole undergoes Friedel Craft's reaction in presence of anhydrous AlCl₃ as a catalyst.

For a first order reaction
\(\\ \\ k=\frac { 2.303 }{ t } log\left( \frac { [{ A }_{ 0 }] }{ [A] } \right) \\ \)   ..(1)
Let[A₀] =100M
When 
t = t_90%; [A] = 10M (given that t_90% = 8hours)
t = t_80%; [A ] = 20M
\(k=\frac { 2.303 }{ { t }_{ 80\% } } \log\left( \frac { 100 }{ 20 } \right) \)
\({ t }_{ 80\% }=\frac { 2.303 }{ K } \log(5)\) ....(2)
Find the value of k using the given data
\(k=\frac { 2.303 }{ { t }_{ 90\% } } \log\left( \frac { 100 }{ 10 } \right) \)
\(k=\frac { 2.303 }{ 8 } \log10\)
\(k=\frac { 2.303 }{ 8 } ...(3)\)
Substitute the value of k in equation (2)
\({ t }_{ 80\% }\frac { 2.303 }{ 2.303/8hours } \log(5)\)
t_80%= 8 hours x 0.6989
t_80%= 5.59 hours

(i) The half life of a reaction is defined as the time required for the reactant concentration to reach one half its initial value.
(ii)   The rate constant for a first order reaction is given by
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(at\quad t={ t }_{ \frac { 1 }{ 2 } };\left[ A \right] =\frac { \left[ { A }_{ 0 } \right] }{ 2 } \)
\(k=\frac { 2.303 }{ t_{ 1/2 } } log\frac { \left[ { A }_{ 0 } \right] }{ \frac { \left[ { A }_{ 0 } \right] }{ 2 } } \)
\(k=\frac { 2.303 }{ { t }_{ 1/2 } } log2\)
\(k=\frac { 2.303\times 0.3010 }{ { t }_{ 1/2 } } =\frac { 0.6932 }{ { t }_{ 1/2 } } \)
\({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
This equation has no concentration term So, the half life of a first order reaction is independent of initial concentration.

(i) Frenkel defect arises due to the dislocation of ions from its crystal lattice.
(ii) The ion which is missing from the lattice point occupies an interstitial position.
(iii) This defect is shown by ionic solids in which cation and anion differ in size.
(iv) Unlike Schottky defect, this defect does not affect the density of the crystal.
For example AgBr, in this case, small Ag⁺ ion leaves its normal site and occupies an interstitial position.

(i) Ionic solids have high melting points.
(ii) These solids do not conduct electricity, because the ions are fixed in their lattice positions.
(iii) They are hard so strong external force can change the relative positions of ions.

(i) Silicones are used for low temperature lubrication and in vacuum pumps, high temperature oil baths etc.
(ii) They are used for making water proofing clothes.
(iii) They are used as insulting material in electrical motor and other appliances.
(iv) They are mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals.

Each halogen combines with other halogens to form a series of compounds are called interhalogen compounds.
Example: AB type: BrF
AB₃ type: ICI₃

A reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. Let us consider the following Cl₂ first order reaction,
A → product
Rate law can be expressed as
Rate = k[A]¹
Where, k is the first order rate constant
\(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^{1}  \)
\(\Rightarrow \frac{-\mathrm{d}[\mathrm{A}]}{[\mathrm{A}]}=\mathrm{kdt}\)  .....(1)
Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial concentration [A₀] to [A] at the later time.
\(\int_{\left[A_{0}\right]}^{[A]} \frac{-d[A]}{[A]}=k \int_{0}^{t} d t  \)
\((-\ln [A])_{\left[A_{0}\right]}^{[A]}=k(t)_{0}^{t}  \)
\(-\ln [A]-\left(-\ln \left[A_{0}\right]\right)=k(t-0) \)
\(-\ln [\mathrm{A}]+\ln \left[\mathrm{A}_{0}\right]=\mathrm{kt}  \)
\(\ln \left(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\right)=\mathrm{kt}\)  .....(2)
This equation is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303.
\(2.303 \log \left(\frac{\left[A_{0}\right]}{[A]}\right)=k t  \)
\(k=\frac{2.303}{t} \log \left(\frac{\left[A_{0}\right]}{[A]}\right)\)   .....(3)

According to Arrhenius, activation energy of the reaction is the minimum energy that a molecule must have to posses to react.
\(\mathrm{k}=\mathrm{A} e^{-\left(\frac{E_{a}}{R T}\right)}\)
Where A is the frequency factor
R is the gas constant
E_a is the activation energy of the reaction
T is the absolute temperature. (in k)
Taking logarithm on both side of the equation (1)
\(\ln \mathrm{k}=\ln \mathrm{A}+\ln \mathrm{e}^{-\left(\frac{E_{a}}{R T}\right)}  \)
\(\ln \mathrm{k}=\ln \mathrm{A}-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\right) \quad(\therefore \ln \mathrm{e}=1) \)
\(\ln \mathrm{k}=\ln \mathrm{A}-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\right)\left(\frac{1}{\mathrm{~T}}\right)\)
At temperature T = T₁; the rate constant k = k₁
\(\ln \mathrm{k}_{1}=\ln \mathrm{A}-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{1}}\right)\)
At temperature T = T₂; the rate constant k = k₂
\(\ln \mathrm{k}_{2}=\ln \mathrm{A}-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{2}}\right)\)
(4)-(3)
\(\ln \mathrm{k}_{2}-\ln \mathrm{k}_{1}=-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{2}}\right)+\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{1}}\right) \)
\(\Rightarrow \ln \left(\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right) \)
\(2.303 \log \left(\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1} \mathrm{~T}_{2}}\right) \)
\(\log \left(\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1} \mathrm{~T}_{2}}\right) \)
\(\ln \mathrm{k}_{2}-\ln \mathrm{k}_{1}=-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{2}}\right)+\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{1}}\right)\)
This equation can be used to calculate E_a from rate constants k₁ and k₂ at temperatures T₁ and T₂.

Victor Meyer's test:
This test is used to distinguish 1°, 2°, and 3° alcohols.
This test consists of the following steps:
(i) Alcohol is converted into alkyl iodide by treatment with P/I₂.
(ii) The alkyl iodide is then converted into nitro alkane by silver nitrate (AgNO₂).
(iii) The nitro alkane is treated with nitrous acid (HNO₂) and then with aqueous KOH.
(iv) The 1°,2° and 3° alcohols are identified from the colour of the product.

(a) Kolbe's reaction:
In this reaction, phenol is first converted into sodium phenoxide which is more reactive than phenol towards electrophilic substitution reaction with CO₂, Treatment of sodium phenoxide with CO₂ at 400K, 4-7 bar pressure followed by acid hydrolysis gives salicylic acid.

(b) Riemer-Tiemann reaction:
On treating phenol with CHCI₃/NaOH, a -CHO group is introduced at ortho position. This reaction proceeds through the formation of substituted benzal chloride intermediate.

Preparation:
As discussed earlier diborane can be prepared by the action of metal hydride with boron. This method is used for the industrial production.
Diborane can also be obtained in small quantities by the reaction of iodine with sodium borohydride in diglyme.
2NaBH₄ + I₂ ⟶ B₂H₆ + 2NaI + H₂
On heating magnesium boride with HCl a mixture of volatile boranes are obtained.
2Mg₃B₂ + 12HCl ⟶ 6MgCl₂ + B4H₁₀ + H₂
B₄H₁₀ + H₂ ⟶ 2B₂H₆

(i) Collision theory is based on the kinetic theory of gases. According to this theory, a chemical reaction occurs as a result of collisions between the reacting molecules.
(ii) Let us understand this theory by considering the following reaction.
A₂(g) + B₂(g) ⟶ 2AB(g)
(iii) If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second.
(iv) Rate ∝ number of molecules colliding per litre per second (collision rate).
(v) The number of collisions is directly proportional to the concentration of both A₂ and B₂.
Collison rate ∝ [A₂][B₂]
Collision rate = Z [A₂][B₂]
(vi) Where, Z is a constant
(vii) A fraction of effective collisions (f) is given by the following expression
\(f={ e }^{ \frac { { -E }_{ a } }{ RT } }\)
(xiv) This fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collides with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.
(viii) The diagram illustrates the importance of proper alignment of molecules which leads to reaction.

(ix) The fraction of effective collisions (f) having proper orientation is given by the steric factor p.
⇒ Rate = p x f x collision rate
\(\Rightarrow Rate=p\times { e }^{ \frac { -Ea }{ RT } }\times Z\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(1)\)
As per the rate law,
Rate = \(k=\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(2)\)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is
\(k=pZ{ e }^{ \frac { -Ea }{ RT } }\)

(a) Elementary reaction
Each and Every single step in a reaction mechanism is called an elementary reaction.
Rate = k[A] [B]
(b)

In bcc unit cell, ΔABC
AC² = AB² + BC²
\(AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)
\(\\ AC=\sqrt { { a }^{ 2 }+{ a }^{ 2 } } =\sqrt { { 2a }^{ 2 } } =\sqrt { 2 } a\)

In ΔACG
AG² = AC² + CG²
\(AG=\sqrt { { AC }^{ 2 }+{ CG }^{ 2 } } \)
\(AG=\sqrt { { \left( \sqrt { 2a } \right) }^{ 2 }+{ a }^{ 2 } } \)
\(AG=\sqrt { { 2a }^{ 2 }+{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } \)
\(AG=\sqrt { 3a } \)
\(\sqrt { 3 } a=4r\)
\(r=\frac { \sqrt { 3 } }{ 4 } a\)
∴ Volume of the sphere with radius 'r' \(=\frac { 4 }{ 3 } { \pi r }^{ 3 }\)
\(=\frac{4}{3}\pi { \left( \frac { \sqrt { 3 } }{ 4 } a \right) }^{ 3 }\)\(=\frac { \sqrt { 3 } }{ 16 } \pi { a }^{ 3 }\)
Number of spheres belong to a unit cell in BCC arrangement is equal to two and hence the total volume of all spheres.
(i) Packing fraction = \(=\frac{Total \quad volume \quad occupied \quad by \quad spheres \quad in \quad a \quad unit \quad cell}{volume \quad of \quad the \quad unit \quad cell}\times100\)
\(\therefore\)Volume of all spheres \(=2\times \left( \frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 16 } \right) =\frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 8 } \)
Packing fraction \(=\frac { \left( \frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 8 } \right) }{ ({ a }^{ 3 }) } \times 100\)
\(=\frac { \sqrt { 3 } \pi }{ 8 } \times 100\)
\(\\ =\sqrt { 3 } \pi \times 12.5\)
= 1.732 x 3.14 x 12.5
= 68%

Zone refining :
1. Zone refining method is based on the principles of fractional crystallisation.
2. When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the molten region. In this process the impure metal is taken in the form of a rod.
3. One end of the rod is heated using a mobile induction heater which results in melting of the metal on that portion of the rod.
4. When the heater is slowly moved to the other end the pure metal crystallises while the impurities will move on to the adjacent molten zone.
5. As the heater moves further away, the molten zone containing impurities also moves along with it.
6. The process is repeated several times by moving the heater in the same direction again and again to get pure metal.
7. This process is carried out in an inert gas atmosphere to prevent the oxidation of metals.
8. Elements such as germanium (Ge), silicon (Si) and galium (Ga) that are used as semiconductor are refined using this process.