Phenyl Acetic acid
Benzyl Magnesium bromide

According to Henderson – Hasselbalch equation,
\(pH=pK_{a}+\log\frac{[salt]}{[acid]}\)
\(p{K_{a}}=-\log K_{a}=-\log(1.8\times10^{-5})=4.74\)
[Salt]=\(\frac{\text {Number of moles of sodium acetate}}{\text {Volume of the solution (litre)}}\)
Number of moles of sodium acetate =\(\frac{\text {mass of sodium acetate}}{\text {molar mass of sodium acetate}}\)
\(=\frac{8.2}{82}=0.1\)
\(\therefore [Salt]=\frac{0.1\ mole}{1/2 \ Litre}=0.2M\)
\([acid]=\frac{(\frac{mass \ of \ CH_{3}COOH}{molar \ mass \ of \ CH_{3}COOH})}{\text{Volume of solution in litre}}\)
=\(\frac{(\frac{6}{60})}{\frac{1}{2}}\)=0.2 M
\(\therefore pH=4.74+log\frac{(0.2)}{(0.2)}\)
pH = 4.74 + log1
pH = 4.74 + 0 = 4.74

A sol is electrically neutral. Hence the medium carries an equal but opposite charge to that of dispersed particles. When sol particles are prevented from moving, under the influence of electric field the medium moves in a direction opposite to that of the sol particles. This movement of dispersion medium under the influence of electric potential is called electro osmosis.

(i) Chloro benzene:
When Chlorobenzene is hydrolysed with 6-8% NaOH at 300 bar and 633K in a closed vessel, sodium phenoxide is formed which on treatment with dilute HCl gives phenol.

(ii) isopropyl benzene:
A mixture of benzene and propene is heated at 523K in a closed vessel in presence of H₃PO₄ catalyst gives cumene (isopropylbenzene). On passing air to a mixture of cumene and 5% aqueous sodium carbonate solution, cumene hydro peroxide is formed by oxidation. It is treated with dilute acid to get phenol and acetone. Acetone is also an important byproduct in this reaction.

(i) Acid catalysed hydration

(ii) Hydroboration

(iii) Hydroxylation using Baeyer's reagent

+ 2H₂O

 

Compound	Name
A	Benzoyl Chloride
B	Benzophenone
C	Ethylbenzoate

(i) Let us consider the reactions between a strong acid, HCI, and a weak base, NH₄OH, to produce a salt, NH₄CI, and water.
HCI_(aq) + NH₄OH_(aq) ⇌ NH₄CI_(aq) + H₂O _(I)
NH₄CI_(aq) ⟶ NH₄⁺ +CI^-_(aq)
(ii) NH₄⁺ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with OH^- from water to produce unionised NH₄OH
NH₄⁺ _(aq) + H₂O_(I) ⇌ NH₄OH_(aq) + H⁺_(aq)
(iii) There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^-]; the solution is acidic and the pH is less than 7.
(iv) The K_h and K_b are related by
\(\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}\)
(Or)
\(\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\)
Degree of hydrolysis (h)
\(\underset{(1-h)}{\mathrm{NH}_{4}^{+}(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(j)} \rightleftharpoons \underset{h} {\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}} +\mathrm{H}^{+} \underset{h}{(\mathrm{aq})}\\ \)
\(\mathrm{K}_{\mathrm{h}} =\frac{\left[\mathrm{NH}_{4} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \\ \)
\(=\frac{\mathrm{hc} \times \mathrm{h}}{(1-\mathrm{h}) \mathrm{c}} \)
\(K_{h}=\frac{h^{2} c}{(1-h)} \)
\(\text{If } \mathrm{h}<<1 ; \mathrm{K}_{\mathrm{h}} \simeq \mathrm{h}^{2} \mathrm{c} \)
\(h^{2}=\frac{K_{h}}{c}\)
\(h=\sqrt{\frac{K_{h}}{c}} \ (or) \ h=\sqrt{\frac{K_{w}}{K_{b} \cdot C}} \quad\left(\because K_{h}=\frac{K_{w}}{K_{b}}\right)\)
Also; \(\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}} \) (or) \(\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}\)
pH = -log [H⁺]

i) Cr² + 2e^- ⟶ Cr
ii) Mn²⁺ + 2e^- ⟶ Mn
iii) Fe²⁺ + 2e^- ⟶ Fe
iv) CO²⁺ + 2e^- ⟶ Co

Formula:
k₁ = A.e^-E_a/RT
Given:
Rate constant, k₁ - 3.0 x 10^-4 s^-1
Frequency factor, A = 6.0 x 10¹⁴ s^-1
Activation energy, E_a = 104.4 kJ
= 104400 J
Temperatures, T₁ = 25^o C; T₂ =∞
Solution:
\({ k }_{ 2 }=A.{ e }^{ -\frac { { E }_{ a } }{ RT } }\)
k₂ = 6.0 x 10¹⁴ x \(\frac { 104400 }{ { e }^{ 8.314\times \infty } } \)
k₂ = 6.0 x 10¹⁴ x 1   \(\left( \therefore e\frac { -{ { E }_{ a } } }{ R\infty } =1 \right) \)
∴ The rate constant at T ⟶ ∞ is 6.0 x 10¹⁴ s^-1

(i) Fluorine is the most reactive element among halogens due to minimum value of F- F bond dissociation energy.
(ii) It can form two types of salts with metals NaF and NaHF₂
(iii) AgF is soluble in water but other AgX are insoluble.
(iv) HF attacks glass while others do not.
(v) Fluorine, does not form any polyhalides (absence of d - orbitals)
(vi) Fluorine exhibit only negative oxidation state (highly electronegative) while other halogens have both +ve and -ve oxidation state.

(i) CaC₂
2 + 2x = 0
2x = -2
x = -1
(ii) H₂CO₃
2 + x + (-6) = 0
x = +4
(iii) HCN
1 + x + (-3) = 0
x = +3 - 1 = +2
(iv) CO
x + (-2) = 0
x= +2

In Ellingham diagram, the graph of \(CO\rightarrow { CO }_{ 2 }\) conversion remains below \(Fe\rightarrow { Fe }_{ 2 }{ O }_{ 3 }\)  upto 1073 K
(for Fe\(\longrightarrow \) FeO). SO, CO (g) act as reducing agent upto this temperature.
\(3{ Fe }_{ 2 }{ O }_{ 3 }+CO\longrightarrow 2{ Fe }_{ 3 }{ O }_{ 4 }\) (i.e. FeO.Fe₂O₃) + CO₂
\({ Fe }_{ 3 }{ O }_{ 4 }+CO\longrightarrow 3Feo+{ CO }_{ 2 }\)
FeO + CO \(\longrightarrow \) Fe + CO₂
Also, graph of C\(\longrightarrow \) CO is below the graph of
Fe \(\longrightarrow \)Fe₂O₃ after 1123 K. So, carbon acts as reducing agent above this temperature.
Fe₂O₃ + C\(\longrightarrow \) 3 CO + 2 Fe.

For a first order reaction
\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 2.303 }{ t } \log\frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ 1 } } \)
V_∞= 58.3 ml.

t(min)	Vt	V∞=Vt	\(k=\frac { 2.303 }{ t } \log\frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ t } } \)
6	19.3	58.3-19.3=39.0	\(k=\frac { 2.303 }{ 6 } \log\left( \frac { 58.3 }{ 39 } \right) =0.0670\) min-1
12	32.6	58.3-32.6=25.7	\(k=\frac { 2.303 }{ 12 } \log\left( \frac { 58.3 }{ 25.7 } \right) =0.06838\) min-1
18	41.3	58.3-41.3=17.0	\(k=\frac { 2.303 }{ 18 } \log\left( \frac { 58.3 }{ 17 } \right) =0.06838\) min-1
24	46.5	58.3-46.5=11.8	\(k=\frac { 2.303 }{ 24 } \log\left( \frac { 58.3 }{ 11.8 } \right) =0.0666\) min-1
30	50.4	58.3 - 50.4 = 7.9	\(k=\frac{2.303}{30} \log \left(\frac{58.3}{7.9}\right)=0.067\) min-1
			Mean value of k = 0.0674 min-1

As the rate constants are constant through out it is a first order reaction. 

(i) Reaction Rate = \(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 }\) ...(1)
When [L] = [4L]
Rate = \(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ 4L \right] }^{ 3/2 }\)
The Reaction Rate = \(8(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 })\) ...(2)
Comparing (1) and (2) rate is increased by 8 times
(ii)  [A] = [2A] and [B] = [2B]
Reaction Rate = \(k{ \left[ 2A \right] }^{ 2 }\left[ 2B \right] { \left[ L \right] }^{ 3/2 }\)
Reaction Rate = \(\\ 8(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 })\)  ...(3)
Comparing (1) and (3) rate is increased by 8 times
(iii) \(\left[ A \right] =\left[ \frac { A }{ 2 } \right] \)
Reaction Rate = \(k{ \left[ \frac { A }{ 2 } \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 }\)
Reaction Rate = \(\frac { 1 }{ 4 } \left( k\left[ { A }^{ 2 } \right] { \left[ B \right] }{ \left[ L \right] }^{ 3/2 } \right) \) ...(4)
Comparing (1) and (4); rate is reduced to (1/4) times.
(iv) \(\left[ A \right] =\left[ \frac { 1 }{ 3 }A \right] and\left[ L \right] =\left[ 4L \right] \)
Rate = \(k{ \left[ \frac { 1 }{ 3 }A \right] }^{ 2 }{ \left[ B \right] }{ \left[ 4L \right] }^{ 3/2 }\)
Rate = \(\left( \frac { 8 }{ 9 } \right) \left( k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 } \right)\) ...(5)
Comparing (1) and (5); rate is reduced to \(\frac { 8 }{ 9 } \) times.

Elemental copper is more stable than Cu²⁺. The electronic configuration of copper is 3d¹⁰ 4s¹ completely filled 3d orbital with stable configuration.
But Cu²⁺ has configuration as 3d⁹. Hence \(\mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{0}\) value is positive for Cu²⁺.

Zirconium is a 4d element and Hafnium is a 5d element. There is an unexpected observation in the atomic radius of 5d elements which have nearly the same atomic radius as that of corresponding 4d elements. This is due to lanthanoid contraction. Hence they exhibit similar properties.

Mn³⁺ has large and negative standard electrode potential E⁰ (-1.18 V) than that of Cr²⁺ which has only -0.91 V. If the standard electrode potential of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily. Hence Mn³⁺ is strongly oxidizing while Cr²⁺ is strongly reducing.

These two are ionisation isomers. [Co(NH₃)₅Cl]SO₄ gives white precipitate with BaCl₂ solution, but not with AgNO₃ solution. [Co(NH₃)₅SO₄]Cl gives curdy white precipitate with AgNO₃ solution but not with BaCl₂ solution.

(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.

1. The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. The coordination number of atoms in a bcc structure is '8'.

(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

A hydride of 2^nd period alkali metal (A) is lithium hydride (LiH).
Lithium hydride (A) reacts with diborane (B) to give lithium borohydride (C) which is acts as a reducing agent.
B₂H₆ + 2 LiH \(\xrightarrow[]{ether}\) 2 LiBH₄
[Diborane (B)]   [Lithium hydride (A)]  [Lithium borohydride (C)]
Result:

Compound	Formula	Name
A	LiH	Lithium hydride
B	B2H6	Diborane
C	LiBH4	Lithium borohydride

(i) Gangue: The ores are associated with nonmetallic impurities, rocky materials and siliceous matter which are collectively known as gangue.
Eg: SiO₂ is the gangue present in the iron ore (Fe₂O₃)
(ii) Slag: In the smelting process, a flux combines with Silica gangue forming slag.
CaO_(s) + SiO_2(s) → CaSiO_3(s)
Flux     + gangue → Slag

S.no	Sol	Gel
(a)	The liquid state of collidal solution	The solid (or) semi solid stage of a colloidal solution.
(b)	Very low viscosity	Very high viscosity
(c)	It does not have definite structure.	It possesses definite structure.

\((E^{0}_{ox})_{Fe^{2+}|Fe} = -0.44\)  and 
\((E^{0}_{red})_{Cu^{2+}|Cu} = 0.34\)
These +ve emf values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

(i) When excess of liquid ammonia is added to an aqueous solution of copper sulphate gives tetra ammine copper (II) sulphate is formed.
CUso₄ + 4NH₃ \(\rightarrow\) [Cu(NH₃)₄]SO₄
(ii) The co-ordination entity is [Cu(NH₃)₄]²⁺

i) Molar conductivity
Given : C = 0.01 M;
\(\kappa=1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \)
\(=1.5 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1} \)
\(\lambda_{\text {cation }}^{0}=248.2 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\lambda_{\text {anion }}^{0}=51.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\Lambda_{m}^{0}=\frac{\kappa \times 10^{-3}}{C} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \)
\(=\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01} \)
\(=1.5 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)
ii) \(\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}\)
\(\Lambda_{\mathrm{m}}^{0}=\lambda_{\text {cation }}^{0}+\lambda_{\text {anion }}^{0} \)
\(=(248.2+51.8) \mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} \)
\(=300 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
\(=300 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}  \)
\(\alpha =\frac{1.5 \times 10^{-3}}{300 \times 10^{-4}}=0.05\)
iii) \(\mathrm{K}_{\mathrm{a}} =\frac{\alpha^{2} \mathrm{C}}{1-\alpha} \)
\(\mathrm{K}_{\mathrm{a}} =\frac{(0.05)^{2} \times(0.01)}{1-0.05}=2.6 \times 10^{-5} \)
(or) 
\(\mathrm{K}_{\mathrm{a}} =\alpha^{2} \mathrm{C}  \)
\(=(0.05)^{2} \times(0.01)  \)
\(\mathrm{K}_{\mathrm{a}} =2.5 \times 10^{-5}\)

(i) The amino acids are linked covalently by peptide bonds. The carboxyl group of the first amino acid react with the amino group of the second amino acid to give an amide linkage between these amino acids. This amide linkage is called peptide bond. The resulting compound is called a dipeptide. Addition an another amino acid to this dipeptide a second peptide bond results in tripeptide.

(ii) Thus we can generate tetra peptide, penta peptide etc... When you have more number of amino acids linked this way you get a polypeptide. If the number of amino acids are less it is called as a polypeptide, if it has large number of amino acids (and preferably has a function) then it is called a protein.
(iii) The amino end of the peptide is known as N - terminal or amino terminal while the carboxy end is called C-terminal or carboxy terminal. In general protein sequences are written from N-Terminal to C-Terminal.The atoms other than the side chains (R-groups) are called main chain or the back bone of the polypeptide.