Ethers having same molecular formula, but alkyl groups attached to the oxygen atom are different. This phenomenon is known as metamerism.
Metamers of 2- methoxy propane
\(\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{O}-\mathrm{C}_{2} \mathrm{H}_{5}-\) diethyl ether 
\(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) methyl n-propyl ether

\((E^{0}_{ox})_{Fe^{2+}|Fe} = -0.44\)  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.

Reaction rate = k[x]^3/2[y]^1/2
(i) Over all order of reaction = (3/2 + 1/2)=2 
i.e., second order reaction.
(ii) Since the rate expression does not contain the concentration of z, the reaction is zero order with respect to z.

Electronic configuration of Ti = 3d²4s²
Electronic configuration of Ti³⁺ =3d¹
Hence number of unpaired electron = 1
Spin only magnetic moment \((\mu)=\sqrt{\mathrm{n}(\mathrm{n}+2)}\)
= \(\sqrt{1(1+2)} \)
= \(\sqrt{3}\) 
=1.732 BM
Electronic configuration of \(\mathrm{Mn}=3 \mathrm{~d}^{5} 4 \mathrm{~s}^{2}\)
Electronic configuration of \(\mathrm{Mn}^{2+}=3 \mathrm{~d}^{5}\)
Hence number of unpaired electrons = 5
Spin only magnetic moment 
\((\mu) =\sqrt{5(5+2)}\)
= 5.92 BM

The structural units of an ionic crystal are cations and anions. They are bound together by strong electrostatic attractive forces. To maximize the attractive force, cations are surrounded by as many anions as possible and vice versa. Hence they are hard and brittle.

Diborane adds on to alkenes and alkynes in ether solvent at room temperature. This reaction is called hydroboration.
\({ B }_{ 2 }{ H }_{ 6 }+6RCH=CHR\longrightarrow 2B(RCH_2-{ CH }R)_{ 3 }B\)

1. This process is called as Hall-Heroult process.
Cathode: In this method, electrolysis is carried out in an iron tank lined with carbon which acts as the cathode.
Anode: The carbon blocks immersed in the electrolyte acts as a anode.
Eletrolyte: A 20% solution of alumina, obtained from the bauxite ore is mixed with molten Cryolite and is taken in the electrolysis chamber.
2. About 10% calcium chloride is also added to the solution.
3. Here Calcium chloride helps to lower the melting point of the mixture.
Temperature: The fused mixture is maintained at a temperature of above 1270 K.
4. The chemical reactions involved in this process as follows
(a) Ionisaiton of alumina: \({ A }l_{ 2 }{ O }_{ 3 }\longrightarrow { 2Al }^{ 3+ }+{ 3O }^{ 2- }\)
(b) Reaction at cathode: \(2{ Al }^{ 3+ }_{(melt)}+{ 6e }^{ - }\longrightarrow { Al }_{ (l) }\)
(c) Reaction at anode: \(6{ O }^{2-}_{(melt)}\longrightarrow { 3O }_{ 2 }+{ 12e }^{ - }\)
5. Since carbon acts as anode the following reaction also takes place
(a) \({ C }_{ (s) }+{ O }^{ 2- }_{(melt)}\longrightarrow CO+{ 2e }^{ - }\)
(b) \({ C }_{ (s) }+{ 2O }^{ 2- }_{(melt)}\longrightarrow { CO }_{ 2 }+{ 4e }^{ - }\)
6. Due to the above two reactions, anodes are slowly consumed during the electrolysis.
7. The pure aluminium is formed at the cathode. The net electrolysis reaction can be written as
\({ 4Al }^{ 3+ }_{(melt)}+{ 6O }^{ 2- }_{(melt)}+{ 3C }_{ (s) }\longrightarrow { 4Al }_{ (l) }+{ 3CO }_{ 2(g) }\)

i) Reducing sugars:
1. Sugars which reduce Tollen's reagent or Fehling's solution or Benedict's solution are called reducing sugars.
2. These contain either α - hydroxyl ketone or cyclical hemi acetal or hemi ketal or structures in equilibrium with open chain forms having a free- CHO or C=O group.
3. E.g. a) All monosaccharide's like D - glucose, D - fructose (aldoses and ketoses)
b) Sugars like Lactose and maltose except sucrose.
ii) Non - reducing sugars:
1. Sugars which do not reduce either Tollen's reagent, Fehling's solution or Benedict's solution are called non-reducing sugars.
2. They contain a stable acetal or ketal structures which cannot be opened into a free carbonyl group.
E.g. Sucrose, starch, cellulose, glycogen, dextrin etc.

(i) HCl cannot be used for making the medium acidic since it reacts with KMnO₄ as follows.
2MnO₄^- + 10 Cl^- + 16H⁺ ⟶ 2Mn²⁺ + 5Cl₂+ 8H₂O
(ii) HNO₃ also cannot be used since it is good oxidising agent and reacts with reducing agents in the reaction.
(iii) However, H₂SO₄ is found to be most suitable since it does not react with potassium permanganate.

(i) Redox reactions involve transfer of electrons from one reactant to another. Such reactions are always coupled, which means that when one substance is oxidised, another must be reduced.
(ii) The substance which is oxidised is a reducing agent and the one which is reduced is an oxidizing agent

(i) The use of 3d electron for formation of I bond increases from Sc to Mn, causing the increase in oxidation state upto +7.
(ii) The reason for Mn having highest oxidation state of +7 is due to the presence of 7 unpaired electrons in its atom.
(iii) As the number of unpaired electrons decrease from Fe to Cu. So there is the decrease in oxidation state.

(i) When the concentration of the reactant is increased by 'n' times, the rate of reaction is also increased by n times. That is, if the concentration of the reactant is doubled, the rate is doubled.
(ii) The unit of rate constant of a first order reaction is sec^-1 or time^-1.
\({ k }_{ 1 }=\frac { rate }{ (a-x) } =\frac { { mol.lit }^{ -1 }{ sec }^{ -1 } }{ { mol.lit }^{ -1 } } \)
(iii) The time required to complete a definite fraction of reaction is independent of the initial concentration, of the reactant if t_1/u is the time of one 'u' th fraction of reaction to take place then from equation.
\({ k }_{ 1 }=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
\(x=\frac { a }{ u } and\quad { t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { a }{ a-\frac { a }{ u } } } ;\)
\({ t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { u }{ (u-1) } } \)
since k₁ = rate constant, t_1/u is independent of initial concentration 'a'.

(i) CI has vacant d-orbitals and hence can show an oxidation state of +3 but F has no d-orbitals, therefore, it cannot show positive oxidation states. Further, since F can show only - 1 oxidation state therefore it forms only CIF of FCI₃.
(ii) Due to larger size CI an accommodate three small F-atoms around it while F being smaller cannot accommodate three bigger sized CI atoms around it.

(i) Ammonia reacts with chlorine and chlorides to give ammonium chloride as a final product.
(ii) The reactions are different under different conditions as given below.
(iii) With excess ammonia
\({ 2NH }_{ 3 }+3{ Cl }_{ 2 }\longrightarrow { N }_{ 2 }+6HCl\)
\(6HCl+6{ NH }_{ 3 }\longrightarrow { 6NH }_{ 4 }Cl\)
(iv) With excess of chlorine ammonia reacts to give nitrogen trichloride, an explosive substance.
\({ 2NH }_{ 3 }+{ 6Cl }_{ 2 }\longrightarrow { 2NCl }_{ 3 }+6HCl\)
\({ 2NH }_{ 3 }(g)+{ HCl }{ (g) }\longrightarrow NH_4Cl(s)\)

Data:
Diffraction angle,
2θ = 14.8⁰ ∴ 8 = 7.4°
Interplanar distance, d = 0.400 nm
Order of reflection, n = 2
Formula:
nλ = 2dsinθ
∴ wavelength,
Solution:
\(\lambda =\frac { 2dsin\theta }{ 2 } \)
= dsinθ
= 0.400 sin 7.4^o
= 0.400 x 1.288 nm
= 0.0512 nm
λ = 0.0512 nm

(i) Compressibility is the ability of the substance to change its shape by applying external pressure.
(ii) Solids form closed packed structure with negligible intermolecular space and possesses very strong intermolecular forces of attraction.
(iii) Thus, they do not change their shape in the presence of external pressure.

(i) Diamond is very hard.
(ii) The carbon atoms in diamond are sp³ hybridised and bonded to four neighbouring carbon atoms by a bonds with a C-C bond length of 1.54 Å.
(iii) This results in a tetrahedral arrangement around each carbon atom that extends to the entire lattice.
(iv) Since all four valance electrons of carbon are involved in bonding there is no free electrons for conductivity.
(v) Being the hardest element, it used for sharpening hard tools, cutting glasses, making bores and rock drilling.

(i) In the concentrated ore, the metal exists in positive oxidation state and hence it is to be reduced to its elemental state.
(ii) From the principles of thermodynamics; that the reduction of oxide is easier when compared to reduction of other compounds of metal and hence, before reduction, the are is first converted into the oxide of metal of interest.

Thus in metal carbonyls, electron density moves from ligand to metal through sigma bonding and from metal to ligand through pi bonding, this synergic effect accounts for strong M \(\longleftarrow \) CO bond in metal carbonyls.

As we move from left to right in a transition metal series, the ionization enthalpy increases as expected. This is due to increase in the nuclear charge corresponding to the filling of d electrons. The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular. The added electron enters (n-1) d orbital and the inner electrons act as a shield and decrease the effect of nuclear charge on valence ns electrons. Therefore, it leads to variation in the ionization energy value.

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) Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as HCl, HNO₃.
(ii) It is used as a drying agent and also used in the preparation of pigments, explosives etc.

Active centres:
The surface of a catalyst is not smooth. It bears steps, cracks and corners. Hence the atoms on such locations of the surface are co-ordinatively unsaturated. So, they have much residual force of attraction. Such sites are called active centres. So, the surface carries high surface free energy. The presence of such active centres increases the rate of reaction (activity) by adsorbing and activating the reactants.
The adsorption theory explains the following:
(i) Increase in the activity of a catalyst by increasing the surface area. Increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.
(ii) The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.
(iii) A promoter or activator increases the number of active centres on the surfaces
Selectivity:
A Catalyst can catalyse a particular type of reaction. Hence they are said to the specific (selectivity) in nature. Enzyme catalysis is highly specific in nature.
\(\mathrm{NH}_{2} \mathrm{CONH}_{2}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { Unease }}{\longrightarrow} 2 \mathrm{NH}_{3}+\mathrm{CO}_{2}\)
The enzyme urease which catalyses their reaction of Urea does not catalyse the reaction of methyl Urea.
\(\mathrm{NH}_{2} \mathrm{CONH} \mathrm{CH_3}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { Urease }}{\longrightarrow} \text { No reaction }\)
Intermediate compound formation theory explains the specificity of a catalyst.

(i) The dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.
(ii) Consider the dissociation of a weak acid, acetic acid whose ionisation is incomplete.
CH₃COOH_(aq) ⇌ H⁺_(aq) + CH₃COO^-_(aq)
(iii) If the salt sodium acetate with common ion CH₃COO^- is added to the above equilibrium, it dissociates completely increasing.
CH₃COONa_(aq)⟶Na⁺_(aq) + CH₃COO^-_(aq)
(iv) Hence, the overall concentration of CH₃COO^- is increased, and the acid dissociation equilibrium is disturbed.
(v) So, in order to maintain the equilibrium, the excess CH₃COO^- ions combines with H⁺ ions to produce much more unionized CH₃COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed.