\(\left( \frac { r_{c^+} }{ r_{A^-} } \right) \) = \(\left( \frac { 0.98 \times 10^{-10 }}{ 1.81 \times 10^{-10} } \right) = 0.54\)
It is in the range of 0.414 to 0.732, hence the coordination number of each ion is 6.

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

i) Silica is an acidic flux which is mixed with Copper pyrites ore and is heated in a reverberatory furnace. The ferrous oxide formed from the ore is basic in nature and it combines with silica to form ferrous silicate. It is removed as slag.
\(\mathrm{FeO}_{(s)}+\mathrm{SiO}_{2(s)} \overset { \Delta }{ \longrightarrow } \mathrm{FeSiO}_{3(s)}\\ gangue \quad flux \quad \quad slag\)
ii) Aluminium oxide does not dissolve in water, but it dissolves in molten cryolite. This is an aluminium compound with a lower melting point than aluminium oxide. The use of cryolite reduces some of the energy costs involved in extracting aluminium.
\(\mathrm{Al}_{2} \mathrm{O}_{3} \overset { \Delta }{ \longrightarrow } 2 \mathrm{Al}^{3+}+3 \mathrm{O}^{2-}\)
iii) Van-Arkel method is based on the thermal decomposition of metal compounds which lead to the formation of pure metals. Zirconium can be purified by using this method. Zirconium metal is heated with iodine at a temperature of 550 K to form zirconium tetraiodide. (ZrI₄). The impurities are left behind
\(\mathrm{Zr}_{(\mathrm{s})}+2 \mathrm{I}_{2(\mathrm{~s})} \frac{550 \mathrm{~K}}{\mathrm{D}} \mathrm{ZrI}_{\text {t(vapour })}\)
The volatile (ZrI₄) on further heating around 1800 K, it gives pure zirconium.
\(\mathrm{ZrI}_{4 \text { (vapour) }} \frac{1800 \mathrm{~K}}{\mathrm{D}} \mathrm{Zr}_{(\mathrm{s})}+2 \mathrm{I}_{2} \)    
iv) In the extraction of lead from galena (PbS) by froth flotation process, if ZnS is present as impurity, Sodium cyanide is added to depresses the flotation property of ZnS. It forms a complex. Na₂[Zn(CN)₄] on the surface of zinc sulphide.

The reaction of CO with hydrogen at a pressure of less than 50 atm using metal catalysts at 500 - 700 K yields saturated and unsaturated hydrocarbons.
\(nCO+(2n+1){ H_2 }\longrightarrow C_{ n }{ H }_{ (2n+2) }+{ nH }_{ 2 }O\)
\(nCO+2n{ H }_{ 2 }\longrightarrow { C }_{ n }{ H }_{ 2n }+{ nH }_{ 2 }O\)

When boric acid or borate salt is heated with ethyl alcohol in presence of conc. sulphuric acid, an ester, trialkylboranes is formed.
The vapour of this ester burns with a green edged flame
This reaction is used to identify the presence of triethyl borate.
\({ H }_{ 3 }{ BO }_{ 3 }+{ 3C }_{ 2 }{ H }_{ 5 }OH\overset { Conc }{ \underset { { H }_{ 2 }{ SO }_{ 4 } }{ \longrightarrow } } B\left( { OC }_{ 2 }{ H }_{ 5 } \right)_3 +3{ H }_{ 2 }O\)
                       Ethyl borate
                        (Green edged flame)

\(\operatorname{Density}(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}} \)
\(\rho=10 \mathrm{~g} \mathrm{~cm}^{-3} ; \mathrm{a}=100 \mathrm{pm}=1 \times 10^{-8} \mathrm{~cm} ; \mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23} ; \mathrm{n}=4 ; \mathrm{M}=? \)
\(M=\frac{\rho \mathrm{a}^{3} \mathrm{N_{A}}}{n} \)
\(=\frac{10 \times\left(1 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}}{4}  \)
\(=\frac{6.023}{4} \)
= 1.505 g /mol
No. of moles \(=\frac{\text { Mass }}{\text { Molar mass }}=\frac{1}{1.505}\)
= 0.664 moles
Hence number of atoms = 0.664 x 6.023 x 10²³ = 3.99 x 10²³ atoms

(i) Many micro organisms (bacteria, fungi and moulds) produce certain chemicals which inhibit the growth or metabolism of some other micro organism. Such chemicals are called antibiotics.
(ii) Eg: Penicillin from the mould penicillium notatum

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

1. The crude metal is refined by electrolysis. It is carried out in an electrolytic cell
Anode : Impure metal to be refined with dilute acid.
Cathode : Thin strips of pure metal
Electrolyte : Aqueous solution of the salts of the metal with dilute acid.
2. The metal dissolves from the anode, pass into the solution.
3. At the same amount of metal ions from the solution will be deposited at the cathode.
4. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud.
Example: Electrolytic refining of silver.
Cathode: Pure silver
Anode: lmpure silver rods
Electrolyte: Acidified aqueous solution of silver nitrate
5. When a current is passed through the electrodes the following reactions will take place
(a) Reaction at anode: \({ Ag }_{ (s) }\longrightarrow { Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\)
(b) Reaction at cathode: \({ Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\longrightarrow { Ag }_{ (s) }\)
6. During electrolysis, at anode silver loses electrons and form silver ions and the silver ions migrate towards the cathode and get discharged and deposited on the cathode.
7. Copper, Zinc etc can also be refined by this process.

Allotropism:
1. Some elements exist in more than one crystalline or molecular forms in the same physical state
(a) In Greek "allos" means ⇒ Another
(b) "trope" means ⇒ Change
2. The different forms of an element are called allotropes.
Allotropy of carbon:
Carbon exists as diamond, graphite, fullerenes, carbon nanotubes and graphene.
Graphite:
1. Graphite is the most stable allotropic form of carbon at normal temperature and pressure.
2. It is soft and conducts electricity.
3. It is composed of flat two dimensional sheets of carbon atoms.
4. Each sheet is a hexagonal.
5. It is "sp²" hybridised.
6. C-C bond length is 1.41 Å
7. Each C-atom forms three σ bonds with three neighbouring carbon atoms using three of its valence electrons and the fourth electron present in the unhybridised p-orbital form a π-bond.
8. The successive C-sheets are held together by weak Vander Waals forces.
9 The distance successive sheet is 3.40 Å.
10. It is used as a lubricant either on its own or as a graphited oil.

Diamond:
1. It is very hard.
2. It is "sp³" hybridised.
3. C-C bond length is 1.54 Å
4. It is used for sharpening hard tools, cutting glasses, making bores and rock drilling.

Fullerenes:
1. These allotropes are discrete molecules such as \(C_{32}, C_{50}, C_{60}, C_{70}, C_{76}\) etc.
2. It has cage like structure
3. The C₆₀ molecules have a "soccer" ball like structure and is called buckminster fullerene or buckyballs.
4. It has a fused ring structure consists of 20 six membered rings and 12 five membered ring.
5. Each carbon atom is "sp²" hybridised.
6. It has three σ bonds and a delocalised π bond giving aromatic character to these molecules.
7. The C-C bond distance is 1.44 Å
8. The C=C bond distance is 1.38 Å.

Carbon nanotubes:
1. Carbon nanotubes, another recently discovered allotropes, have graphite like tubes with fullerene ends.
2. Along the axis, these nanotubes are stronger than steel and conduct electricity.
3. These have many applications in nanoscale electronics, catalysis, polymers and medicine.
Graphene:
 It has a single planar sheet of "sp²" hybridised carbon atoms that are densely packed in a "honeycomb crystal" lattice.

Metal excess defect:
(i) It arises due to the presence of more number of metal ions as compared to anions.
(ii) Examples: NaCl, KCl
(iii) The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the presence of extra cation.

(iii) For example, when NaCI crystals are heated in the presence of sodium vapour, Na⁺ ions are formed and are deposited on the surface of the crystal.
(iv) Chloride ions (Cl^-) diffuse to the surface from the lattice point and combines with Na⁺ ion.
(v) The electron lost by the sodium vapour diffuse into the vacancy created by the Cl^- ions.
(vi) Such anionic vacancies which are occupied by unpaired electrons are called F centers. Hence, the formula of NaCl can be written as Na_1+xCl.
Metal deficiency defect:

(i) Metal deficiency defect arises due to the presence of less number of cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states.
(ii) For example, In FeO crystal, some of the Fe²⁺ ions are missing from the crystal lattice. To maintain the electrical neutrality, twice the number of other Fe²⁺ ions in the crystal is oxidized to Fe³⁺ ions. In such cases, overall number of Fe²⁺ and Fe³⁺ ions is less than the O^2- ions.

Narcotic drugs: [Opioids]
(i) These are narcotic analgesics. These relieve pain and produce sleep. These drugs are addictive. In poisonous dose, these produces coma and ultimately death.
(ii) They are used for either short-term or long term relief of severe pains. They are mainly used for post operative pain, pain of terminal cancer.
(iii) Eg: Morphine, Codeine
Non-narcotic drugs:
(i) These are non- narcotic analgesics. These reduce the pain without causing impairment of consciousness.
(ii) They alleviate pain by reducing local inflammatory responses.
(iii) These are used for short-term pain relief and for modest pain like headache, muscle strain, bruising or arthritis.
(iv) These drugs have many other effects such as reducing fever (antipyretic) and preventing platelet coagulation (eg. aspirin).
(v) These reduce fever by causing the hypothalamus to over ride a prostaglandin - induced increase in temperature.
(vi) Eg. Aspirin, Paracetamol, Ibuprofen. There are two types in it.
a) Anti inflammatory: eg: Aspirin, Paracetamol
b) Non steroidal anti -inflammatory drugs: (NSAIDs) eg: Ibuprofen

(v) CH₂ = CH – CH(Cl)–O–CH₃ : 3 - methoxy,3- choloro propene (mixed ether)
(vi) dibenzyl ether CH₆CH₅ -O-CH₂-C₆H₅ : benzoxy toluene + Simple ether
(vii) vinyl allyl ether : ethoxy  2 - propene
CH₂ = CH –O– CH₂–CH–CH₂(mixed ether)

1. Reduction of oxides of active metals such as sodium, potassium etc., by carbon is thermodynamically not feasible.
2. Such metals are extracted from their ores by using electrochemical methods.
3. In this technique, the metal salts are taken in a fused form or in solution form.
4. The metal ion present can be reduced by some suitable reducing agent or by electrolysis.
5. Gibbs free energy for the electrolysis process is given by
Here,
 \(\Delta\)G^o = -nFE^o
 \(\Delta\)G^o - Standard Gibb's free energy change
n - number of electrons involved,
F - Faraday,
E^o - Standard electrode potential
6. If E^o is positive then the \(\Delta\)G^o is negative and the reduction is spontaneous
7. Hence a redox reaction is planned in such a way that the e.m.f of the net redox reaction is positive.
8. When a more reactive metal is added to less reactive metal salt solution, the more reactive metal will go into the solution.
Example:
\({ Cu }_{ (s) }+2{ Ag }^{ + }_{ (aq) }\longrightarrow { Cu }^{ 2+ }_{ (aq) }+2{ Ag }_{ (s) }\)
\({ Cu }^{ 2+ }_{ (aq) }+{ Zn }_{ (s) }\longrightarrow { Cu }_{ \left( s \right) } +{ Zn }^{ 2+ }_{ (aq) }\)

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

1. A double salt which contains fourth-period alkali metal (A) is potash alum
K₂SO₄ Al₂(SO₄)₃ - 24 H₂O
2. On heating potash alum (A) 500 k give anhydrous potash alum (or) burnt alum (B).
\(\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O} \stackrel{500 \mathrm{~K}}{\longrightarrow} \mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+24 \mathrm{H}_{2} \mathrm{O}\)
[Potash alum (A)]                         [Burnt alum (B)]
3. Aqueous solution of burnt alum, has sulphates ion, potassium ion and aluminium ion. Sulphate ion reacts with BaCl₂ to form a white precipitate of Barium Sulphate
(SO₄)₂ + BaCl₂ → BaSO₄ + 2 Cl¯
Aluminium ion reacts with alizarin solution to give a red colour compound.

\(\operatorname{Density}(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}} \)
\(\mathrm{n}=2, \mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23} ; \mathrm{a}=288 \mathrm{pm}=2.88 \times 10^{-8} \mathrm{~cm}, \rho=7.2 \mathrm{~g} \mathrm{~cm}^{-3} \)
\(\therefore M=\frac{\rho \times a^{3} \times N_{A}}{n} \)
\(=\frac{7.2 \times\left(2.88 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}}{2} \)
\(=517.95 \times 10^{-1} \)
\(=51.795 \mathrm{~g} \mathrm{~mol}^{-1} \)
Number of moles (n) \(=\frac{\text { Mass }}{\text { Molar mass }}=\frac{208}{51.795}\)=4.02 moles
No. of atoms = No. of moles \(\times\) Avogadro number
=n \(\times\) N_A
\(=4.01 \times 6.023 \times 10^{23} \)
\(=24.15 \times 10^{23} \text { atoms }\)

For cubic closed packed structure
\(r =\frac{a \sqrt{2}}{4}  \)
\(\therefore a =\frac{4 r}{\sqrt{2}}  \)
\(=\frac{4 \times 1.25 \times 10^{-8}}{1.414}=3.53 \times 10^{-8} \mathrm{~cm} \)
= 353 pm