BaSO₄ ⇌ Ba²⁺ + SO₄^2-
K_sp = (s) (s)
K_sp = (s)²
= (2.42 × 10^-3gL^-1)²
\(=\frac{2.42 \times 10^{-3} gL^{-1}}{233 \text{ g mol}^{-1}}\)
= (0.01038 x 10^-3)²
= (1.038 x  x 10^-5)²
= 1.077 x 10^-10
= 1.08 × 10^-10mol²L^-2

\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A₀] = 100: [A] = 25
\(6.909=\frac { 2.303 }{ t } \log\frac { \left[ {100 } \right] }{ \left[ 25\right] } \)
\(t =\frac { 2.303 }{ 6.909 } \log(4)\)
\(t =\frac { 1 }{ 3 } \log(2^2)\)
\(= \left( \frac { 2 }{ 3 } \right) \log2\)

Packing efficiency = 68%
\(\therefore\)empty space percentage = 100 - 68 = 32%

a) Fe²⁺  b) Fe³⁺  c) Fe⁰

Cr ⇒ [Ar]3d⁵4s¹

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

The galvanic cell is represented by a cell diagram, for example, Daniel cell is represented as
\({ Zn }_{ (s) }|{ Zn }_{ (aq) }^{ 2+ }||{ Cu }_{ (aq) }^{ 2+ }|{ Cu }_{ (s) }\)
(i) A single vertical bar (|) represents a phase boundary
(ii) The double vertical bar (||) represents the salt bridge.
(iii) The anode half cell is written on the left side of the salt bridge and the cathode half cell on the right side.
(iv) The anode and cathode are written on the extreme left and extreme right, respectively.
(v) The emf of the cell is written on the right side after cell diagram.

(a) CH₃COONa is a salt of weak acid
(CH₃COOH) and a strong base (NaOH).
Hence, the solutions is alkaline due to hydrolysis.
\(CH_{3}COO^{-}_{(aq)}+H_{2}O_{(aq)}\rightleftharpoons CH_{3}COOH_{(aq)}+OH^{-}_{(aq)}\)
(i)\(h=\sqrt{\frac{K_{w}}{K_{a}\times C}}\)
Given that pK_a =4.74
pK_a = -log K_a
ie., K_a = antilog of (-pK_a)
= antilog of (-4.74)
= antilog of (-5 + 0.26)
= 10^-5 \(\times\) 1.8 = 1.8 \(\times\) 10^-5
[antilog of 0.26 = 1.82 \( \simeq\) 1.8]
\(\therefore\) h=\(\sqrt{\frac{1\times10^{-14}}{1.8\times10^{-5}\times0.1}}\)
h = 7.5 x 10^-5
(ii) \(K_{h}=\frac{K_{w}}{K_{a}}=\frac{1\times10^{-14}}{1.8\times10^{-5}}\)
\(=5.56\times10^{-10}\)
iii) \(pH=7+\frac{pK_{a}}{2}+\frac{logC}{2}\)
= \(7+\frac{4.74}{2}+\frac{log0.1}{2}\)
= 7 + 2.37 - 0.5
= 8.87

\(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightleftharpoons 3 \mathrm{Ca}^{2+}_{(aq)}+2 \mathrm{PO}_{4_{(aq)}}^{3-}\\\)
(s)                           (3s)             (2s)
\(K_{sp}=[Ca^{2+}]^{3}[PO_{4}^{3-}]^{2}\)
\(K_{sp}=(3s)^{3}(2s)^{2}\)
\(K_{sp}=27s^{3}.4s^{2}\)
\(K_{sp}=108s^{5}\)
(or)
\(K_{s p} =m^{m} \cdot n^{n} \cdot(s)^{m+n} \)
\(K_{\text {sp }} =3^{3} \cdot 2^{2} \cdot(s)^{3+2} \)
\(K_{s p} =27 \times 4 \times(s)^{5} \)
\(=108(s)^{5}=108 s^{5}\)

Let the concentration of the reactant be [A] = a
Rate of reaction R = k [A]²
R=ka²
(i) R = k (2a)²
= 4 ka²
= 4R
∴ The rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half [A] = \(\frac { 1 }{ 2 } \) a than the rate of reaction would be
\(R=k\left( \frac { 1 }{ 2 } a \right) 2\)
\(=\frac { 1 }{ 4 } k{ a }^{ 2 }\)
\(=\frac { 1 }{ 4 } kR\)
∴ The rate of the reaction would reduced by \({ \frac { 1 }{ 4 } }^{ th }\)

(i) Cr₂O₇^2- + 2OH^- ⟶ 2CrO₄^2- + H₂O
(ii) MnO₄^- + 4H⁺ + 3e^- ⟶ MnO₂ + 2H₂O

(i) The intermolecular forces of attraction that are present in solids are very strong.
(ii) The constituent particles of solids cannot: move from their positions. They have fixed positions.
(iii) However, they can oscillate about their mean positions.
(iv) This is the reason solids are rigid.

At high temperatures it forms higher boranes liberating hydrogen.
\(5{ B }_{ 2 }{ H }_{ 6 }\overset { 388K }{ \underset { U-tube }{ \longrightarrow } } 2{ B }_{ 5 }{ H }_{ 11 }+4{ H }_{ 2 }\)
\(2{ B }_{ 2 }{ H }_{ 6 }\overset { 198-373K }{ \longrightarrow } { B }_{ 4 }{ H }_{ 10 }+{ H }_{ 2 }\)
\(5{ B }_{ 2 }{ H }_{ 6 }\overset { 373K }{ \underset { sealed\ tube }{ \longrightarrow } } { B }_{ 10 }{ H }_{ 14 }+8{ H }_{ 2 }\)
\(5{ B }_{ 2 }{ H }_{ 6 }\overset { 473-523K }{ \longrightarrow } 2{ B }_{ 5 }{ H }_{ 9 }+6{ H }_{ 2 }\)
\(10{ B }_{ 2 }{ H }_{ 6 }\overset { 523K }{ \longrightarrow } 2{ B }_{ 5 }{ H }_{ 9 }+2{ B }_{ 5 }{ H }_{ 10 }+11{ H }_{ 2 }\)
\({ B }_{ 2 }{ H }_{ 6 }\overset { Red \ hot }{ \longrightarrow } 2B+3{ H }_{ 2 }\).

\(Eu (63)-[\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{0} 6 \mathrm{~s}^{2}, \mathrm{Eu}^{2+}-[\mathrm{Xe}] 4 \mathrm{f}^{7} \)
\(\mathrm{Ce}(58)-[\mathrm{Xe}] 4 \mathrm{f}^{1} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}, \mathrm{Ce}^{2+}-[\mathrm{Xe}] 4 \mathrm{f}^{1} 5 \mathrm{~d}^{1} \)
Eu²⁺ has exactly half filled stable electronic configuration. Hence Europium (II) is more stable than Cerium (II).

(i) Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice. This effect does not change the stoichiometry of the crystal.
(ii) Ionic solids in which the cation and anion are of almost of similar size show schottky defect.
Example: NaCl.
(iii) Presence of large number of schottky defects in a crystal, lowers its density.
(iv) Presence of Schottky defect in the crystal provides a simple way by which atoms or ions can move within the crystal lattice.

Fusion of urea with B(OH)_3' in an atmosphere of ammonia at 800 - 1200 K gives boron nitride.
B(OH)₃ + NH₃ \(\overset { \Delta }{ \longrightarrow } \) BN+ 3H₂O

Self condensation of two molecules of alkyl nitrile (containing α-H atom) in the presence of sodium to form iminonitrile.

With oxygen it forms a toxic carbonyl chloride Hence it is not used an an anaesthetic.

Condensation polymers are formed by the reaction between functional groups of an adjacent monomers with the elimination of simple molecules like H₂O, NH₃ etc....

Chemically soap is the sodium or potassium salt of higher fatty acids. Soaps are made from animal fats or vegetable oils. They contain glyceryl esters of long chain fatty acids:

Drugs that are used to reduce fever are called antipyretic.
Eg: Asprin, paracetamol

(i) Antagonists drugs: The drug that binds to the receptor site and inhibits its natural function are called antagonists.
(ii) Agonist drugs: The drugs which mimic the natural messenger by switching on the receptor are called agonist.

Proteins.

In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pair of bases. Cytosine forms hydrogen bond with guanine while adenine forms hydrogen bonds with thymine. As a result the two strands are complementary to each other.

Lactose is a dissaccharide so on hydrolysis gives I two monosaccharides.
\(\underset { Lactose }{ { C }_{ 12 }{ H }_{ 22 }{ O }_{ 11 }+{ H }_{ 2 }O } \overset { + }{ \longrightarrow } \underset { D(+)G1ucoseD(+)Galactose }{ { C }_{ 6 }{ { H }_{ 12 }{ O }_{ 6 } }+{ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 } } \)

(i) This is also called as salt isomerism.
(ii) This type of isomers arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms. In the below mentioned examples, the nitrite ion is bound to the central metal ion Co³⁺ through a nitrogen atom in one complex and through oxygen atom in other complex.
\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{NO}_{2}\right)\right]^{2+}\)

\(\text { Density }(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}}  \)
\(\mathrm{n}=4, \mathrm{M}=\text { Molar mass of } \mathrm{KF}=58.1 \mathrm{~g} / \mathrm{mol}  \)
\(\rho=2.48 \mathrm{~g} \mathrm{~cm}^{-3}  \)
\(\mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23} \)
\(a^{3} =\frac{n M}{\rho N_{A}}=\frac{4 \times 58.1}{2.48 \times 6.023 \times 10^{23}}  \)
\(a^{3} =15.55 \times 10^{-23}  \)
\(a^{3} =0.1555 \times 10^{-21}  \)
\(a =\sqrt[3]{0.1555 \times 10^{-21}}  \)
\(a =0.5375 \times 10^{-7} \mathrm{~cm}=5.375 \times 10^{-8} \mathrm{~cm}=537.5 \mathrm{pm}  \)
\(d =\frac{a}{\sqrt{2}}(\text { for fcc })  [\therefore r = \frac{a\sqrt{2}}{4}]\)
\(=\frac{537.5}{1.414}=380.13 \mathrm{pm}\)
\(\therefore\) The distance between K⁺ and F^- in KF = 380.13 pm

CO acts as a strong reducing agent.
Example: 3CO + Fe₂O₃ \(\longrightarrow \)  2Fe + 3CO₂
It reduces metallic. oxides into metals.

(i) Free radical polymerisation.
(ii) Styrene polymerises to polystyrene when it is heated to ionic with a peroxide initiator. The mechanism involves the following steps.

(iii) Propagation step

The stabilized radical attacks another monomer molecule to give an elongated radical.

Chain growth will continue with the successive addition of several thousands of monomer units.
(iv) The above chain reaction can be stopped by stopping the supply of monomer or by coupling of two chains or reaction with an impurity such as oxygen.

(i) (A) that gives turbidity within 5-10 min on reaction with anhydrous ZnCl₂/HCl means it must be secondary alcohol.
\({ CH }_{ 3 }\underset { \overset { | }{ OH } }{ CH } -{ CH }_{ 3 }\overset { anhydrous\ ZnCl_{ 2 }/HCl }{ \longrightarrow } { CH }_{ 3 }-\underset { \overset { | }{ CH } }{ CH } -{ CH }_{ 3 }\) 
(ii) Iso propyl alcohol on treatment with sodium hypochlorite, undergoes oxidation to give acetone compound (B).
\({ CH }_{ 3 }-\underset { \overset { | }{ \underset { (A) }{ OH } } }{ C } H-{ CH }_{ 3 }\overset { (O) }{ \longrightarrow } { CH }_{ 3 }-\underset { \overset { | }{ \underset { (B) }{ O } } }{ C } -{ CH }_{ 3 }\) 

i) In increasing order of solubility in water:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}  \)
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\)
ii) In increasing order of basic strength:
a. Aniline, p - toluidine and p - nitro aniline
p - toluidine > aniline >p - nitro aniline

 b.  \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}, \mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2}\\ 2^{o} \text { amine } \quad \quad \quad \quad   e^{\ominus} \text { with drawing (group) }\)
(iii) In decreasing order of basic strength in gas phase:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{NH},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N} \text { and } \mathrm{NH}_{3}  \)
\(\mathrm{NH}_{3}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{~N}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NH}\\ \quad \quad \quad 1^{0} \text { amine } \quad 2^{0} \text { amine } \quad \quad \quad 3^{0} \text { amine }\)
(iv) In increasing order of boiling point:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)
\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \\ 2^{0} \text { amine } \quad \quad1^{0} \text { amine }\)
Generally amines have lower boiling point than alcohol. Due to comparable molecular mass and weaker H-bonds in Amines.
(v) In decreasing order of the pK_b values:
\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\mathrm{CH}_{3} \mathrm{NH}_{2}<\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} \)
\(\quad \quad \quad 2^{0} \text { amine } \quad 1^{0} \text { amine } 1^{0} \text { amine } \quad 2^{0} \text { amine } \)
\(\mathrm{PK}_{b}: \quad 3.00\quad < \quad 3.29 \quad < \quad 3.38 \quad<\quad 9.30\)
pK_b ,Due to + 1 effect of C₂H₅ group. Higher the value of pK_b lower is the basicity
(vi) Increasing Order of basic strength:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2}>\mathrm{CH}_{3} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \\ \left(\mathrm{pK}_{6}: 9.38 \quad > \quad \quad 8.92 \quad \quad > \quad 3.38 \quad > \quad \quad 3.00\right)\)
Due to + 1 effect of C₂H₅ group.
In decreasing order of basic strength:

(i) Only the ions that have unpaired electrons in d- orbital and in which d - d transition is possible will be coloured.
(ii) The ions in which d - orbitals are empty or completely filled will be colourless as no d -d transition is possible in those configurations.
(iii) From the above ions, it can be easily observed that only Sc³⁺ has an empty d - orbital and Cu⁺ has completely filled d-orbitals.
(vi) All other ions, except Sc³⁺ and Cu⁺, will be coloured in aqueous solution because of d - d transition.

AAAA type of three dimensional packing:

1. This is simple cubic arrangement.
2. Three dimensional packing arrangement can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions.
3. Spheres in one layer sitting directly on the top of in the previous layer so that all layers are identical.
4. All spheres of different layers of crystal are perfectly aligned horizontally and also vertically.
5. In simple cubic packing, each sphere is in contact with 6 neighbouring spheres
6. Four in its own layer, one above and one below and hence the coordination number of the sphere in simple cubic arrangement is 6.
ABABA type of three dimensional packing:

(i) This is body centered cubic arrangement.
(ii) The spheres in the first layer are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A.
(iii) The third layer is a repeat of the first.
(iv) This pattern ABABAB is repeated throughout the crystal.
(v) Each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.
ABCABC type of three dimensional packing:

(i) This is face centered cubic arrangement.
(ii) In this arrangement (FCC) second layer spheres are arranged at the dips of first layer. Third layer spheres are arranged in a manner such that it cover the octahedral void.
(iii) Then no longer third layer is similar to first or second layer.
(iv) Third layer gives different arrangement. Fourth layer spheres are similar to first layer.
(v) If the first, second and third layer are represented as A, B, C then this type of packing gives the arrangement of layers as ABCABC.. and the sequence is repeated.