H₂O + H₂O ⇌  H₃O⁺ + OH^-
acid 1  base 1  acid 2   base 2
HF + H₂O ⇌  H₃O⁺ + F^-
acid 1  base 1  acid 2   base 2
∴ Conjugate bases are OH^- and F^- respectively.

T₁= 200 K ; k = k₁
T₂ =400 K ; k₁ = k₂ = 2k₁
log\(\frac {{ k }_{ 2 } }{ { k }_{ 1 } } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] \)
log \( [\frac {{2 k }_{ 1 } }{ { k }_{ 1 } }]\)
\(= \left[ \frac { E_{a} }{ 2.303\times 8.314 JK^{-1} mol^{-1} } \right] \)
\(= \left[ \frac { 400 k - 200K }{200 k \times 400 k } \right] \)
\(E_{a} = \frac{ 0.3010 \times 2.303 \times 8.314 JK^{-1}mol^{-1} \times 200K \times 400 K}{200 K}\)
E_a = 2305 J mol^-1
E_a = 2.305 kJ mol^-1

The complex is [M(H₂O)₅Cl]Cl₂
1000 ml of 1 M solution of the complex gives 2 moles of Cl^- ions 1000 ml of 0.01 M solution of the complex will give 
\(\frac{100 ml \times 0.01M \times 2Cl^-}{1000 ml \times 1M}\)
= 0.002 moles of Cl^- ions

Structure of amines:
(i) Like, ammonia, nitrogen atom of amines is trivalent and carries a lone pair of electron and sp³ hybridised, out of the four sp³ hybridised orbitals of nitrogen, three sp³ orbitals overlap with orbitals of hydrogen (or) alkyl groups of carbon, the fourth sp³ orbital contains a lone pair of electron.
(ii) Hence, amines possess pyramidal geometry. Due to presence of lone pair of electron C-N-H (or) C-N-C bond angle is less than the normal tetrahedral bond angle 109.5°.
For example, the C-N-C bond angle of trimethylamine is 108° which is lower than tetrahedral angle and higher than the H-N-H bond angle of 107°.
(iii) This increase is due to the repulsion between the bulky methyl groups.

(i) Alcohols are weakly basic in nature so easily in presence of strong acids.
(ii) Due to the presence of +ve charge on the oxygen atom, C-O in protonated alcohols became weak hence easily cleaved by halide ions to form alkyl halides.
(iii) Where as phenols are much weaker bases due to delocalisation of the lone pair of electrons on the oxygen atom due to resonance and so are not easily protonated.
(iv) The C-O bond in phenols has some double bond character; hence not easily cleaved by halide ions to form respective alkyl halides.

Oxidation half cell : A metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.
Reduction half cell : A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.

(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

Mass = 0.44 g
No. of moles = Given Volume/Molar Volume = 112/22400
Molar Mass =  0.44/112 x 22400 = 88g
C_nH_2n+1 + OH = 88
12n+ (1)(2n+1) + 16 +1 = 88
14n + 18 = 88
14n = 88 - 18
n =70/14 = 5

                                                          Pentanoicacid

(i) 2-methyl propan-2-o1 < Butan-2-ol < Butan-1-ol

(ii) propan-2-ol < propan-1-ol < propan -1,3- diol < propan-1,2,3-triol

(i) The highest oxidation state in oxides and fluorides is due to small size and high electro negativity of F and O.
(ii) Many Cu⁺ compounds are unstable in aqueous solution and undergo disproportionation.
2Cu⁺ ⟶ Cu²⁺+ Cu
This suggest that in aqueous solution Cu+_(aq), converts into Cu²⁺_(aq) which is due to much more negative Δ_hyd H^- of Cu²⁺_(aq) than Cu⁺, which compensates more for the second ionisation enthalpy of Cu.

(i) Gold, one of the expensive and precious metals. It is used for coinage, and has been used as standard for monetary systems in some countries.
(ii) It is used extensively in jewellery in its alloy form with copper.
(iii) It is also used in electroplating to cover other metals with a thin layer of gold which are used in watches, artificial limb joints, cheap jewellery, dental fillings and electrical connectors.
(iv) Gold nanoparticles are also used for increasing the efficiency of solar cells and also used an catalysts.

Let \(\left[A_{0}\right]=100 \%\), t = 50 minutes
Then [A]=100 - 40 = 60 %
(1) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{\mathrm{o}}\right]}{[\mathrm{A}]}\)
\(=\frac{2.303}{50} \log \left(\frac{100}{60}\right) \)
\(=\frac{2.303}{50} \log 1.667  \)
\(=\frac{2.303}{50} \times 0.2219 \)
\(\mathrm{k}=0.010216 \mathrm{~min}^{-1}  \)
\(\mathrm{k}=1 \times 10^{-2} \mathrm{~min}^{-1}\)
(2)  \( t =\frac{2.303}{0.010216} \log \left(\frac{100}{20}\right) \)
\(t =\frac{2.303}{{0.010216}}\times 0.6990\)
= 225.43 \(\times \) 0.6990
t = 157.58 min.
The time at which the reaction will be 80% complete is 157.58 min.

(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.

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

(i) Anode: Zinc amalgamated with mercury
(ii) Cathode: HgO mixed with graphite
(iii) Electrolyte: Paste of KOH and ZnO

(ii) benzoic acid

(iii) α - hydroxyphenylaceticacid.

(i)  According to Lewis theory a species that donates a pair of electron is called Lewis base.
(ii) Nitrogen more in NH₃ is less electro negative than oxygen in water. So the non - bonded electron pair on nitrogen is more available for sharing than a non - bonded electron pair on oxygen atom. So NH₃ is a stronger lewis base than H₂O.

The reaction is a zero order reaction whose rate is independent on the concentration of reactants.

(i) The electron gain enthalpy for oxygen is less negative because of its small size due to which the electron repulsions in the relatively small 2p-subshell are comparatively large.
(ii) Hence the Incoming electrons are not accepted with the same ease as in case of sulphur as it has relatively large size.

(i) The stereoisomers of a coordination compound have the same chemical formula and connectivity between the central metal atom and the ligands.
(ii) But they differ in the spatial arrangement of ligands in three dimensional space. They can be further classified as geometrical isomers and optical isomers.

SiO₂ is used in the metallurgy of copper to remove FeO as FeSiO₃ (slag) (i.e) acidic flux is used to remove basic impurities.
\(FeO+{ SiO }_{ 2 }\longrightarrow { FeSiO }_{ 3 }\)

(i) An organic compound which has the smell of bitter almonds is benzaldehyde C₆H₅CHO and it is (A).
(ii) Benzaldehyde reacts with acetic anhydride in the presence of sodium acetate gives Cinnamic acid.
\(\underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO } +\left( { CH }_{ 3 }CO \right) _{ 2 }O\overset { { CH }_{ 3 }COONa }{ \longrightarrow } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }CH=CHCOOH } +{ CH }_{ 3 }COOH\) 
(iii) Benzaldehyde on treatment with alcoholic KCN gives Benzoin a dimer and it is (C).

Electrophoresis:
(i) When electric potential is applied across two platinum electrodes dipped in a hydrophilic sol, the dispersed particles move toward one or other electrode.
(ii) This migration of sol particles under the influence of electric field is called electrophoresis or cataphoresis.
(iii) If the sol particles migrate to the cathode, then they posses positive (+) charges, and if the sol particles migrate to the anode then they have negative charges(-).
(iv) Thus from the direction of migration of sol particles we can determine the charge of the sol particles.
(v) Hence electrophoresis is used for detection of presence of charges on the sol particles

Factors affecting adsorption
Qualitatively, the extent of surface adsorption depends on
(i) Nature of adsorbent
(ii) Nature of adsorbate
(iii) Pressure
(iv) Concentration at a given temperature.
1. Surface area of adsorbent:
As the adsorption is a surface phenomenon it depends on the surface area of adsorbent. i.e., higher the surface area, higher is the amount adsorbed.
2. Nature of adsorbate:
The nature of adsorbate can influence the adsorption. Gases like SO₂, NH₃, HCl and CO₂ are easily liquefiable as have greater vander waal's force of attraction. On the other hand, permanent gases like H₂, N₂ and O₂ cannot be liquefied easily. These permanent gases are having low critical temperature and adsorbed slowly, while gases with high critical temperature are adsorbed readily.
3. Effect of temperature:
When temperature is raised chemisorption first increases and then decreases. whereas physisorption decreases with increases in temperature.
4. Effect of Pressure:
Chemical adsorption is fast with increase in pressure, it cannot alter the amount of adsorption. In physisorption, the extend of adsorption increases with increase in pressure.

On the basis of magnetic properties, materials can be broadly classified as
(a) paramagnetic materials
(b) diamagnetic materials, besides these there are ferromagnetic and antiferromagnetic materials
(i) Materials with no elementary magnetic dipoles are diamagnetic, in other words a species with all paired electrons exhibits diamagnetism.
(ii) This kind of materials are repelled by the magnetic field because the presence of external magnetic field, a magnetic induction is introduced to the material which generates weak magnetic field that oppose the applied field
(iii) Paramagnetic solids having unpaired electrons possess magnetic dipoles which are isolated from one another.
(iv) In the absence of external magnetic field, the dipoles are arranged at random and hence the solid shows no net magnetism.
(v) But in the presence of magnetic field, the dipoles are aligned parallel to the direction of the applied field and therefore, they are attracted by an external magnetic field.
(vi) Ferromagnetic materials have domain structure and in each domain the magnetic dipoles are arranged.
(vii) But the spin dipoles of the adjacent domains are randomly oriented.
(viii) Some transition elements or ions with unpaired d electrons show ferromagnetism.

(i) O₂ molecules are held together by weak Vander Waal's force because of small size and high electronegativity of oxygen.
(ii) In contrast, sulphur shows catenation and forms stronger S-S bonds.
(iii) Due to catenation, sulphur forms octa-atomic S₈ molecules having eight membered puckered ring structure.
(iv) Because of its bigger size the force of attraction holding S₈ molecules are much stronger.
(v) Hence sulphur is a solid at room temperature or in other words, that is why there is a large difference between the boiling point (also melting points) of the two elements.

Crystal defects are classified as follows
(i) Point defects
(ii) Line defects
(iii) Interstitial defects
(iv) Volume defects
Point defects are further classified as follows

(i) Diamminesilver(I) chloride
(ii) Potassiumpentacyanidonitrosylferrate(II)
(iii) Ptricarbonyltriphenylphosphanechromium(O)
(iv) diamminesilver(I) ion
(v) Hexafluoridoferrate(II) ion.