According to Saytzeff rule.

The incorrect statement is option (n)
Physisorption is an exothermic process. Hence increases in temperature decreases the Physisorption

HPO₄²⁻ can have the ability to accept a proton to form H₂PO₄^-
It can also have the ability to donate a proton to form PO₄^-3

x ml of 0.1 M NaOH + x mL of 0.01 M HCI
No. of moles of NaOH = 0.1 x X x 10^-3
= 0.1 X x 10^-3
No. of moles of HCI = 0.01 x X x 10^-3
= 0.01 X x 10^-3
No. of moles of NaOH after mixing
= 0.1 X x 10^-3 - 0.01 X x 10^-3
= 0.09 X x 10^-3
Concentration of NaOH\(= (\frac{0.09x \times 10^{-3}}{2x \times 10^{-3}}) = 0 .045\)
[OH^-] = 0.045
pOH =-log (4.5 x 10^-2)
= 2 -log 4.5
= 2 - 0.65 = 1.35
pH = 14 - 1.35 = 12.65

	A	⟶	B	C	D
Initial rate (M s-1)	a		0	0	0
Reaction number	x	-	-	-	-
After time t	(a - x)		x	x	x
Total number of moles	= (a + 2x)

Given:
At low pressure the reaction follows first order therefore,
Rate α [reactant]¹
Rate α (surface area)
At high pressure due to the complete coverage of surface area, the reaction follows zero order.
Rate α [reactant]⁰
Therefore the rate is independent of surface area.

let edge length = a
\(\sqrt{2a} = 4r\)
\(a = \frac{4 \times 300}{\sqrt {2}}\)
a = 600 x 1.414
a = 848.4 pm

For [MA₄B₂]⁴⁺ complexes - geometrical isomerism is possible
{[Co(NH₃)₄Br₂]CI,[Co(NH₃)₄Br CI]Br} - ionisation isomers

[\( \overset{+}{Fe}\)(H₂O)₅\( \overset{+}{NO}\)]²⁺ SO^2-₄
+1 and +1 respectively

As we move from La to Lu, their metallic behaviour because almost similar to that of aluminium.

MnO^-₄ + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

(B) \({ CH }_{ 3 }-\overset { \underset { | }{ OH } }{ CH } -{ CH }_{ 3 }\)  Isopropyl alcohol
(C)  \(CH_{ 3 }-\overset { \underset { | }{ OH } }{ CH } -{ CH }_{ 3 }\) Isopropyl chloride 
(D) CH₃ - CH - CH₂  Propylene

Hexamethylene tetramine formed by the reaction of formaldehyde and ammonia is called urotropine.

The migration of colloidal particles under the influence of an electric field is termed cataphoresis or 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) The migration of sol particles under the influence of electric field is called electrophoresis or cataphoresis. 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(-). Thus from the direction of migration of sol particles we can determine the charge of the sol particles. Hence electrophoresis is used for detection of presence of charges on the sol particles.

In a colloidal, the substance present in larger I amount is called dispersion medium and the substance present in less amount is called dispersed phase.

This redox process which causes the deterioration of metal is called corrosion.

Conductivity, molar conductance and equivalent conductance increases with dilution whereas Conductance (C) decreases.

(i) 2-Methoxy ethanol;
(ii) Dimethoxy methane and
(iii) Phenyl methanol

According to Lewis concept of Acid and bases, any species capable of accepting an electron pair is an acid. BF₃ is electron deficient so accepts a pair of electron, Hence termed as acid

	(i)	(ii)
Conjugate Acid	\({ NH }_{ 4 }^{ + }\)	\({ H }_{ 2 }{ SO }_{ 4 }\)
Conjugate Base	\({ NH }_{ 2 }^{ - }\)	\(^{ }{ SO }_{ 4 }^{ 2- }\)

Molar conductance = Λ_{m =} \( \frac{k\ (Sm^{-1})\times10^{-3}}{M} mol^{-1}m^{3} \)
\(= \frac{(12.04 \times 10^{-2} Sm^{-1}) \times 10^{-3} (mol^{-1}m^{3})} {0.025}\)
= 481.6 x 10^-5 Sm²mol^-1

Electrochemical reaction at cathode is Ag⁺ + e^-  → Ag (reduction)
m = ZIT
Z = \(\frac{\text {molar mass of Ag}}{(96500)}\) = \(\frac{108}{1 \times 96500} \)
I = 2A
t = 20 x 60S = 1200 S
It = 2A  x 1200S = 2400C
m = \(\frac{108 gmol^{-1}}{96500 C mol^{-1}} \times\) 2400C
m = 2.68g

\(\underset{0.001M}{HCl}\overset{H_{2}O}{\rightleftharpoons }\underset{0.001M}{H_{3}O^{+}}+\underset{0.001M}{Cl^{-}}\)
H₃O⁺ from the auto ionisation of H₂O (10^-7M) is negligible when compared to the H₃O⁺ from 10^-3M HCl.
Hence [H₃O⁺] = 0.001 mol dm^-3
pH = -log₁₀ [H₃O⁺]
= -log₁₀^(0.001)
= -log₁₀(10^-3) = 3

[H₃O⁺] = 4 \(\times\) 10^-5M
pH = - log₁₀[H₃O⁺]
pH=-log₁₀[4 \(\times\) 10^-5]
pH = -log₁₀[4] - log₁₀[10^-5]        log₁₀ 10 = 1
pH = -log 4 + 5log₁₀10
= 5 - log 4
= 5 - 0.6021
=4.3979
Since pH is less than 7, the solution is acidic.

Rate = K[X]ⁿ[y]^m
4 x 10^-3 mol L^-1s^-1= 2 x 10^-2s^-1(0.2mol L^-1)ⁿ(0.2mol L^-1)^m
\(\frac { 4\times { 10 }^{ -3 }mol\quad { L }^{ -1 }{ s }^{ -1 } }{ 2\times { 10 }^{ -2 }{ s }^{ -1 } } =\) (0.2)^n+m(mol L^-1)^n+m
0.2(mol L^-1) = (0.2)^n+m(mol L^-1)^n+m
Comparing the powers on both sides
The overall order of the reaction n + m = 1

Permanganate ion has tetrahedral geometry in which the central Mn⁷⁺ is Sp³ hybridised.

(i) Silver in its +1 oxidation state, exhibits 4d¹⁰5s⁰configuration.
(ii) But in some compounds, it also shows +2 oxidation state, so the configuration becomes 4d⁹5s⁰
(iii) Here, d-orbital is not completely filled. Therefore silver is a transition element.

(i) I Order reaction
(ii) S^-1
(iii) \({ t }_{ \frac { 1 }{ 2 } }\) = 0.693/k

Rate of a reaction at any time depends on the concentration of the reactants which keeps on decreasing with time.

Element of group 15 have stable half-filled p-orbitals hence large amount of energy is required to remove electrons as compared to group 16, which has an incomplete p subshell.

Yellow phosphorus reacts with alkali on boiling in an inert atmosphere liberating phosphine. Phosphorus acts as a reducing agent.
\({ P }_{ 4 }+NaOH+{ H }_{ 2 }O\longrightarrow \underset { Sodiumhypophosphite }{ { 3NaH }_{ 2 }{ PO }_{ 2 } } +\underset { Phosphine }{ { PH }_{ 3 } } \uparrow \)

The total number. of atoms per unit cell in bcc.
\(=\frac { { N }_{ c } }{ 8 } +\frac { { N }_{ b } }{ 1 } =\frac { 8 }{ 8 } +\frac { 1 }{ 1 } =1+1=2\)
N_b = Number of atoms inside the body; Nc is the number of atoms at the comers.

(i) The intermolecular forces of attraction that are present in solids are very strong.
(ii) The constituent particles of solids have fixed position.
(iii) Hence, solids have a definite volume.

(i) Anhydrous AlCl₃ is an electro-deficient compound while hydrated AICl₃ is not.
(ii) Therefore, anhydrous AICl₃ is more soluble in diethyl either because the oxygen atom of ether donates a pair of electrons to the vacant p-orbital on the Al atom in AlCl₃ forming a co-ordinate bond.
(iii) In case of hydrated AlCl₃, Al is not electron deficient, since, H₂O has already donated a pair of e^- to it.

(i) It is used for purification of water
(ii) It is also used for water proofing and textiles.
(iii) It is used in dyeing, paper and leather tanning industries.
(iv) It is employed as a styptic agent to arrest bleeding.

In this method, the concentrated ore is oxidised by heating it with excess of oxygen in a suitable furnace below the melting point of the metal.

The change in Gibbs energy is related to the equilibrium constant K as \(\Delta \)G = - RT Ink. A room temperature all reactants and product of the given reaction are in solid state. As a result, equilibrium does not take place at room exist between the reactants and the products. Hence the reaction does not take place at room temperature. Certain amount of activation energy is essential even for reactions which are thermodynamically feasible

The coordination compounds with same formula, but have different connections among their constituent atoms are called structural isomers or constitutional isomers.

Tetrahedral, paramagnetic.

An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are usually non-stoichiometric compounds. Transition metals form a number of interstitial compounds such as TiC, ZrH_1.92, Mn₄N etc.
Properties of interstitial compound
(i) They are hard and show electrical and thermal conductivity.
(ii) They have high melting points higher than those of pure metals.
(iii) Transition metal hydrides are used as powerful reducing agents 
(iv) Metallic carbides are chemically inert.

In Gd⁺³ there are 64 electrons. Hence electronic configuration will be [Xe]4f⁷ 5d¹ 6s². Hence no electrons are there in outer d - orbital. Due to this it is colourless.

(i) \(\mathrm{Mg}\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{2-} \rightleftharpoons \mathrm{Mg}^{2+}+\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{2-} (2 ions)\)
(ii) \(\begin{aligned} {\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3\left[\mathrm{CoF}_6\right]_2 \rightleftharpoons 3\right.} & {\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} } \\ & (5 \text { ions }) \end{aligned}\)\(+2\left[\mathrm{CoF}_6\right]^{3-}\)
(iii) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]= \text{ No ions}\)
If no of ions increases, molar conductivity increases molar conductivity of the complex also INCREASES.
\(\therefore\) The order of the given compound is
[Cr(NH₃)₃Cl₃]<Mg[Cr(NH₃)₃Cl₃] < [Cr(NH₃)₅Cl₃] [CoF₆]₂
  (No ion)          (2 ions)                    (5 ions)

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

(i) In heavier post transition metals, the outers electrons (ns) have a tendency to remain inert and show reluctance to take part in the bonding, which is known as inert pair effect.
(ii) This effect is also observed in groups 14, 15 and 16.

Catenation is an ability of an element to form chain of atoms.
The conditions for catenation.
(a) The valency of element is greater than or equal to two.
(b) Element should have an ability to bond with itself
(c) The self bond must be as strong as Its bond with other elements
(d) Kinetic inertness of catenated compound towards other molecules.
(e) Carbon possesses all the above properties and forms a wide range of compounds with itself and with other elements such as H, O, N, S and halogens.

In p-block elements, the first member of each group differs from the other elements of the corresponding group. The following factors are responsible for this anomalous behaviour
(i) Small size of the first member.
(ii) High ionisation enthalpy and high electronegativity.
(iii) Absence of d-orbitals in their valance shell.

Ellingham diagram for the formation of ZnO and CO₂ intersects around 1200 K. Above this temperature ZnO lies above Carbon which indicates Carbon is the better reducing agent CO is more effective reducing agent only below 983 K.

This method is commonly used to concentrate sulphide ores such as galena (PbS), zinc blende (ZnS) etc.
In this method, the metallic ore particles which are preferentially wetted by oil can be separated from gangue.

(i) An organic compound (A) with formula C₇H₆O form bisulphite means it must be benzaldehyde C₆H₅CHO.
(ii) Benzaldehyde on treatment with KCN form benzoin.
\(\underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO+{ C }_{ 6 }{ H }_{ 5 }CHO } \overset { KCN }{ \longrightarrow } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }-\underset { \overset { | }{ HO } }{ CH } -\underset { \overset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 } } \) 
(iii) Benzaldehyde on treatment with sodium acetate and acetic anhydride form an acid, cinnamic acid it is (C).
\(\underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO } +\left( { CH }_{ 3 }{ CO } \right) O\overset { { CH }_{ 3 }COONa }{ \longrightarrow } \underset { (C) }{ { C }_{ 6 }{ H }_{ 5 }CH } =CH-COOH+CH_{ 3 }COOH\)

Compound	Compound Name	Formula
A	Benzaldehyde	C6HsCHO
B	Benzoin	C6H5-CH-OHCO-C6H5
C	Cinnamic acid	C6HsCH=CHCOOH

a.	Ethanol	Phenol
i.	It does not give violet colour with neutral FeCI3	It gives violet colour with netural FeCI3
ii.	It does not declourise bromine water (no electroplilic substitution).	It declourises Br/H2O and the product formed is 2, 4, 6-tribromo (electrtophilic substitution).

a.	Phenol	Acetic acid
i.	It gives violet colour with neutral FeCI3.	It does not give violet colour with netural FeCI3.
ii.	It undergoes electrophilic substitution reaction at ortho & para position.	It does not undergo electrophilic substitution

a.	Phenol	Aniline
i.	It does not undergo diazotisation reaction at all temperature	It undergo diazotisation reaction at 0 C0.
ii.	It undergoes condensation polymerisation with HCHO to produce bakelite	It does not undergo condensation polymerisation reaction with HCHO.

a.	Phenol	Anisole
i.	It gives violet colour with neutral FeCI3.	It does not give violet colour with neutral FeCI3.
ii.	It undergoes coupling reaction with ben- zene diazonium chlo ride to give 'red dye'.	It does not undergo coupling reaction with benzene diazonium chloride.

Freundlich adsorption isotherm:
According to Freundlinch
\(\frac { x }{ m } =kp^{ \frac { 1 }{ n } }\) 
where x is the amount of adsorbate or adsorbed on 'm' gm of adsorbent at a pressure of p. K and n are constants Value is always less than unity.
This equation is applicable for adsorption of gases on solid surfaces. The same equation becomes  \(\frac { x }{ m } =Kc^{ \frac { 1 }{ n } }\) when used for adsorption in solutions with c as concentration.
These equation quantitively predict the effect of pressure(or concentration) on the adsorption of gases(or adsorbates) at constant temperature.
Taking log on both sides of equation \(\frac { x }{ m } ={ kp }^{ \frac { 1 }{ n } }\) 
\(log\frac { x }{ m } =logK+\frac { 1 }{ n } logP\) 

Hence the intercept represents the value of log k and the slope \(\frac { b }{ q } \) gives \(\frac { 1 }{ n } \)
This equation explains the increase of \(\frac { x }{ m } \) with increase in pressure. But experimental values show the deviation at low pressure.

E^o_cell = 1.56V [Zn²⁺]
= 0.1 M [Ag⁺] = 10 M
Formula:
\([{ E }_{ cell }={ E }_{ cell }^{ o }-\frac { RT }{ nF } In\frac { [{ Zn }^{ 2+ }] }{ [{ Ag] }^{ 2 } } ]\)
Solution:
= 1.56 - 0.02955 log 0.001;
= 1.56 - (- 0.08865)
= 1.56 + 0.08865 = 1.6486 V
E_cell = 1.6486 V.

pH = - log [H₃O⁺]
(i) pH - log [10⁴]
pH = log 1 - log 10⁴
pH =  - 4
(ii) pH = - log [10^-7]
pH = log 1 - log 10^-7.
pH = 7
(iii) pH = - log [H₃O⁺]
pH = - log [6.8 x 10^-3]
pH = log 1 - log 6.8 - log 10^-3
= 3 - 0.8325
pH = 2.17 (or) 2.2
(iv) pH = - log [3.2 x 10^-5]
pH = log 1 - log 3.2 - log 10^-5
= 5 - 0.5051
pH = 4.49 (or) 4.5
(v) pH = - log [0.035]
pH = log 1 - log 0.035
= 2 - 0.5441
pH = 1.46 (or) 1.5
(vi) pH = - log [0.25]
pH = log 1 - log 0.25
= 1 - 0.3979
pH = 0.602 (or) 0.60
(vii) pH = -log [H₃O⁺]
pH = log 1 - log 5.4 - log 10^-9
= 7 - 0.7324
pH = 8.267 (or) 8.3
(viii) pH = - log [7.1 x 10^-7]
pH = log 1 - log 7.1 - log 10^-7
= 7 - 0.8513
pH = 6.2.

Kohlraush's law:
(i) At infinite dilution, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. i.e., the molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.
(ii) For a uni - univalent electrolyte such as NaCl, the Kohlraush's law is expressed as
\(\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaCl}}=\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Na}^{+}}+\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Cl}}^{-}\)
(iii) In general, according to Kohlraush's law, the molar conductivity at infinite dilution for a electrolyte represented by the formula A_x B_y, is given below.
\(\left(\Lambda_{m}^{0}\right)_{A_{x} B_{y}}=x\left(\lambda_{m}^{0}\right)_{A^{y+}}+y\left(\lambda_{m}^{0}\right)_{B^{x-}}\)
b) Calculation of molar conductance at infinite dilution for weak electrolytes experimentally.
(i) However, the same can be calculated using Kohlraush's Law. For example, the molar conductance of CH₃COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa.
\(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{0} \) ..........(1)
\(\Lambda_{\mathrm{HCl}}^{0}=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0} \) .........(2)
\(\Lambda_{\mathrm{NaCl}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\) ...........(3)
(ii) Equation (1) + Equation (2) - Equation (3) gives,
\(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}\right)+\left(\Lambda_{\mathrm{HCl}}^{0}\right)-\left(\Lambda_{\mathrm{NaCl}}^{0}\right)=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}-}^{0} \)
\(=\Lambda_{\mathrm{CH}_{2} \mathrm{COOH}}^{0}\)

[R]₀ = 5g, [R] = 3g
K = 1.15 x 10^-3 s^-1
As the reaction is of first order
K = \(\frac { 2.303 }{ t } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
\(t=\frac { 2.303 }{ 1.15\times { 10 }^{ -3 }{ s }^{ -1 } } \log { \frac { 5g }{ 3g } } \)
= 200 x 10³ (log 1.667)s
= 20 x 10³ x 0.22 19 s
= 443.8 s
= 444 s (approximately)

(i) It is generally expected a steady decrease in atomic radius along a period as the nuclear charge increases and the extra electrons are added to the same sub shell.
(ii) But for the 3d transition elements, the expected decrease in atomic radius is observed from Sc to V, thereafter up to Cu the atomic radius nearly remains the same.
(iii) As we move from Sc to Zn in 3d series the extra electrons are added to the 3d orbitals, the added 3d electrons only partially shield the increased nuclear charge and hence the effective nuclear charge increases slightly.
(iv) However, the extra electrons added to the 3d sub shell strongly repel the 4s electrons and these two forces are operated in opposite direction and as they tend to balance each other, it leads to constancy in atomic radii.
(v) At the end of the series, d - orbitals of Zinc contain 10 electrons in which the repulsive interaction between the electrons is more than the effective nuclear charge and hence, the orbitals slightly expand and atomic radius slightly increases.

(i) The element which occupies group number 15 and period number 3 is phosphorus. Hence, A is phosphorus This is further supported by the fact that A exhibits allotropy and A is tetra atomic.
(ii) Phosphorus reacts with caustic soda to form phosphine as below:
4P + 3 NaOH + 3H₂O \(\longrightarrow \) PH₃ + 3 NaH₂PO₂
Hence, B is phosphine PH₃ which has rotten fish odour.
(iii) Phosphorus reacts with dry chlorine to form phosphorus trichloride which has a pungent odour.
P₄ + 6 Cl₂\(\longrightarrow \) 4PCl₃
Thus, A = Phosphorus
B = Phosphine
C = Phosphorus trichloride

(i) The colour develops because of the presence of electrons in the 8 anionic sites.
(ii) These electron absorb energy from the visible region of radiation and get excited. 
(iii) For example when crystals of NaCl are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the crystal and the deposited Na atoms.
(iv) During this process, the Na atoms on the surface lose electrons to form Na⁺ ions 
(v) These electrons get excited by absorbing energy from the visible light and impart yellow colour to the crystals.

DIAMOND	GRAPHITE
C is sp3 hybridised.	C is sp2 hybridised.
Three dimensional, tetrahedral structure.	Two dimensional, sheet like structure.
Crystalline, transparent with extra brilliance.	Crystalline, opaque and shiny substance.
It is hard with high density and high melting point.	It is soft with low density and high melting point.
Bad conductor of and electricity.	Good conductor of heat and electricity.

(i) Alumina is separated from silica in a bauxite ore through Baeyer's process, in which bauxite ore is concentrated by the method of leaching or chemical separation.
(ii) Chemical method is employed in case where the ore is to be in a very pure form, e.g., aluminium extraction. Bauxite (Al₂O₃), an ore of aluminium, contains SiO₂ and Fe₂O₃ as impurities. When bauxite ore is treated with NaOH, the Al₂O₃ goes into solution as sodium meta aluminate leaving behind the undissolved impurities [Fe₂O₃, SiO₂, Fe(OH)_3' etc.], which are then filtered off.
\({ Al }_{ 2 }{ O }_{ 3 }+{ 2NaOH }\longrightarrow \underset { Sod.meta.aliminate\\ (In \ solution \ form) }{ { 2NaAlO }_{ 2 }+{ H }_{ 2 }O } \)
(iii) The filtrate (containing sodium meta aluminate) on dilution, and stirring gives a precipitate of aluminium hydroxide, which is filtered, and ignited to get pure alumina.
\({ NaAlO }_{ 2 }+2{ H }_{ 2 }O\longrightarrow \underset { Ppt }{ { Al(OH) }_{ 3 } } +NaOH\)
\(2Al\left( OH \right) _{ 3 }\overset { \Delta }{ \longrightarrow } \underset { Pure }{ { Al }_{ 2 }{ O }_{ 3 } } +3{ H }_{ 2 }O\)

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

a) i) Ellingham diagram for the formation of Al₂O₃ and MgO intersects around 1500^oC. Above this temp Mg lies above the Aluminium line. Hence only above 1500^oC Aluminium might be excepted to reduce magnesia.
ii) Ellingham diagram for the formation of MgO lies below the formation of Al₂O₃. Hence MgO is more stable than Al₂O₃. Hence Magnesium could reduce Alumina.
1. Below 983K, formation of CO line lies below many of the metal oxide formation in Ellingham diagram, hence CO is more effective reducing agent than Carbon.
2. But above this temperature Carbon lies below other metal oxides.
b) Around 1200K Carbon lies below the formation of Fe₂O₃. Hence it is possible to reduce Fe₂O₃ by coke at 1200K.

When alkylcyanides are reduced using SnCl₂ HCl, imines are formed, which on hydrolysis gives corresponding aldehyde.
\({ CH }_{ 3 }-C\equiv N\overset { { SnCl }_{ 2 }/HCl }{ \longrightarrow } { CH }_{ 3 }-CH={ NH }\overset { { H }_{ 3 }{ O }^{ + } }{ \longrightarrow } { CH }_{ 3 }-CHO\)

1-Butanol, being primary alcohol gets oxidised by dilute KMnO₄. The brown precipitate is due to the formation of manganese dioxide.

Oxidation of alcohols with acidified K₂Cr₂O₂ gives different oxidation products.

(i) Fuller's earth and silica gel are used for refining process.
(ii) Sugar prepared from molasses is decolourised to remove coloured impurities by adding animal charcoal which acts as decolourising material.
(iii) Catalysis is an important branch of surface chemistry which is based on the phenomenon of adsorption of materials on the catalyst surface.

During recharge process, the role of anode and cathode is reversed and H₂SO₄ is regenerated.
Oxidation occurs at the cathode ( now act as anode)
\(\stackrel {+2}{{ PbS }O}_{ 4(s) }+2{ H }_{ 2 }O(l)\rightarrow \stackrel{+4}{ Pb }O_{ 2(S) }+{ SO }_{ 4(aq) }^{ 2- }\)
Reduction occurs at the anode (now act as cathode)
\({ PbS }O_{ 4(s) }^{ }+2{ e }^{ - }\rightarrow { Pb }_{ (s) }+{ SO }_{ 4(aq) }^{ 2- }\)
Overall reaction:
\({ 2PbS }O_{ 4(s) }^{ }+2{ H }_{ 2 }O(l)\rightarrow { Pb }_{ (s) }+{ Pb }O_{ 2(S) }+{ 4H }_{ (aq) }^{ + }+{ SO }_{ 4(aq) }^{ 2- }\)
Thus, the overall cell reaction is exactly the reverse of the redox reaction which takes place while discharging

According to Arrhenius, an acid is a substance that dissociates to give hydrogen ions in water. For example, HCl, H₂SO₄ etc., are acids. Their dissociation in aqueous solution is expressed as
\({ HCl }_{ (g) }\overset { { H }_{ 2 }O }{ \rightleftharpoons } { H }_{ (aq) }^{ + }+{ Cl }_{ (aq) }^{ - }\)
Similarly a base is a substance that dissociates to give hydroxyl ions in water. For example, substances like NaOH, Ca(OH)² etc., are bases.
\({ Ca(OH) }_{ 2 }\overset { { H }_{ 2 }O }{ \rightleftharpoons } { Ca }_{ (aq) }^{ 2+ }+{ 2OH^- }_{ (aq) }\)

(i) In lyophillic colloids or sols definite attractive force or affinity exists between dispersion medium and dispersed phase. Examples: sols of protein and starch. They are more stable and will not get precipitated easily. 
(ii) In a lyophobic colloids, no attractive force exists between the dispersed phase and dispersion medium. They are less stable and precipitated readily, but cannot be produced again by just adding the dispersion medium. 
Examples: sols of gold, silver, platinum and copper.

HCN is a weak acid
\({\left[\mathrm{H}^{+}\right] } =\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}  \)
\(=\sqrt{10^{-9} \times 0.4}  \)
\(=\sqrt{4 \times 10^{-10}}   \)
\(=2 \times 10^{-5} \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}^{+}\right]  \)
\(=-\log _{10}\left(2 \times 10^{-5}\right)  \)
\(=-\left[\log _{10} 2-5 \log 10\right]  \)
\(=5-\log 2\)
= 5-0.3010 = 4.6990

Let [A₀] = 100;
When t = t_99.9%; [A] = (100-99.9) = 0.1
\(k=\frac { 2.303 }{ t } \log\left( \frac { [{ A }_{ 0 }] }{ [A] } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log\left( \frac { 100 }{ 0.1 } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log1000\)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } (3)\)
\({ t }_{ 99.9\% }=\frac { 6.909 }{ K } \)
\({ t }_{ 99.9\% }=10\times \frac { 0.69 }{ K } \)
\({ t }_{ 99.9\% }={ 10 } t_{ 1/2 }\)

\(2{ NH }_{ { 3 }_{ (g) } }\overset { pt }{ \longrightarrow } { N }_{ { 2 }_{ (g) } }+3{ H }_{ { 2 }_{ (g) } }\)
Rate = \(-\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { N }_{ 2 } \right] }{ dt } \)
However, it is given that the reaction is of zero order. Therefore
\(-\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { N }_{ 2 } \right] }{ dt } k\)
= 2.5 x 10^-4 mol L^-1 S^-1
The rate of production of N₂ is
\(\frac { d\left[ { N }_{ 2 } \right] }{ dt } \)=  2.5 x 10^-4 mol L^-1 S^-1
The rate of production of H₂ is
\(\frac { d\left[ { N }_{ 2 } \right] }{ dt } \)= 3 x 2.5 x 10^-4 mol L^-1 s^-1
= 7.5 x 10^-4 mol L^-1 s^-1.

(i) Atomic size : The atomic size in 3d transition series decrease from Sc to Mn and then Fe, CO, Ni have almost same atomic size while copper has bigger size. It is because number of unpaired electrons in d- orbitals increase in the beginning till Mn. Therefore effective nuclear charge increases hence atomic size decreases then pairing of electrons in d- orbitals takes place, so the atomic size remains the same and finally it increases due to repulsion between paired electrons in d- orbitals which leads to decrease in effective nuclear charge.
(ii) They have high enthalpy of atomisation due to strong metallic bonds and additional covalent bonding due to the presence of unpaired electrons in d- orbitals.

Structure of ammonia
(i) Ammonia molecule is pyramidal in shape N-H bond distance is 1.016Å and H-H bond, distance is 1.645Å with a bond angle 107°.

(ii) The structure of ammonia may be regarded as a tetrahedral with one lone pair of electrons in one tetrahedral position hence it has a pyramidal shape.

Simple Cubic Lattice
100,101,111

\({ d }_{ hkl }=\frac { a }{ \sqrt { { h }^{ 2 }+{ k }^{ 2 }+{ l }^{ 2 } } } \)
\({ d }_{ 100 }=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+0^{ 2 }+0^{ 2 } } } =1\)
\({ d }_{ 101 }=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+0^{ 2 }+0^{ 2 } } } =\frac { 1 }{ \sqrt { 2 } } \)
\({ d }_{ 111 }=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+1^{ 2 }+1^{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \)
\(\\ { d }_{ 100 }:{ d }_{ 101 }:{ d }_{ 111 }=1:\frac { 1 }{ \sqrt { 2 } } :\frac { 1 }{ \sqrt { 3 } } (or)\)
=1:0.707:0.577

(i) Boron trifluoride is obtained by the treatment of calcium fluoride with boron trioxide in presence of conc. sulphuric acid.
B₂O₃ + 3CaF₂ + 3H₂SO₄ \(\overset { \triangle }{ \longrightarrow } \) 2BH₃ + 3CaSO₄ + 3H₂O
(ii) It can also be obtained by treating boron trioxide with carbon and fluorine.
B₂O₃ + 3C + 3F₂ ⟶ 2BF₃ + 3CO
(iii) In the laboratory pure BF₃ is prepared by the thermal decomposition of benzene diazonium tetrafluoro borate.
PhN₂BF₄ \(\overset { \triangle }{ \longrightarrow } \) BF₃ + PhF + N₂

(i) Copper is used for making coins and ornaments along with gold and other metals.
(ii) Copper and its alloys are used for making wires, water pipes and other electrical parts.

(i) Coordination compounds which possess chirality exhibit optical isomerism similar to organic compounds.
(ii) The pair of two optically active isomers which are mirror images of each other are called enantiomers.
(iii) Their solutions rotate the plane of the plane polarised light either clockwise or anticlockwise and the corresponding isomers are called 'd' (dextrorotatory) and 'I'(levorotatory) forms respectively.
(iv) The octahedral complexes of type I [M(xx)₃]^n±, [M(xx)₂AB]^n± and [M(xx)₂B₂]^n± exhibit optical isomerism.
(v) Examples:
The optical isomers of [Co(en)₃]³⁺

(i) The half life of a reaction is defined as the time required for the reactant concentration to reach one half its initial value.
(ii)   The rate constant for a first order reaction is given by
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(at\quad t={ t }_{ \frac { 1 }{ 2 } };\left[ A \right] =\frac { \left[ { A }_{ 0 } \right] }{ 2 } \)
\(k=\frac { 2.303 }{ t_{ 1/2 } } log\frac { \left[ { A }_{ 0 } \right] }{ \frac { \left[ { A }_{ 0 } \right] }{ 2 } } \)
\(k=\frac { 2.303 }{ { t }_{ 1/2 } } log2\)
\(k=\frac { 2.303\times 0.3010 }{ { t }_{ 1/2 } } =\frac { 0.6932 }{ { t }_{ 1/2 } } \)
\({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
This equation has no concentration term So, the half life of a first order reaction is independent of initial concentration.

(i) The elements in which the extra electron enters (n-2) f orbitals are called f-block elements. These elements are called as inner transition elements because they form a transition. Series within the transition elements.
(ii) The f-block elements are also called as rare earth elements. They are divided into lanthanoid series (4f block elements) and actinoid series (5f block elements).

1. [Ni(CN)₄]^2- is a low spin square planar complex as it contains strong field CN^- ligand in it.
2. Oxidation state of Ni in complex is +2 Electronic Configuration of Ni²⁺ is 3d⁸ 4s⁰ 

Crystal field splitting in square planar complex. 
1. No unpaired electrons, so the complex is diamagnetic.
2. [NiCl₄]^2- is a high spin tetrahedral complex as it contains weak field Cl^- ligand in it.
3. Oxidation state of Ni in complex in +2
4. Electronic Configuration of Ni²⁺ is 3d⁸ 4s⁶.

It has two unpaired electrons, so the complex is paramagnetic.

\(\text {(a) }\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}:{ }_{24} \mathrm{Cr} \Rightarrow{ }_{18}[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{4} \)
\({ }_{21} \mathrm{Cr}^{3+} \Rightarrow{ }_{18}[\mathrm{Ar}] 3 \mathrm{~d}^{3}\)

(i) d²sp³ hybridisation (octahedral)
(ii) It has three unpaired electrons (n = 3)
(iii) So it is paramagnetic
(iv) Magnetic moment \(\left(\mu_{\mathrm{s}}\right)=\sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{BM}\)
\(=\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}\)
\((b) \ \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-:}{ }_{28} \mathrm{Ni} \Rightarrow[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{8} ;{ }_{26} \mathrm{Ni}^{2+} \Rightarrow{ }_{18}[\mathrm{Ar}] 3 \mathrm{~d}^{8}\)

(i) dsp² hybridisation
(ii) Geometry - square planar
(iii) No unpaired electrons- It is Diamagnetic
(iv) Magnetic moment (μ_s) = 0.

(i) Helium is used to provide inert atmosphere in electric-arc welding of metals.
(ii) Helium has lowest boiling point hence used in cryogenics.
(iii) It is much less denser than air and hence used for filling air balloons.