Scattering of light

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

In 140days ⇒ initial concentration reduced to (1/2) g
In 280 days ⇒ initial concentration reduced to (1/4) g
In 420 days ⇒ initial concentration reduced to (1/8) g
In 560 days ⇒ initial concentration reduced to (1/16) g

Ti⁴⁺ (d⁰ ⇒  0BM)
Co²⁺ (d⁷ spain free ⇒  t_2g⁵, e²_g; n = 3; μ = 3.9BM)
Cu²⁺  (d⁹ Low spain ⇒  t_2g⁶, e³_g; n = 1; μ = 1.732BM)
Ni²⁺  (d⁸ Low spain ⇒  t_2g⁶, e²_g; n = 2; μ = 2.44 BM)

Mn²⁺ ⇒ 3d⁵ contains 5 unpaired electrons
n = 5,
\( \sqrt{n(n+ 2)} \) BM
\(= \sqrt{5(5+ 2)} = \sqrt{35} = 5.92 BM\)

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.

The feasibility of a redox reaction can be predicted with the help of the electrochemical series.
(i) The net emf of the cell reaction Ecellcan be calculated from the expression,
E^o_cell= E^o_cathode - E^o_anode
(ii) In general, if
E^o_cell = +ve, the reaction is feasible.
E^o_cell = -ve, the reaction is not feasible.

(a) Alcohols are soluble in water because they form intermolecular hydrogen bonding with water. Lower members are completely miscible with water and the higher members are not. This is because of the increase in size of hydrophobic (water repelling) alkyl group in the alcohol.
(b) Strong basic substances like organo metallic compounds (RMgX-Grignard reagent) are decomposed by alcohol.
R - OH + CH₃MgBr ⟶ R - O - Mg - Br + CH₄
Hence, alcohols cannot be used as a solvent for Grignard reagent.

Acetaldehyde - CH₃-CHO

 

(i) When potassium dichromate is heated with any chloride salt in the presence of Conc. H₂SO₄, orange red vapours of chromyl chloride (CrO₂CI₂) is evolved.
(ii) This reaction is used to confirm the presence of chloride ion in inorganic qualitative analysis

(iii) The chromyl chloride vapours are dissolved in sodium hydroxide solution and then acidified with acetic acid and treated with lead acetate. A yellow precipitate of lead chromate is obtained

(i) The given defect is Stoichiometric since the number of cations and anions are missing from lattice sites unequal.
(ii) The defect is Schottky defect.
These defects are vacancy defects shown by ionic solids in which the anion and cation are nearly of the same size.
Eg: NaCI and KCI shows Schottky defects

(i) The chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product.
(ii) The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.
Example:
Let us compare the following two reactions that you carried out in volumetric analysis.
1) Redox reaction between ferrous Ammonium Sulphate (FAS) and KMnO₄.
2) Redox reaction between oxalic acid and KMnO₄.
(i) The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe²⁺. In fact heating is required for the reaction between KMnO₄ and Oxalate ion and is carried out at around 60^oC.
(ii) The physical state of the reactant also plays an important role to influence the rate of reactions.
(iii) Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants.
Ex : Na_(s) + I₂(vap)   [Faster]
Na_(s) + I₂(s)   [Slower]
KI_(aq) + Pb(NO3)_2(aq) → PbI₂ (yellow) [Faster]
KI_(s) + Pb(NO3)_2(s) → PbI₂ (yellow)  [Slower]

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:

Kohlrausch deduced the following empirical relationship between the molar conductance (\({ \Lambda }_{ m }\)) and the concentration of the electrolyte (C).
\(​​{ \Lambda }_{ m }={ \Lambda }_{ m }^{ o }-k\sqrt { C } \)

(i) 
                                             n-Nitro benzoate (Major Product)

(ii) 

\(\mathrm{Hg}_{2} \mathrm{Cl}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Hg}_{2}^{2+} \text { (aq) }+2 \mathrm{Cl^-}_{(\mathrm{aq})}\\ s \quad \quad \quad \quad \quad \quad s \quad \quad \quad \quad \quad 2s\)
\(\mathrm{K}_{\mathrm{sp}} =\left[\mathrm{Hg}_{2}^{2+}\right]{\left[\mathrm{Cl}^{-}\right]^{2}}  \)
\(=(\mathrm{s})(2 \mathrm{~s})^{2} \)
\(\mathrm{~K}_{\mathrm{sp}} =4 \mathrm{~s}^{3}\)

N₂O₅ \(\overset { { k }_{ 1 } }{ \longrightarrow } \) 2NO₂ + \(\frac { 1 }{ 2 } \) O₂
(i) At time t = 0, the volume of oxygen liberated is zero.
(ii) Let V, and V_∞ be the measured volumes of oxygen liberated after the reactant has reacted in 't' time and at completion (t = ∞).
(iii) Initial concentration of N₂O₅ is proportional to total volume of oxygen liberated (i.e.,) (V_∞).
(iv) (V_∞ - V_t) is proportional to undecomposed. N₂O₅ at time 't'.
(v) ∴ k₁ = \(\frac { 2.303 }{ t } \log { \frac { { V }_{ \infty } }{ \left( { V }_{ \infty }-{ V }_{ t } \right) } } { sec }^{ -1 }\)

(i) l₂ < F2 < Br₂ < Cl₂·
(ii) HF < Hq < HBr < HI.

(i) Nitric acid is used as a oxidising agent and in the preparation of aqua regia.
(ii) Salts of nitric acid are used in photography (AgNO₃) and gun powder for fire arms (NaNO₃)·

Distillation, liquation, electrolytic refining, zone refining, vapour phase method, van -Arkel method

Magnetic moment (μ) = \(\sqrt { n(n+2) } \) BM
Fe (z = 26) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s⁰

∴μ =  \(\sqrt { n(n+2) } \) = \(\sqrt { 4(4+2) } =\sqrt { 24 } \) 
= 4. 89 BM  

(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).

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.

(i) An acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton form other substance.
(ii) In other words, an acid is a proton donor and a base is a proton acceptor.
(iii) When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCI behaves as an acid and H₂O is base. The proton transfer from the acid to base can be represented as
HCI + H₂O ⇌ H₃O⁺ + Cl^-
(iv) When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia (NH₃) acts as a base and H₂O is acid. The reaction is represented as
H₂O + NH₃ ⇌ NH₄⁺ + OH^-
(v) Let us consider the reverse reaction following equilibrium.
\(\underset { proton\ donar\\ \quad \quad \ (acid) }{ HCl } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad (base) }{ { H }_{ 2 }O } \leftrightharpoons \underset { Proton\ donar\\ \quad \quad \quad \quad \ (acid) }{ { H }_{ 2 }{ O }^{ + } } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad \ (base) }{ { Cl }^{ - } } \) 
H₃O⁺ donates a proton to Cl^- to form HCI i.e., the products also behave as acid and base.
(vi) In general, Lowry - Bronsted (acid - base) reaction is represented as 
Acid₁ + Base₂ ⇌ Acid₂ + Base₁
(vii) The species that remains after the donation of a proton is a base (Base₁) and is called the conjugate base of the Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid - base pairs.

(i) The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-1)d orbital and ns orbital as the energy difference between them is very small.
(ii) At the beginning of the series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.
(iii) The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases.
(iv) Hence, the first and last elements show less number of oxidation states and the middle elements with more number of oxidation states.
(v) For example, the first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.
(vi) The relative stability of different oxidation - states of 3d metals is correlated with the extra stability of half filled and fully filled electronic configurations. Example: Mn²⁺(3d⁵) is more stable than Mn⁴⁺(3d³).

(i) Average rate of reaction between the time interval, 30 to 60 seconds
\(=\frac { d\left[ Ester \right] }{ dt } \)
\(=\frac { 0.31-0.17 }{ 60-30 } =\frac { 0.14 }{ 30 } \)
= 4.67 x 10^-3 mol L^-1 s^-1.
(ii) For a pseudo first order reaction,
\(k=\frac { 2.303 }{ t } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
For, t = 303
\({ k }_{ 1 }=\frac { 2.303 }{ t } \log { \frac { 0.55 }{ 0.31 } } \)
For, t = 60 s
\({ k }_{ 2 }=\frac { 2.303 }{ 60 } \log { \frac { 0.55 }{ 0.17 } } \)
For, t = 90 s
\({ k }_{ 3 }=\frac { 2.303 }{ 90 } \log { \frac { 0.55 }{ 0.085 } } \)
= 2.075 x 10^-2 s^-1
The average rate constant,
\(k=\frac { { k }_{ 1 }+{ k }_{ 2 }+{ k }_{ 3 } }{ 3 } \)
\(=\frac { \left( 1.911\times { 10 }^{ -2 } \right) +\left( 1.957\times { 10 }^{ -2 } \right) +\left( 2.075\times { 10 }^{ -2 } \right) }{ 3 } \)
= 1.98 x 10^-2 s^-1.

(I) Since the byproduct of roasting to sulphide ore is SO₂ It turns lime water milky.
Therefore, gas 'B' must be SO₂
(ii) As the gas 'B' is obtained when amorphous solid 'A' burns in air therefore, amorphous solid 'A' must be sulphur S₈
\(\underset { (A) }{ { S }_{ g } } +{ 8O }_{ 2 }\overset { \Delta }{ \longrightarrow } \underset { (B) }{ { 8SO }_{ 2 } } \)
(iii) Gas (B) reduces acidified aqueous KMnO₄ solution and reduces Fe³⁺ to Fe²⁺ salts as shown below:
\(\underset { (yellow) }{ { 2MnO }_{ 4 } } ^{ - }+\underset { (b) }{ { SO }_{ 2 } } +2{ H }_{ 2 }O\longrightarrow { 2Fe }^{ 2+ }+\underset { (Green) }{ { SO }_{ 4 }^{ 2- } } +{ 4H }^{ + }\)
(iv) Thus, solid 'A' is S₈ and gas 'B' is SO₂

The stability of a complex or co-ordination compound refers to the extent up to which it exists in a solution as co-ordination sphere.
(i) Change on the central metal ion: Greater the charge on the central metal ion, greater the stability of complex.
(ii) Nature of the metal ion: Group 3 and 6 and inner transition elements form stable complexes when donor atoms of the ligands are N, O and F. The elements after group 6 of the transition metals form stable complex when the donor atoms of the ligands are the heavier members of N, O and F family.
(iii) Basic nature of the ligands: Greater the basic strength, greater is the stability of the complex.
(iv) Presence of chelate rings: Its presence increases the stability of the complex. it is called chelate effect. It is maximum for the 5 and 6 membered rings.
(v) Effect of multidentate cyclic ligand: If the ligands are IT multidentate and cyclic without any steric effect the stability of the complex get increased.