Solution

(a) It selectively prevents one of the sulphide ore from coming to the froth.

(b) Helps in converting Zr into its volatile compound ZrI₄.

(c) Provides flux to remove impurities.

Solution

(a) Physiosorption has weak van der Waal forces while Chemisorption has strong chemical bonds.

(b) Physiosorption is favorable at low temperature while Chemisorption Increases till a certain temperature and then decreases afterwards.

(c) In Physiosorption enthalpy of adsorption is low while in Chemisorption enthalpy of adsorption is high.

Solution

Solution

(a) (i) Cu⁺¹(3d¹⁰) compounds are white because of absence of unpaired electrons while Cu⁺²(3d⁹) compounds are coloured due to unpaired e- which shows d-d transition.

(ii) Chromates (CrO₄²⁻) in the acidic solution are converted into dichromates and thus the colour becomes (Cr₂O₇²⁻) orange.

(iii) Transition elements are characterized by partially filled (n-1) d subshells. Zn, Cd and Hg are not regarded as transition elements as they have completely filled (n-1)d subshell. They possess completely filled d-orbitals in their ground state as well as in oxidized state.

(b) Co = [Ar]4s²3d⁷
Co⁺² =[Ar] 3d⁷
<p><img alt="" src="" style="height:184px; width:300px" /></p>

or
Lanthanoids
(1) Most of them are not radioactive.
(2) Don’t show a wide range of oxidation state.
(3) Most of their ions are colourless.
Actinoids
(1) All are radioactive.
(2) Show a wide range of oxidation states.
(3) Most of their ions are coloured.
(b) (i) Sc³⁺, because of absence of unpaired electron it is diamagnetic.
Sc= 1s² 2s² 2p⁶ 3s³ 3p⁶ 3d¹ 4s²
Sc³⁺= 1s² 2s² 2p⁶ 3s³ 3p⁶
(ii) Cr, because of presence of greater number of unpaired electrons leading to strong intermetallic bonding than Cu it has high melting and boiling point.
<p><img alt="" src="" style="height:207px; width:300px" /></p>

Solution

(a) t-butyl alcohol will undergo acid catalysed dehydration faster because the tertiary carbonium ion formed with t-butyl alcohol is more stable than primary carborium ion formed with n-butanol.

(b) <p><img alt="" src="" style="height:641px; width:500px" /></p>
or

<p><img alt="" src="" style="height:205px; width:350px" /></p>

<p><img alt="" src="" style="height:301px; width:350px" /></p>

Solution

(a) Zn (s) + Cu²⁺ (aq) —>Zn²⁺ (aq) + Cu (s).
Given : E° for Zn²⁺/Zn= E°ᵃⁿᵒᵈᵉ = – 0·76 V and
E° for Cu²⁺/Cu = E°ᶜᵃᵗʰᵒᵈᵉ = +0·34 V
(a) E°cell =E°c– E°a
= 0.34 – (-0.76)
= 1.10V
𝚫G°= -nFE°
= -2 × 1.10 × 96500
= -212300 J/mol or -212.3 kJ/mol

Advantage : -(1) A fuel cell works with an efficiency of 70 %.
(2) These are no objectionable by-products and therefore, its a pollution free source of energy.
or

(a) (i) Silver wire at 30°C because as temperature decreases, resistance decreases so conduction increases.
(ii) 0.1 M CH₃COOH, because on dilution degree of ionization increases hence conduction increases.
(iii) KCl solution at 50°C, because at high temperature mobility of ions increases and hence conductance increases.

● Electrochemical cell
(1) Anode -ve
Cathode +ve
(2) Convert chemical Energy to electrical energy.
● Electrolytic cell
(1) Anode +ve
Cathode -ve
(2) Convert electrical Energy to chemical energy.