(i) Cell 'B' will act as electrolytic cell as it has lower emf.
\(\therefore\) The electrode reactions will be
Zn²⁺ + Ze-  \(\longrightarrow\) Zn(s) at cathode.
Cu \(\longrightarrow\) Cu²⁺ + Ze^- at anode.
(ii) Cell 'B' will act as galvanic cell as it has higher emf and will supply electrons into cell 'A'.
The electrode reaction will be
At anode: Zn \(\longrightarrow\) Zn2⁺ + 2e^-
At cathode: Cu2⁺ 2e^- \(\longrightarrow\) Cu(s)

Those collisions which lead to the formation of product molecules are called effective collisions.
Rate of reaction = f x Z
where 'Z' = collision frequency, f = fraction of collisions (effective).

First order reaction

In some complexes, the two unpaired 3d electrons pair up and the hybridization is dsp² . Thus, no unpaired electron is left. In some other complexes, one unpaired 3d electron is excited to 4p. Hybridization is again dSp²  but there are two unpaired electrons present.

potassium hexanitratocobaltate.

\(A_m={1000\times k\over M}\)
138.9 S cm² mol^-1=\(1000\times k\over 1.5\)
\(k={138.9\times 1.5\over 1000}={208.05\over 1000}={2.0805}\times 10^{-1}\ S\ cm^{-1}\)

(i) For the reaction \(A \rightarrow B\),
Rate of reaction \((r)=k[A]^2\)
If the concentration of reactant increased to three times.
Rate of reaction \(\left(r^{\prime}\right)=k[3 A]^2\)
Thus, on dividing equations (1) and (2)
\( \frac{r}{r^{\prime}}=\frac{k[A]^2}{k[3 A]^2} =\frac{1}{9}\)
Therefore, rate of formation of B increases to nine times.
(ii) Pseudo first order reaction: The reaction which is bimolecular but order is one is called pseudo first order reaction. This happens when one of the reactants is in large excess. E.g., acidic hydrolysis of ester (ethyl acetate).

Let the rate law in terms of rate of formation of $D$ be
\(\frac{d[D]}{d t}=k[A]^4 \quad B^b\)
1. \(6.0 \times 10^{-3}=k(0.1)^a(0.1)^b\) ....(i)
2. \(7.2 \times 10^{-3}=k(0.3)^a(0.2)^b\) ...(ii)
3. \(2.88 \times 10^{-2}=k(0.3)^a(0.4)^b\) ...(iii)
4. \(2.40 \times 10^{-2}=k(0.4)^a(0.1)^b\) ...(iv)
Divide Eq. (iv) by Eq. (i), we get
\(4=(4)^a\)
 Therefore a = 1
Divide Eq. (iii) by Eq. (ii), we get 
\(4 =(2)^b\)
\(2^2  =(2)^b\)
b=2
Order with respect to A=1
Order with respect to B=2
\(\text { Rate law }=\frac{d[D]}{d t}=k[A][B]^2\)
On putting the values of ' A ' and ' B ' in any equation, say (i)
\( 6.0 \times 10^{-3} \mathrm{M}_{\min ^{-1}}=k(0.1 \mathrm{M})(0.1 \mathrm{M})^2\)
\( \therefore \quad k=6 \mathrm{M}^{-2} \mathrm{~min}^{-1}\)

\((i) \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \\ x+(0) \times 5+(-1) \times 1+(-1) \times 2=0 \\ x+0-3=0 \\ x=+3 \)
So. the name of the complex is pentaarnminechloridocobalt (Ill) chloride.

Let the oxidation state of Pd is x.
(+1)2 +x+(-1)4 = 0
2+x-4 = 0
⇒ x = +2
So, the name of the complex is potassiumtetrchloridopalladate (Il).

(i) [Co(en)₃] Cl₃
IUPAC name:
Tris - (ethane - l,2- diamine) cobalt (III) chloride.

(ii) Pt(NH₃)₂CI₂
(ill) [Fe(NH₃)₄CI₂]Cl
IUPACname
Tetraamminedichloridoiron (II) chloride
Isomers Geometrical isomers (cis and trans)

6.747 x 10¹⁰

200 ohm^-1 cm² eq^-1

(c) : 75% of reaction is completed in two half-lived  i.e., 2 \(\times\) t ₁₂ = 32 min or t _1/2 = 16 min

(i) (a) Metal carbonyls: Metal carbonyls are the compounds in which carbon monoxide (CO) acts as the ligand. The metal-carbon bond in metal carbonyls possesses both \(\sigma\) and \(\pi\) characters, e.g. N(CO)₄[nickeltetracarbonyl]. Its IUPAC name is tetra carbonyl nickel(0).
(b) Chelate: An inorganic metal complex in which there is a close ring of atoms caused by attachment of a ligand to a metal atom at two points. An example is the complex ion formed between ethylene diamine and cupric ion,[Cu(NH₂CH₂CH₂NH₂)₂]²⁺.

(ii) (a) Add BaCl₂ solution. [Co(NH₃)]₅Cl]SO₄ will give white ppt. of BaSO₄.  
 \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{SO}_{4}+\mathrm{BaCl}_{2}(a q) \)\( \longrightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}+\mathrm{BaSO}_{4} \downarrow \)
                                                                    \(\text { (White ppt.) }\)
Add AgNO₃(aq) solution [Co(NH₃)₅SO₄] Cl will give white ppt. of AgCl.
\( {\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Cl}+\mathrm{AgNO}_{3}(a q)} \) \(\longrightarrow \mathrm{AgCl}+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{NO}_{3} \)
                                                                                          \(\text { (White ppt) }\)
(c) Fe³⁺ (4s⁰3d⁵) In strong field \( \Delta_{0}>P t_{2 g}^{5} e g^{0} \)
In weak field \(\Delta_{0}

(a) (i) Limiting molar conductivity \(\left(\Lambda_{m}^{\circ}\right)\) : It is defined as the maximum molar conductance of an electrolyte when solution is infinitely dilute, i.e. concentration approaches zero when electrolyte solution is kept in cell, unit distance apart having large area of cross-section to hold enough electrolyte.
(ii) Fuel cell is a cell in which chemical energy of a fuel is converted into electrical energy.
(b) R = 100 ohms, M = 0.1 mol L^-1 of KCI
\(\kappa=\frac{1}{R} \times \frac{l}{A}\) 
\(\Rightarrow 1.29 \times 10^{-2} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}=\frac{1}{100 \mathrm{ohm}} \times \frac{l}{A}\) 
\(\Rightarrow \ \frac{l}{A}=1.29 \mathrm{~cm}^{-1}\) 
Again,  \(\kappa=\frac{1}{R} \times \frac{l}{A}=\frac{1}{520} \times 1.29\) 
= \(\frac{1.29}{5.20} \times 10^{-2}\) 
= 2.48 x 10^{- 3} ohm^-1 cm^-1
Now, M = 0.02 mol L^-1
\(\Lambda_{m}=\frac{1000 \kappa}{\mathrm{M}}=\frac{1000 \times 2.48 \times 10^{-3}}{0.02}\) 
= 124 S cm² mol^-1

Degree of dissociation is the ratio of molar conductivity at a specific concentration to the molar conductivity at infinite solution.
Degree of dissociation = \(\frac { { \Lambda }_{ m }^{ c } }{ { \Lambda }_{ m }^{ 0 } } \)
(b) \({ E }_{ cell }^{ ° }={ E }^{ ° }{ Ag }^{ + }/Ag-{ E }^{ ° }{ Ni }^{ 2+ }/Ni\)
= 0.80V - 0.25V
= 0.55V
\({ log }_{ { K }_{ c } }=\left( \frac { n{ E }^{ ° }cell }{ 0.059 } \right) \)
\(=\frac { 2\times 0.55 }{ 0.059 } \)
\({ log }_{ { K }_{ c } }=18.644\)
\({ K }_{ c }=Antilog\quad 18.644{ K }_{ c }=4\times { 10 }^{ 18 }\)
\({ \Delta }_{ G }^{ ° }=nFE°cell\)
\(=-2\times 96500C{ mol }^{ -1 }\times 0.55V\)
\(=-106,150J{ mol }^{ -1 }\)
Max.work = +106150 J mol^-1

Temperature has a great effect on reaction rates or reaction rate constants. In general, an increase in temperature increases the rates of almost all reactions. For homogeneous chemical reactions, the rate of the reaction or rate constant (k) becomes almost double for every 10° rises in temperature.According to collision theory, the rate of reaction depends upon,
(i) the collision frequency of the reacting molecules.
(ii) the fraction of effective collisions.
The frequency of collisions between reacting molecules depends upon the average velocity of the molecules. With the increase in temperature, the average velocity of the molecules increases which results into increase in collision frequency. However, an increase of 10° rise of temperature increases the number of collisions by only 1.016 times. Further, we know that all the collisions are not effective. For molecules to undergo effective collisions, they must possess a certain minimum amount of energy called threshold energy. The molecules which possess energy equal to or greater than threshold energy will result in the formation of products. However, the fraction of such molecules capable of effective collisions is very small. As the temperature has increased the fraction of molecules possessing energies greater than threshold energy increases and so the rate of the reaction increases.
Let us consider the effect of increase of temperature on the number of effective collisions (having energies greater than threshold energy). The energy. distribution, of molecules at two temperatures \({ T }_{ 1 }\) and \({ T }_{ 2 }\) (where \({ T }_{ 2 }={ T }_{ 1 }+10°\)) is shown in figure. It is clear from the figure that the curve at higher temperature gets shifted towards the right indicating that at higher temperature, the molecules have higher energies. Further, the curve at higher temperature is flatter than that at lower temperature which also indicates that the number of molecules with higher energy content have increased.
The minimum energy required for the effective collisions also shown in the diagram and the number of molecules possessing energies equal to or greater than E, is proportional to area abed at temperature \({ T }_{ 1 }\) and area above at temperature \({ T }_{ 2 }\). In the figure, the area abef is roughly twice as large as abcd. Since the rate of reaction depends upon the number of molecules which possess energies larger than activation energy (for effective collisions), it may be interpreted that the fraction of molecules possessing activation energy has increased approximately two times and thereby, increases the rate of reaction by two times for a rise of 10 degrees.
Thus, we may conclude that increase in the rate of reaction with the rise in temperature is mainly due to the increase in number of effective collisions.

\({ \triangle }_{ t }=\left( \frac { 4 }{ 9 } \right) { \triangle }_{ 0 }\).  Thus,  \({ \triangle }_{ t }\) is smaller than \({ \triangle }_{ 0 }\).  Hence, less energy (higher wavelength) is absorbed by tetrahedral complexes than by octahedral complexes of the same metal and ligands.  Therefore, the observed colour are different.

As.ΔH = - 33 kJ mol^-1, the reaction is exothermic. The activation energy diagram will be as shown in fig.
ΔH = E_f - E_b= - 33 kJ
k_f = Ae^-E_f/RT
kb = Ae^-E_b/RT
\(K_c={k_f\over k_b}=e^{(E_b-E_f)/RT}\)
In \(K_c={E_b-E_f\over RT}\ or\ log\ K_c={E_b-E_f\over 2.303RT}={30000\ J\ mol^{-1}\over 2.303(8.314JK^{-1}mol^{-1})300K}=5.2227\)
K_c = Antilog 5·2227 = 1·67 x 10⁵
Substituting \(E_b={31\over 20}E_f,\)We get
\(E_f-{31\over 20}E_f=-33\ or\ {-{11\over 20}}E_f=-33\ or\ E_f={33\times20\over 11}=60kJ\ mol^{-1}\)
E_b = E_f + 33 = 60 + 33 = 93 kJ mol^-1