The cell potential is determined by the difference between the standard electrode potentials of the two half-reactions. The general formula for the cell potential is:
\[ E_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

In this case, the standard electrode potential for the Sn\(^{4+}\)/Sn\(^{2+}\) couple is +0.15 V (cathode), and for the Cr\(^{3+}\)/Cr couple is -0.74 V (anode). Substituting these values:
\[ E_{cell} = 0.15 \, V - (-0.74 \, V) = 0.15 + 0.74 = 1.19 \, V \]

Therefore, the cell potential is +1.19 V, corresponding to option (A).
Quick Tip: The cell potential is calculated by subtracting the anode potential from the cathode potential: \(E_{cell} = E_{cathode} - E_{anode}\).

The spin-only magnetic moment (\(\mu\)) for a transition metal ion can be calculated using the formula:
\[ \mu = \sqrt{n(n+2)} \, B.M. \]

where \(n\) is the number of unpaired electrons. For the Cr\(^{3+}\) ion, which has 3 unpaired electrons (since Cr has atomic number 24 and the electron configuration for Cr\(^{3+}\) is [Ar] 3d\(^3\)), \(n = 3\). Substituting into the formula:
\[ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, B.M. \]

Thus, the spin-only magnetic moment of Cr\(^{3+}\) is approximately 3.47 B.M., corresponding to option (C). Quick Tip: The magnetic moment for an ion can be calculated using the formula \(\mu = \sqrt{n(n+2)}\), where \(n\) represents the number of unpaired electrons.

When solute molecules undergo association, they tend to aggregate, forming larger particles such as dimers or polymers. As a result, the number of free particles in the solution decreases. This leads to a higher observed molar mass, as the aggregates contribute to the overall mass without increasing the number of individual solute particles. Hence, the abnormal molar mass increases in such cases.


Final Answer: (A) increase
Quick Tip: When molecules associate, they form aggregates, which causes an increase in the molar mass due to fewer free particles in solution.

In an SN2 reaction, which stands for bimolecular nucleophilic substitution, the nucleophile attacks the electrophilic carbon from the opposite side of the leaving group. This backside attack leads to the displacement of the leaving group and results in an inversion of the configuration at the carbon center undergoing substitution. The inversion is a characteristic feature of SN2 reactions due to the simultaneous attack and departure of the leaving group in a single step.


Final Answer: (C) inversion of configuration
Quick Tip: In SN2 reactions, the nucleophile attacks the carbon center from the opposite side of the leaving group, causing an inversion of configuration.

The boiling point of a compound is directly related to its molecular size and intermolecular forces, primarily van der Waals forces. Larger molecules with more carbon atoms generally have higher boiling points due to stronger intermolecular forces. The correct order of boiling points is:

(iii) \(<\) (i) \(<\) (ii). The order follows from the increasing size of the alkyl chain and molecular weight. Compound (ii) has the largest chain, followed by (i), and (iii) has the smallest chain.
Quick Tip: Boiling points increase with increasing molecular size and the number of carbon atoms in a compound due to stronger intermolecular van der Waals forces.

To determine the IUPAC name, we first need to assign oxidation states to the central metal atom, platinum.

- The ammine group (NH\(_3\)) is neutral, and the chloride ions (Cl\(^-\)) each have a charge of -1.

- Since the complex has a charge of +2, we can determine that platinum must be in the +4 oxidation state to balance the charges.

- Therefore, the IUPAC name is Diamminedichloridoplatinum (IV).
Quick Tip: When naming coordination complexes, the oxidation state of the metal ion is crucial in determining the IUPAC name.

Propyl magnesium bromide (C\(_3\)H\(_7\)MgBr), a Grignard reagent, reacts with carbon dioxide (CO\(_2\)) to form a carboxylate ion intermediate.

- After acid hydrolysis, this intermediate is converted into propanoic acid (C\(_3\)H\(_7\)COOH).

- The overall reaction is a nucleophilic addition of the Grignard reagent to CO\(_2\), followed by protonation to yield the acid.

- Thus, the acid formed is propanoic acid.
Quick Tip: Grignard reagents react with CO\(_2\) to form carboxyl acids upon hydrolysis, and they are important in organic synthesis for forming carbon-carbon bonds.

Potassium permanganate (KMnO\(_4\)) in acidic solution is a strong oxidizing agent.

- When it reacts with sulphite ions (SO\(_3^{2-}\)), it oxidizes them to sulfate ions (SO\(_4^{2-}\)).

- This is a typical redox reaction where MnO\(_4^-\) is reduced and sulphite is oxidized to sulfate.

- Therefore, the correct product is sulfate (SO\(_4^{2-}\)).
Quick Tip: KMnO\(_4\) is a strong oxidizing agent and can oxidize a variety of ions, including sulphite to sulfate in acidic conditions.

The acidity of phenol (C\(_6\)H\(_5\)OH) is higher than that of alcohols (ROH) and water (H\(_2\)O) because the phenoxide ion (C\(_6\)H\(_5\)O\(^-\)) formed after deprotonation is stabilized by resonance with the aromatic ring.

- Water and alcohols do not benefit from this stabilization. Thus, the order of acid strength is:
C\(_6\)H\(_5\)OH \(>\) ROH \(>\) H\(_2\)O.
Quick Tip: In phenol, the negative charge on the oxygen in the phenoxide ion is stabilized by resonance with the benzene ring, making it more acidic than alcohols and water.

In a concentrated salt solution, the water inside the mango moves out towards the higher concentration of solutes (salt) outside the mango through the process of reverse osmosis.

- This results in the mango shrinking and shriveling due to the loss of water.

- Osmosis typically moves water from low to high solute concentration, but in this case, because the solution is concentrated, the water moves out from the mango, causing it to shrivel.
Quick Tip: Reverse osmosis occurs when water moves from an area of lower solute concentration to an area of higher solute concentration through a semipermeable membrane.

To reduce propanamide to propanamine, LiAlH\(_4\) (Lithium aluminium hydride) is the best reagent. LiAlH\(_4\) is a strong reducing agent that can reduce amides to amines. Other reagents like Br\(_2\) in NaOH or iodine with red phosphorus do not reduce amides to amines. Thus, LiAlH\(_4\) is the correct choice.
Quick Tip: LiAlH\(_4\) is a strong reducing agent used to reduce amides to amines, aldehydes to alcohols, and other reductions in organic synthesis.

- Glucose is an aldohexose, meaning it is a six-carbon sugar with an aldehyde group.
- On heating with HI (hydriodic acid), glucose undergoes reduction to form n-hexane.
- Glucose exists in both the pyranose and furanose forms in its cyclic structures.
- Glucose gives Schiff’s test because it has an aldehyde group. Schiff’s reagent turns pink in the presence of aldehydes. Thus, the statement "It does not give Schiff’s test" is false.
Quick Tip: Glucose can exist in both the pyranose and furanose forms depending on the reaction conditions and the position of the anomeric hydroxyl group.

- Assertion (A) is true because \(\alpha\)-amino acids (except glycine) are optically active due to the presence of a chiral center at the \(\alpha\)-carbon. Glycine is the only amino acid that does not have a chiral center.

- Reason (R) is also true because most naturally occurring amino acids have the L-configuration, which is commonly found in proteins.

- Reason (R) correctly explains Assertion (A) because the L-configuration is directly related to the optical activity of these amino acids. Thus, both the assertion and reason are correct, and the reason is the correct explanation.
Quick Tip: Glycine is the only naturally occurring amino acid that is not optically active because it does not have a chiral center.

- Assertion (A) is true because all naturally occurring amino acids, except glycine, are optically active due to the chiral center at the \(\alpha\)-carbon. Glycine is the exception.
- Reason (R) is true because glycine lacks a chiral center, making it non-optically active.
- Therefore, Reason (R) correctly explains why glycine is the only amino acid that is not optically active. Thus, both Assertion (A) and Reason (R) are true, and the reason is the correct explanation.
Quick Tip: The presence of a chiral center in a molecule is what makes it optically active. Glycine does not have a chiral center, so it is not optically active.

- Assertion (A) is correct because, typically, the boiling points of alkyl halides increase with the size of the halogen atom. Iodine being the largest atom has the highest boiling point, and fluorine has the lowest due to its smaller size and weak van der Waals forces.

- Reason (R) is also true, as alkyl halides' boiling points increase due to stronger intermolecular forces (e.g., dipole-dipole interactions, London dispersion forces). However, the reasoning does not explain why the order of boiling points follows the pattern RI \(>\) RBr \(>\) RCl \(>\) RF. Thus, Reason (R) does not correctly explain Assertion (A).
Quick Tip: When studying boiling points of halides, remember that the size of the halogen atom plays a crucial role in determining the strength of van der Waals forces and the boiling point.

- Assertion (A) is correct because both [Cr(H\(_2\)O)\(_6\)]Cl\(_2\) and [Fe(H\(_2\)O)\(_6\)]Cl\(_2\) are homoleptic complexes, meaning that all the ligands attached to the metal ion are the same (in this case, water molecules).

- Reason (R) is also correct because, in homoleptic complexes, the same type of ligand (like H\(_2\)O) is attached to the central metal atom. This directly explains why the complexes are homoleptic.
Quick Tip: Homoleptic complexes have only one type of ligand attached to the central metal atom. This can be contrasted with heteroleptic complexes, which have different types of ligands.

- Benzaldehyde is less reactive in nucleophilic addition reactions compared to propanal.

- This is because the phenyl group in benzaldehyde is an electron-withdrawing group through resonance. This decreases the electron density on the carbonyl carbon, making it less susceptible to nucleophilic attack.

- In contrast, propanal does not have an electron-withdrawing group attached to the carbonyl carbon, making it more reactive in nucleophilic addition reactions.
Quick Tip: Electron-withdrawing groups like the phenyl group in benzaldehyde reduce the electrophilicity of the carbonyl carbon, making the compound less reactive in nucleophilic addition reactions.

(a) The oxidation half-reaction involves the reduction of MnO\(_4^-\) (permanganate ion) to Mn\(^{2+}\), and the oxidation half-reaction involves the oxidation of thiosulfate (S\(_2\)O\(_3^{2-}\)) to sulfate (SO\(_4^{2-}\)). The balance of this equation is:

8MnO\(_4^-\) + 3S\(_2\)O\(_3^{2-}\) + H\(_2\)O \(\rightarrow\) 8Mn\(^{2+}\) + 3SO\(_4^{2-}\) + 2H\(^+\).


(b) The chromium in Cr\(_2\)O\(_7^{2-}\) is reduced from +6 to +3, and the tin in Sn\(^{2+}\) is oxidized to Sn\(^{4+}\). The balanced equation is:

Cr\(_2\)O\(_7^{2-}\) + 3Sn\(^{2+}\) + 14H\(^+\) \(\rightarrow\) 2Cr\(^{3+}\) + 3Sn\(^{4+}\) + 7H\(_2\)O.
Quick Tip: Always balance the atoms and charge when balancing redox reactions, ensuring that the number of atoms of each element and the total charge are the same on both sides of the equation.

(a) Cooking is faster in a pressure cooker because the pressure inside the cooker is higher. This increases the boiling point of water, allowing the food to cook at a higher temperature. As a result, the cooking time decreases.


(b) When the volume of the resulting solution decreases upon mixing, it indicates a negative deviation from Raoult’s law. This typically occurs when there are strong intermolecular interactions between the two components, such as hydrogen bonding or dipole-dipole interactions, which cause the components to interact more strongly than they would with themselves. The temperature will decrease as the mixing of the liquids leads to a decrease in the system’s internal energy.
Quick Tip: When studying Raoult’s law deviations, remember that negative deviations occur when the solution exhibits stronger intermolecular interactions than in pure liquids, leading to a decrease in vapor pressure and volume contraction.

An azeotrope is a mixture of two or more liquids that behaves as a single substance during distillation, i.e., it has a constant boiling point and the composition of the liquid phase and vapor phase is the same.

- A positive deviation from Raoult's law occurs when the intermolecular forces between unlike molecules are weaker than those between like molecules, leading to an increase in vapor pressure and a boiling point that is lower than the boiling points of the individual components.
- A negative deviation occurs when the intermolecular forces between unlike molecules are stronger than those between like molecules, leading to a decrease in vapor pressure and a higher boiling point.

- An example of a negative deviation azeotrope is the mixture of water and hydrochloric acid (HCl), which boils at a constant temperature and has the same composition in both the liquid and vapor phases.
Quick Tip: Azeotropes are formed when two liquids interact strongly enough to maintain a constant composition during distillation, and they do not follow Raoult’s law due to the non-ideal behavior of the mixture.

- In part (a), the reaction of CH\(_3\)CH\(_2\)Cl with NaCN gives A, which is CH\(_3\)CH\(_2\)CN (ethyl cyanide). Hydrogenation (H\(_2\) / Ni) of this compound will reduce the nitrile group to an amine group, resulting in B, which is CH\(_3\)CH\(_2\)NH\(_2\) (ethylamine).

- In part (b), the reaction of C\(_6\)H\(_5\)NH\(_2\) (aniline) with NaNO\(_2\) in the presence of HCl at low temperatures forms a diazonium salt (A), which upon treatment with an acid (H\(^+\)) forms B, which is aniline again (C\(_6\)H\(_5\)NH\(_2\)).
Quick Tip: When dealing with diazonium salts, remember that they are typically formed in the presence of nitrous acid (generated from NaNO\(_2\) and HCl) and react with nucleophiles like phenols or aryl groups under acidic conditions.

- (a) The hydrolysis of sucrose (C\(_{12}\)H\(_{22}\)O\(_{11}\)) results in the formation of glucose (C\(_6\)H\(_{12}\)O\(_6\)) and fructose (C\(_6\)H\(_{12}\)O\(_6\)).

- (b) The hydrolysis of lactose (C\(_{12}\)H\(_{22}\)O\(_{11}\)) produces glucose (C\(_6\)H\(_{12}\)O\(_6\)) and galactose (C\(_6\)H\(_{12}\)O\(_6\)).
Quick Tip: To identify hydrolysis products, remember that sucrose is a disaccharide made of glucose and fructose, while lactose consists of glucose and galactose. Hydrolysis breaks the glycosidic bond in both sugars.

Henry's law states that the concentration of a gas in a liquid is directly proportional to the pressure of the gas above the liquid: \[ C = k_H \times P \]
Where:
- \(C\) is the concentration of the gas (mol/L),
- \(k_H\) is Henry's law constant (Pa),
- \(P\) is the partial pressure of the gas (Pa).


Given:
- \(k_H = 1.67 \times 10^8\) Pa,
- \(P = 2.53 \times 10^5\) Pa,
- Volume = 500 mL = 0.5 L,

Using Henry’s law: \[ C = (1.67 \times 10^8) \times (2.53 \times 10^5) = 4.23 \times 10^{13} \, mol/L \, Pa \]
Now, to calculate the moles of CO\(_2\), use the relation: \[ moles of CO_2 = C \times V = (4.23 \times 10^{-8} \, mol/L) \times (0.5 \, L) = 2.12 \times 10^{-8} \, mol \] Quick Tip: In Henry’s law, remember to use the pressure in the same units as the constant to ensure correct results.

First, calculate the cell potential \(E^\circ_cell\) using the formula: \[ E^\circ_cell = E^\circ_{cathode} - E^\circ_{anode} \]
Here, Cr is oxidized (anode) and Cd\(^{2+}\) is reduced (cathode), so: \[ E^\circ_cell = (-0.40) - (-0.74) = 0.34 \, V \]

Next, calculate the Gibbs free energy change \(\Delta_r G^\circ\) using the relation: \[ \Delta_r G^\circ = -nFE^\circ_cell \]
Since the reaction involves 6 electrons (from 2 Cr\(^{3+}\) to Cr and 3 Cd\(^{2+}\) to Cd), we get: \[ \Delta_r G^\circ = -6 \times (96500) \times (0.34) = -197220 \, J/mol = -197.22 \, kJ/mol \]

To calculate \(K_C\), use the relation: \[ \Delta_r G^\circ = -RT \ln K_C \] \[ K_C = e^{-\Delta_r G^\circ / RT} = e^{(197220) / (8.314 \times 298)} = e^{80} \approx 10^{38} \]

Thus, log \(K_C\) = 38.
Quick Tip: To convert from Gibbs free energy to the equilibrium constant, use the equation: \[ \Delta_r G^\circ = -RT \ln K_C \] This equation relates the spontaneity of a reaction to the equilibrium constant.

The relation between the rate constant and temperature is given by the Arrhenius equation: \[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Where:
- \(k_1\) and \(k_2\) are the rate constants at temperatures \(T_1\) and \(T_2\),
- \(E_a\) is the activation energy,
- \(R\) is the gas constant,
- \(T_1 = 293\) K, \(T_2 = 313\) K.


Given that the rate quadruples, \[ \frac{k_2}{k_1} = 4 \Rightarrow \ln 4 = \frac{E_a}{8.314} \left(\frac{1}{293} - \frac{1}{313}\right) \] \[ 0.602 = \frac{E_a}{8.314} \left(\frac{1}{293} - \frac{1}{313}\right) = \frac{E_a}{8.314} \times \left(0.00342 - 0.00319\right) \] \[ 0.602 = \frac{E_a}{8.314} \times 0.00023 \] \[ E_a = \frac{0.602 \times 8.314}{0.00023} = 21936 \, J/mol = 21.94 \, kJ/mol \] Quick Tip: For a temperature increase that leads to a change in reaction rate, use the Arrhenius equation to calculate the activation energy. Remember that rate constants are related to temperature exponentially.

(a) The major monohalo product of this reaction is the bromoalkane formed by the addition of Br₂ to the alkene, resulting in a dibromo product. The Br will add to the double bond in an anti fashion.

(b) The reaction of CH₃–HBr will produce a bromoalkane, where the hydrogen atom will add to one of the carbon atoms of the double bond, and the Br will add to the other.

(c) The reaction of HO–H₂C with HCl and heat will lead to the formation of an alkene through an elimination reaction, where water is removed from the alcohol.
Quick Tip: In halogenation reactions, the halogen adds to the alkene based on the stability of the intermediate carbocation or radical.

(a) Chlorobenzene can be converted to biphenyl using the Wurtz-Fittig reaction, which involves coupling two chlorobenzene molecules in the presence of a sodium metal catalyst.

(b) Propene can be converted to 1-Iodopropane via an electrophilic addition of iodine, where iodine adds across the double bond.

(c) 2-Bromobutane can be converted to but-2-ene via a dehydrohalogenation reaction, in which a base (like KOH) eliminates the hydrogen atom from one carbon and the bromine atom from the adjacent carbon, forming a double bond.
Quick Tip: Understanding the mechanism behind electrophilic additions and eliminations helps in predicting products accurately in organic transformations.

(a) Copper has a high \(E^\circ\) value in the +2 oxidation state because it has a relatively stable electronic configuration, and the reduction potential is high due to the stability of the Cu\(^{2+}\) ion.

(b) The element that is a strong reducing agent in the +2 oxidation state is Zinc (Zn). This is because zinc readily loses its 2 electrons to become Zn\(^{2+}\), which has a stable electronic configuration.

(c) Zn\(^{2+}\) salts are colourless because Zn\(^{2+}\) has no unpaired electrons, and hence, it does not absorb visible light, which is necessary for the salt to appear coloured.
Quick Tip: To understand redox behavior, always look for the stability of the ion formed after the electron transfer.

(a) The order of boiling points based on hydrogen bonding and molecular mass is: \[ (CH_3)_2NH < CH_3CH_2NH_2 < CH_3CH_2OH \]
(CH₃)₂NH has the lowest boiling point because it has fewer hydrogen bonds than the other two. CH₃CH₂OH has the highest boiling point due to the presence of strong hydrogen bonding.

(b)
(i) Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis because the aryl group in the aromatic primary amine makes it less reactive in nucleophilic substitution.

(ii) Amides are less basic than amines because the lone pair of electrons on the nitrogen in amides is delocalized into the carbonyl group, which reduces its availability for protonation, making amides less basic than amines.
Quick Tip: When comparing boiling points, always consider the effect of hydrogen bonding and molecular size.

(a) A native protein refers to a protein that is in its natural, functional form, with its specific three-dimensional structure intact. This form allows the protein to perform its biological function.

(b) A nucleotide is a basic building block of nucleic acids (DNA and RNA). It consists of a nitrogenous base, a sugar molecule, and a phosphate group. Nucleotides link together to form the backbone of nucleic acids.

(c) An essential amino acid is an amino acid that cannot be synthesized by the body and must be obtained through the diet. Examples include lysine, leucine, and tryptophan.
Quick Tip: In biochemistry, understanding the structure and function of proteins, nucleotides, and amino acids is crucial for exploring metabolic processes.

The rate-determining step is the slowest step in a multi-step reaction. It controls the overall rate of the reaction as the entire reaction cannot proceed faster than its slowest step. The rate constant of this step determines the rate of the entire reaction.
Quick Tip: In a multi-step reaction, identify the slowest step (rate-determining step) to determine the overall reaction rate.

When phenol reacts with bromine in carbon disulfide (CS\(_2\)), it undergoes electrophilic substitution at the ortho and para positions relative to the hydroxyl group. Bromine substitutes the hydrogen atoms at these positions, resulting in the formation of 2,4,6-tribromophenol.
Quick Tip: In the presence of a halogen like Br\(_2\), phenol undergoes halogenation predominantly at the ortho and para positions due to the electron-donating effect of the OH group.

To convert ethanol to But-2-enal, the following steps are involved:


1. Oxidation of Ethanol to Acetaldehyde:
Ethanol (CH\(_3\)CH\(_2\)OH) can be oxidized to acetaldehyde (CH\(_3\)CHO) by using an oxidizing agent such as potassium dichromate (K\(_2\)Cr\(_2\)O\(_7\)) or potassium permanganate (KMnO\(_4\)). The reaction proceeds as:
\[ CH_3CH_2OH \xrightarrow{Oxidizing agent} CH_3CHO + H_2O \]


2. Aldol Condensation to Form But-2-enal:
Acetaldehyde undergoes an aldol condensation reaction in the presence of a base (such as NaOH) to form But-2-enal (CH\(_3\)CH=CHCHO). The aldol condensation reaction proceeds as follows:
\[ 2 CH_3CHO \xrightarrow{NaOH} CH_3CH=CHCHO + H_2O \]

Thus, ethanol is first oxidized to acetaldehyde and then undergoes aldol condensation to form But-2-enal.
Quick Tip: Aldol condensation is commonly used for the formation of \(\beta\)-unsaturated aldehydes, like But-2-enal, by coupling aldehyde molecules.

1. Identification of (A):
The molecular formula of (A) is C\(_8\)H\(_{16}\)O\(_2\). A compound with this formula could be 1-octanol (CH\(_3\)(CH\(_2\))\(_6\)CH\(_2\)OH), as it fits the molecular weight and structure.

2. Reaction of (A) with dilute sulphuric acid:
When 1-octanol reacts with dilute sulphuric acid, it undergoes hydrolysis, breaking the molecule into a carboxylic acid and an alcohol. In this case, it forms octanoic acid (CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_2\)COOH) and 1-propanol (CH\(_3\)CH\(_2\)OH) as products.

3. Oxidation of (C) (1-propanol) with chromic acid:
1-Propanol (C) is oxidized by chromic acid (H\(_2\)CrO\(_4\)) to form propanoic acid (CH\(_3\)CH\(_2\)COOH), which is compound (B). The oxidation reaction is:
\[ CH_3CH_2OH + [O] \rightarrow CH_3CH_2COOH \]

4. Dehydration of (C) (1-propanol) to But-1-ene:
On dehydration, 1-propanol undergoes elimination to form But-1-ene. The reaction proceeds as follows:
\[ CH_3CH_2OH \xrightarrow{H_2SO_4} CH_3CH=CH_2 \]

Thus, the identified compounds are:
- (A) = 1-Octanol
- (B) = Propanoic acid
- (C) = 1-Propanol Quick Tip: Dehydration reactions of alcohols generally lead to the formation of alkenes, with the elimination of a water molecule.

1. [Fe(H\(_2\)O)\(_6\)]\(^{2+}\):
- Hybridization: The central iron ion (Fe\(^{2+}\)) is surrounded by six water molecules. This corresponds to an octahedral geometry, which is formed due to sp\(^3\)d\(^2\) hybridization.
- Shape: Octahedral.
- Magnetic property: Since Fe\(^{2+}\) has unpaired electrons, the complex is paramagnetic.

2. [NiCl\(_4\)]\(^{2-}\):
- Hybridization: The central nickel ion (Ni\(^{2+}\)) is surrounded by four chloride ions, resulting in a tetrahedral geometry. This geometry arises from sp\(^3\) hybridization.
- Shape: Tetrahedral.
- Magnetic property: Ni\(^{2+}\) has unpaired electrons, so this complex is paramagnetic. Quick Tip: Octahedral complexes often have sp\(^3\)d\(^2\) hybridization, while tetrahedral complexes typically use sp\(^3\) hybridization. Paramagnetism is common in complexes with unpaired electrons.

1. [Co(H\(_2\)O)(CN)(en)\(_2\)]\(^{2+}\):
This is a cobalt complex with water, cyanide, and ethylenediamine (en) ligands. The IUPAC name is:
Aqua-cyanido-bis(ethylenediamine)cobalt(III) ion.

2. [PtCl\(_4\)]\(^{2-}\):
This is a platinum(II) complex with four chloride ions. The IUPAC name is:
Tetrachloroplatinate(II) ion.

3. [Cr(NH\(_3\))\(_4\)Cl(ONO)]\(^+\):
This is a chromium(III) complex with ammonia, chloride, and nitrito (ONO) ligands. The IUPAC name is:
Tetraamminochloro(nitrito)chromium(III) ion. Quick Tip: In naming coordination compounds, the ligands are named first, followed by the central metal and its oxidation state in parentheses. Prefixes are used for indicating the number of each type of ligand.

The cell consists of two half-cells:

1. Anode half-reaction (oxidation):
The oxidation takes place at the tin electrode:
\[ Sn(s) \rightarrow Sn^{2+}(aq) + 2e^- \]
The concentration of Sn\(^{2+}\) is 0.004 M.

2. Cathode half-reaction (reduction):
The reduction occurs at the hydrogen electrode:
\[ 2H^+(aq) + 2e^- \rightarrow H_2(g) \]
The concentration of H\(^+\) is 0.02 M, and the pressure of H\(_2\) gas is 1 bar.

The overall cell reaction is: \[ Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g) \]

To calculate the e.m.f. of the cell, we use the Nernst equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]

Where:
- \( E^\circ_{cell} \) is the standard cell potential,
- \( n \) is the number of electrons transferred,
- \( Q \) is the reaction quotient.

The standard cell potential is the difference between the standard electrode potentials of the cathode and anode: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00 \, V - (-0.14 \, V) = 0.14 \, V \]

The reaction quotient \( Q \) is given by: \[ Q = \frac{[Sn^{2+}]}{[H^+]^2} = \frac{0.004}{(0.02)^2} = \frac{0.004}{0.0004} = 10 \]

Now, we can substitute these values into the Nernst equation:
\[ E_{cell} = 0.14 - \frac{0.0591}{2} \log 10 \]

Since \( \log 10 = 1 \), we get:
\[ E_{cell} = 0.14 - \frac{0.0591}{2} \times 1 = 0.14 - 0.02955 = 0.11045 \, V \]

Thus, the e.m.f. of the cell is approximately 0.110 V.
Quick Tip: The Nernst equation is useful for calculating the cell potential at non-standard conditions, such as varying concentrations or pressures.

The difference occurs due to the E\(^0\) values. The E\(^0\) value for oxygen is +1.23 V, while the E\(^0\) value for chlorine is +1.36 V. Because chlorine has a higher E\(^0\) value, it is more readily oxidized in the electrolysis of NaCl, even though oxygen should theoretically be produced according to E\(^0\) values.
Quick Tip: Higher E\(^0\) values indicate that a substance is more easily reduced or oxidized. In electrolysis, the substance with a higher E\(^0\) value is usually produced at the anode.

In a lead storage battery, the reactions are as follows:

1. Anode reaction (oxidation):
\[ Pb(s) + SO_4^{2-} \rightarrow PbSO_4(s) + 2e^- \]

2. Cathode reaction (reduction):
\[ PbO_2(s) + 4H^+ + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l) \]

3. Overall cell reaction:
\[ Pb(s) + PbO_2(s) + 4H^+ + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l) \] Quick Tip: Lead storage batteries use lead and lead oxide as electrodes, with sulfuric acid as the electrolyte, undergoing charge and discharge reactions.