General Chemistry

The elevation in boiling point is given by the formula:

Where:
- = elevation in boiling point
- = molal elevation constant
- = molality of the solution
Step 1: Calculate elevation in boiling point.

Step 2: Calculate number of moles of solute.
Mass of solute
Molar mass of solute

Step 3: Calculate mass of solvent in kilograms.
Mass of solvent
Step 4: Calculate molality of solution.

Step 5: Calculate boiling point elevation constant.


Thus, the boiling point elevation constant for the liquid is .

Both complexes contain Fe(II) (d ), but the ligands differ in field strength. CN is a strong-field ligand and produces a large octahedral splitting ( ), giving a low-spin configuration , while H O is a weak-field ligand with smaller , often giving higher-spin populations.
Because is different, the energy of the electronic transition (and thus the wavelength of light absorbed) differs for the two complexes. They therefore absorb different parts of the visible spectrum and appear different in colour (e.g., [Fe(CN) ] is very pale/yellowish, while [Fe(H O) ] is pale green). Hence, colour difference arises from differing crystal-field splitting due to ligand field strength.

Gabriel-Pthalimide synthesis is preferred for synthesizing primary amines because it offers a selective and reliable method for their preparation. The key advantage of this method is that it exclusively yields primary amines, avoiding the formation of secondary and tertiary amines, which are often byproducts in other amination reactions. In the Gabriel-Pthalimide synthesis, phthalimide (an imide derivative) is reacted with an alkyl halide in the presence of a base (such as potassium hydroxide, ) to perform an nucleophilic substitution. This reaction results in the formation of an N-alkylphthalimide intermediate. The next step involves the hydrolysis of this intermediate to release the primary amine and regenerate phthalic acid (which is typically removed). The reaction proceeds as follows: This method provides good yields of primary amines and is favored because it avoids the formation of unwanted amine byproducts. Additionally, Gabriel-Pthalimide synthesis is relatively simple and provides a high degree of purity for the primary amines produced.

This reaction is a reduction of benzoic acid (C H COOH) with an amine (ethylamine) in the presence of an acid ( ) to form phenylethylamine.

The reaction involves the Friedel-Crafts alkylation of benzene using ethyl chloride in the presence of anhydrous AlCl . This results in the formation of ethylbenzene.

In this reaction, the mercuric ion ( ) with sulfuric acid ( ) catalyzes the hydration of the alkyne to form an alcohol. The product of the reaction is propan-2-ol.

(x) :
The IUPAC name of this complex is pentaamminechloridocobalt(III) chloride.

Explanation:
- The complex ion consists of cobalt as the central metal ion, five ammonia molecules (NH₃) and one chloride ion (Cl) as ligands.
- The oxidation state of cobalt is +3, and the complex is neutralized by two chloride ions (Cl⁻) outside the coordination sphere.

(y) :
The IUPAC name of this complex is potassium tetracyanidonickelate(II).

Explanation:
- The complex ion consists of nickel as the central metal ion, four cyanide (CN⁻) ligands.
- The oxidation state of nickel is +2, and the complex is neutralized by two potassium ions (K⁺).

DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are both nucleic acids but differ in their structural and functional characteristics. Below are the key differences:
Structural Differences:
1. Sugar Molecule:}
- DNA contains deoxyribose sugar, which lacks one oxygen atom (at the 2' carbon) compared to ribose.
- RNA contains ribose sugar, which has a hydroxyl group (-OH) attached to the 2' carbon atom.
2. Strand Type:}
- DNA is typically double-stranded, forming a double helix structure.
- RNA is usually single-stranded and can form various structures like hairpins.
3. Nitrogenous Bases:}
- DNA contains the bases adenine (A), thymine (T), cytosine (C), and guanine (G).
- RNA contains the bases adenine (A), uracil (U), cytosine (C), and guanine (G). Note that RNA uses uracil instead of thymine.
4. Helical Structure:}
- DNA has a right-handed double helix structure.
- RNA is single-stranded and may adopt various secondary structures.
5. Length and Stability:}
- DNA is generally longer and more stable than RNA, designed to store genetic information.
- RNA is usually shorter and less stable, designed for transient functions in the cell.
Functional Differences:
1. Genetic Material:}
- DNA carries the genetic blueprint for the synthesis of proteins and is involved in the storage and transmission of genetic information.
- RNA plays a role in transcribing the genetic code from DNA and is involved in the synthesis of proteins (as mRNA, rRNA, and tRNA).
2. Types of RNA:}
- DNA does not have different types but is used in the formation of mRNA, rRNA, and tRNA in RNA transcription.
- RNA has several types: mRNA (messenger RNA), rRNA (ribosomal RNA), and tRNA (transfer RNA), each playing a role in protein synthesis.
3. Replication and Transcription:}
- DNA is self-replicating, meaning it can duplicate itself during cell division.
- RNA is transcribed from DNA and is not capable of replication itself but can be synthesized through transcription.
4. Function in Protein Synthesis:}
- DNA stores the instructions for protein synthesis, while RNA is involved in the direct synthesis of proteins in the cytoplasm.

The formula for resistance (R) of a solution is given by:

R = ρL / A

Where:

• R is the resistance,
• ρ is the resistivity,
• L is the length of the column,
• A is the cross-sectional area of the column.

We are given:

• Resistance R = 6.5 × 10³ Ω,
• Length L = 50 cm = 0.50 m,
• Diameter d = 1 cm = 0.01 m.

Step 1: Calculate the cross-sectional area.

The cross-sectional area A of the column is given by the area of a circle:

A = π (d/2)² = π (0.01/2)² = 7.854 × 10^-5 m²

Step 2: Calculate the resistivity.

Rearranging the formula for resistance, we can solve for resistivity:

ρ = (R × A) / L = (6.5 × 10³ × 7.854 × 10^-5) / 0.50 = 1.02 Ω·m

Thus, the resistivity is ρ = 1.02 Ω·m.

Step 3: Calculate the conductivity.

Conductivity σ is the reciprocal of resistivity:

σ = 1 / ρ = 1 / 1.02 = 0.980 S/m

Thus, the conductivity is σ = 0.980 S/m.

Step 4: Calculate the molar conductivity.

Molar conductivity Λ_m is given by:

Λ_m = κ / C

Where:

• κ is the conductivity,
• C is the concentration of the solution in mol·L⁻¹.

Given C = 0.10 mol·L⁻¹, we can calculate the molar conductivity:

Λ_m = 0.980 / 0.10 = 9.80 S·m²/mol

Thus, the molar conductivity is Λ_m = 9.80 S·m²/mol.

Molality is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per kilogram of solvent. The formula for molality is: Where: - is the number of moles of solute (in this case, ethanoic acid),
- is the mass of the solvent (benzene) in kilograms.
We are given:
- Mass of ethanoic acid ,
- Mass of benzene ,
- Molar mass of ethanoic acid .
Step 1: Calculate the number of moles of ethanoic acid (solute).
The number of moles is calculated by the formula: Step 2: Calculate molality using the formula.
Now, we can use the formula for molality: Thus, the molality of the solution is .

Faraday's First Law: The mass of a substance liberated (or deposited) at an electrode is directly proportional to the total electric charge passed through the electrolyte.
where is mass deposited, is charge, and is the electrochemical equivalent.
Faraday's Second Law: When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their chemical (equivalent) weights.
Numerical Part: In CuSO electrolysis, copper is deposited by For a metal ion requiring electrons, where , , , , .

Step 1: Nature of forces in butane.
Butane (C H ) is a hydrocarbon and non-polar in nature. The only intermolecular forces present in butane are weak Van der Waals (London dispersion) forces. These forces require relatively little energy to overcome, so butane has a low boiling point.
Step 2: Nature of forces in propanol.
Propanol (C H OH) is an alcohol containing an –OH group. The hydroxyl group allows molecules of propanol to form intermolecular hydrogen bonds. Hydrogen bonds are much stronger than Van der Waals forces. Breaking these bonds requires more energy, which raises the boiling point of propanol significantly.
Step 3: Comparison.
Because propanol forms hydrogen bonds while butane does not, the boiling point of propanol is much greater than that of butane.

Nucleic acids are complex biomolecules that store and transmit genetic information in living organisms.
They are polymers of nucleotides, which consist of a nitrogenous base, a pentose sugar, and a phosphate group.
The two main types of nucleic acids are:
- DNA (Deoxyribonucleic acid)
- RNA (Ribonucleic acid)
Two important properties:
1. Genetic Information Storage: DNA carries hereditary information that controls the development, functioning, and reproduction of all living organisms.
2. Self-Replication and Protein Synthesis: Nucleic acids can replicate themselves and guide protein synthesis through transcription (DNA → RNA) and translation (RNA → protein).
Thus, nucleic acids are essential molecules of life.

For a galvanic cell at standard state, and . Hence, .
Given: .
Overall reaction involves transfer of electrons.
Step 1: Calculate standard Gibbs energy change.
Step 2: Calculate equilibrium constant.
Answer: .

We are given 30 mass percent benzene in the solution. This means:
- Mass of benzene = 30 g
- Mass of carbon tetrachloride (CCl ) = 70 g
Molar masses:
- Molar mass of benzene (C H ) = 78 g/mol
- Molar mass of carbon tetrachloride (CCl ) = 154 g/mol
Step 1: Calculate moles of benzene.
Step 2: Calculate moles of carbon tetrachloride.
Step 3: Calculate mole fraction of benzene.
Thus, the mole fraction of benzene is 0.458.

Raoult's Law states that the partial vapor pressure of a component in a liquid solution is directly proportional to its mole fraction in the solution. Where:
- = partial vapor pressure of component A in solution,
- = mole fraction of component A in solution,
- = vapor pressure of pure component A.
For a binary solution (components A and B), the total vapor pressure is: Significance: - Raoult's Law helps in understanding colligative properties like relative lowering of vapor pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
- It is valid for ideal solutions where intermolecular interactions between unlike molecules are similar to those between like molecules.

Uses of Carbon Tetrachloride (CCl ):
Used as a cleaning agent for degreasing metals and machinery.
Acts as a solvent for oils, fats, and varnishes.
Earlier used in fire extinguishers (though avoided now due to toxicity).
Plays a role in the manufacture of refrigerants and propellants.
Uses of Chloroform (CHCl ):
Widely used as a solvent in laboratories and industries.
Employed in the preparation of chlorofluorocarbons (CFCs).
Historically used as an anesthetic, though not common now due to harmful effects.
Helps in extraction and purification of alkaloids and other organic compounds.

Step 1: Compare Ethanol and Phenol.
- Ethanol ( ) is a neutral alcohol with very weak acidic character because the bond is not easily ionizable.
- Phenol ( ) is more acidic than ethanol because the phenoxide ion formed after deprotonation is stabilized by resonance in the benzene ring.
Step 2: Compare Phenol and Acetic acid.
- Acetic acid ( ) is much more acidic than phenol because the acetate ion is stabilized by resonance involving two oxygen atoms.
Step 3: Compare Acetic acid and Chloroacetic acid.
- Chloroacetic acid ( ) is stronger than acetic acid because the electron-withdrawing chlorine atom shows the (inductive) effect, which stabilizes the conjugate base (chloroacetate ion).
Final Order:
Thus, the correct option is (C).

Step 1: Recall what Fehling’s test detects.
Fehling’s solution test is used to detect reducing sugars. Reducing sugars have a free aldehyde group ( ) or a free ketone group ( ) that can be oxidized. On heating with Fehling’s solution, such sugars reduce Cu ions to Cu O, producing a red precipitate.
Step 2: Analyze each option.
- Glucose is an aldohexose and a reducing sugar. It contains a free aldehyde group that can reduce Fehling’s solution, giving a positive test (red precipitate of Cu O).
- Sucrose is a non-reducing sugar because the glycosidic bond involves the reducing groups of both glucose and fructose units, so it does not give a positive Fehling’s test.
- Fats are esters of glycerol and fatty acids; they do not have free aldehyde or ketone groups, so they do not respond to Fehling’s test.
- Proteins contain amino acids linked by peptide bonds; they do not act as reducing sugars, hence no reaction with Fehling’s solution.
Step 3: Final Answer.
Only glucose gives a positive Fehling’s solution test.

The colour of transition metal compounds arises due to d–d electronic transitions within partially filled d-orbitals. When light falls on such a compound, electrons in lower energy d-orbitals absorb specific wavelengths and get excited to higher energy d-orbitals, producing characteristic colours.
Now, let us analyze the given ions:
1. : Electronic configuration = [Ar] 3d . Since the d-orbitals are partially filled, d–d transitions are possible, so compounds of Cr are coloured.
2. : Electronic configuration = [Ar] 3d . With partially filled d-orbitals, d–d transitions occur, hence Co compounds are coloured.
3. : Electronic configuration = [Ar] 3d . This ion also has partially filled d-orbitals, allowing d–d transitions, making Ni compounds coloured.
4. : Electronic configuration = [Ar] 3d . Here, the d-orbitals are completely filled. Since no d–d transitions are possible, compounds of Zn are colourless.
Therefore, the correct answer is option (D) Zn .

A gaseous solution refers to a homogeneous mixture in which the solvent is a gas. The solute may be solid, liquid, or gas, but the solvent must be a gas.
Let us analyze the given options one by one:
1. Solution of camphor in nitrogen: Here nitrogen is the solvent (gas) and camphor particles get dispersed in it. This forms a gaseous solution. This is the correct example.
2. Solution of hydrogen in palladium: This represents a solid solution, since palladium is a metal and hydrogen gas occupies interstitial sites in its lattice. This is not a gaseous solution.
3. Amalgam of mercury with sodium: This is a solid–liquid type solution, where sodium dissolves in mercury to form an amalgam. It is not a gaseous solution.
4. Oxygen dissolved in water: This is a gas–liquid solution, not a gaseous solution.
Hence, the correct answer is option (A) Solution of camphor in nitrogen.

Kohlrausch's Law:
Kohlrausch's Law of limiting molar conductivity states that the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductances of the individual ions that make up the electrolyte. Mathematically, it is expressed as: Where: - is the limiting molar conductivity of the electrolyte,
- is the limiting molar conductivity of the cation,
- is the limiting molar conductivity of the anion.
For , the cation is and the anion is . From the given data:
- ,
- .
Using Kohlrausch's Law: Substituting the values: Thus, the limiting molar conductivity of is 195.3 S cm mol .

Effect of Dilution on Conductivity:
The conductivity of a solution decreases with dilution because as the solution is diluted, the number of ions in the solution per unit volume decreases. This leads to a decrease in the overall current-carrying capacity of the solution. The key factors involved are: 1. Ion Concentration:
Conductivity ( ) of a solution depends on the concentration of ions. As the concentration of ions decreases with dilution, the conductivity also decreases. The relationship is given by: Where: - is the conductivity,
- is the concentration of the ion,
- is the molar conductivity of the ion. As dilution decreases the concentration , the conductivity decreases as well.
2. Ion Pairing:
At higher concentrations, ions tend to form ion pairs, which reduce the number of free ions in the solution, thus decreasing the conductivity. When a solution is diluted, ion pairing is minimized, but the total number of free ions decreases due to the lower concentration.
3. Decrease in Collision Frequency:
In concentrated solutions, ions are closer to each other and collide more often, which increases the chances of electrical conduction. As the solution is diluted, the ions are further apart, leading to fewer collisions and lower conductivity. Thus, with dilution, the total ion concentration and the ion mobility decrease, leading to a reduction in the overall conductivity of the solution.

Transition metals form colored compounds due to the electronic transitions between their d-orbitals. The color is a result of these transitions when electrons move from one d-orbital to another under the influence of visible light.
- In transition metals, the d-orbitals are split into two energy levels under the influence of a ligand field. When light is absorbed, electrons are excited from lower to higher energy d-orbitals, and the energy corresponding to the absorbed light corresponds to the color observed in the complex. This is why many transition metal compounds, like those of copper, chromium, and cobalt, show distinct colors.

Water-soluble vitamins are vitamins that dissolve in water and are not stored in the body for long periods. They need to be replenished regularly through diet. Here are two examples of water-soluble vitamins: 1. Vitamin C (Ascorbic acid):
- Function: Vitamin C plays a crucial role in collagen synthesis, wound healing, and maintaining the integrity of skin, blood vessels, and bones. It also acts as an antioxidant, protecting cells from damage.
- Disease due to deficiency: Scurvy. Scurvy is characterized by symptoms such as bleeding gums, joint pain, weakness, and skin rashes. This condition arises due to the lack of collagen synthesis, which Vitamin C is vital for.
- Symptoms: Swollen and bleeding gums, loose teeth, fatigue, and bruising.
2. Vitamin B1 (Thiamine):
- Function: Thiamine plays an essential role in carbohydrate metabolism, as it is a coenzyme in the decarboxylation of alpha-keto acids, and it is vital for nerve function and energy production.
- Disease due to deficiency: Beriberi. There are two types of beriberi: wet and dry. Wet beriberi affects the cardiovascular system, leading to swelling, shortness of breath, and heart failure. Dry beriberi affects the nervous system, causing weakness, numbness, and motor impairment.
- Symptoms: Weakness, fatigue, nerve damage, and swelling in the body.
Water-soluble vitamins are important for overall health, and their deficiency can lead to serious conditions.

Raoult’s Law states that the partial vapor pressure of each volatile component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. Mathematically, it is represented as: Where: - is the partial vapor pressure of component ,
- is the mole fraction of component in the solution,
- is the vapor pressure of the pure component .
Given: - Vapor pressure of chloroform ,
- Vapor pressure of dichloromethane ,
- Mass of chloroform ,
- Mass of dichloromethane ,
- Molar mass of chloroform ,
- Molar mass of dichloromethane .
Step 1: Calculate moles of each component.
For chloroform: For dichloromethane: Step 2: Calculate total moles in the solution.
Total moles = .
Step 3: Calculate mole fractions.
Mole fraction of chloroform: Mole fraction of dichloromethane: Step 4: Apply Raoult’s Law to calculate the total vapor pressure.
Vapor pressure of the solution is the sum of the partial pressures: Thus, the vapor pressure of the solution is .

Alcohols are more soluble in water compared to hydrocarbons of similar molecular weight due to the presence of the hydroxyl group (-OH), which allows alcohols to form hydrogen bonds with water molecules. The following points explain this in detail:
1. Hydrogen Bonding:
Alcohols contain a hydroxyl group (-OH), which can form hydrogen bonds with water molecules. Water is a polar solvent, and the hydroxyl group in alcohols is also polar, making it capable of forming hydrogen bonds. These interactions increase the solubility of alcohols in water. On the other hand, hydrocarbons are non-polar and cannot form hydrogen bonds with water, making them much less soluble.
2. Polarity:
The hydroxyl group in alcohols makes them polar, while hydrocarbons are non-polar. Water is a polar solvent, and polar substances tend to dissolve more easily in polar solvents. This is the main reason why alcohols are more soluble in water than hydrocarbons of similar molecular weight.
3. Hydrophilic and Hydrophobic Parts:
In alcohols, the -OH group is hydrophilic (water-loving) and can interact with water molecules, whereas the hydrocarbon chain is hydrophobic (water-repelling). However, the hydrophilic nature of the hydroxyl group overcomes the hydrophobic nature of the hydrocarbon part, making alcohols more soluble than hydrocarbons in water.
4. Size of the Alcohol:
The solubility of alcohols in water decreases with increasing size of the hydrophobic hydrocarbon chain. Larger alcohols, such as octanol, are less soluble in water compared to smaller alcohols, like methanol or ethanol. However, alcohols are still more soluble than hydrocarbons of comparable molecular weight due to the presence of the -OH group.
5. Hydrogen Bond Formation:
In water, the hydroxyl group of alcohols can form hydrogen bonds not only with other alcohol molecules but also with water molecules. This significantly increases the solubility of alcohols. Hydrocarbons, lacking the ability to form hydrogen bonds with water, are much less soluble in water. Thus, alcohols are more soluble in water because of their ability to form hydrogen bonds with water molecules, whereas hydrocarbons lack this ability.

Fructose is considered the sweetest of all the common sugars. It is a monosaccharide that is sweeter than glucose, sucrose, and lactose. Fructose is found in honey, fruits, and root vegetables, and is often used as a sweetener in foods and beverages.
- Glucose is sweet but not as sweet as fructose.
- Lactose is a disaccharide found in milk and has a less sweet taste than fructose.
- Sucrose (table sugar) is a disaccharide consisting of glucose and fructose but is not as sweet as fructose alone.
Thus, the correct answer is option (D) Fructose.

The solubility of gases in liquids is generally inversely proportional to temperature, meaning that as the temperature of the liquid increases, the solubility of the gas decreases. This can be explained through the following points: 1. Increased Kinetic Energy of Gas Molecules
As the temperature increases, the kinetic energy of the gas molecules also increases. The molecules move faster, and as a result, they are more likely to overcome the intermolecular forces that hold them in the liquid. This leads to a higher tendency for the gas molecules to escape from the liquid into the gas phase. Therefore, the gas becomes less soluble in the liquid.
2. Le Chatelier’s Principle
Le Chatelier’s principle states that if a system at equilibrium is disturbed, the equilibrium shifts in the direction that minimizes the effect of the disturbance. For gas solubility, when temperature increases, the equilibrium shifts in such a way that more gas molecules escape the liquid phase, reducing the solubility. This is because the dissolution of gas is an exothermic process, and according to Le Chatelier’s principle, increasing the temperature causes the system to favor the gas phase, thus decreasing solubility.
3. Vapor Pressure of the Gas
At higher temperatures, the vapor pressure of the gas increases. This means that the gas is more likely to escape from the liquid to the surface and enter the gas phase. Thus, with an increase in temperature, the solubility of gases decreases because more gas molecules are found in the vapor phase. 4. Entropy Considerations
As temperature increases, the disorder or entropy of the system increases. At higher temperatures, the gas molecules have more freedom of motion and are less likely to stay dissolved in the liquid. This also leads to a decrease in solubility.
In conclusion, at higher temperatures, gas molecules gain enough energy to escape from the liquid, and the solubility decreases.

The Hinsberg’s Test is the standard method for distinguishing primary, secondary and tertiary amines. It uses benzenesulfonyl chloride (C H SO Cl) in the presence of aqueous alkali (NaOH/KOH). Step 1: Reaction with primary amine.
Primary amines react with benzenesulfonyl chloride to give sulfonamides which are soluble in alkali due to the presence of an acidic hydrogen atom. This product dissolves in alkali, forming a soluble salt. Step 2: Reaction with secondary amine.
Secondary amines react to form sulfonamides which have no acidic hydrogen and hence remain insoluble in alkali. The product is insoluble in alkali. Step 3: Reaction with tertiary amine.
Tertiary amines do not react with benzenesulfonyl chloride under these conditions. They remain unreacted and can be recovered unchanged. Conclusion:
- Primary amines: give soluble sulfonamide salts.
- Secondary amines: give insoluble sulfonamides.
- Tertiary amines: do not react.

Step 1: Intermolecular forces in amines.
Amines can form intermolecular hydrogen bonds because of the presence of a nitrogen atom with a lone pair. However, N–H bonds are less polar than O–H bonds, and nitrogen is less electronegative than oxygen. Thus, the hydrogen bonding in amines is weaker compared to alcohols and carboxylic acids. Step 2: Intermolecular forces in alcohols.
Alcohols have O–H bonds, which are more polar and form stronger hydrogen bonds than N–H bonds. Hence, alcohols have higher boiling points than corresponding amines. Step 3: Intermolecular forces in carboxylic acids.
Carboxylic acids form very strong dimeric hydrogen bonds in which two molecules are linked via two O–H…O bonds. This gives them the highest boiling points among the three classes of compounds. Conclusion:
Boiling point order for comparable molecular masses is:

A solid solution in which the solute is a gas typically refers to a gas dissolved in a solid (usually a metal). In this case, hydrogen gas can be absorbed by palladium metal to form a solid solution, called palladium hydride. This is a well-known example of a gas being absorbed into a metal to form a solid solution. Palladium is particularly known for its ability to absorb hydrogen gas into its structure at certain conditions, forming a hydrogen-rich alloy.
The other options are not examples of solid solutions involving a gas as the solute. For example:
- Oxygen dissolved in water is a gas dissolved in a liquid, not a solid solution.
- Solution of camphor in nitrogen is a liquid-solute in gas-solvent solution.
- Glucose dissolved in water is a solid dissolved in liquid, not a solid solution.
Thus, the correct answer is option (C) Solution of hydrogen in palladium.

The magnetic moment of an ion is calculated using the formula:
Where is the number of unpaired electrons in the ion. For a bivalent ion with atomic number 25, we need to determine the electron configuration and the number of unpaired electrons. The electron configuration for an atomic number 25 element (Manganese) is:
In the bivalent state , the electron configuration becomes:
In this configuration, there are 5 unpaired electrons in the orbitals. Therefore, . Now, we can substitute into the formula for magnetic moment:
Hence, the magnetic moment of the bivalent ion in aqueous solution is 2.83 BM. Thus, the correct answer is option (B) 2.83 BM.

(i) Zwitter ion:
A zwitter ion is a dipolar ion that contains both a positive and a negative charge within the same molecule but is overall electrically neutral. Amino acids in aqueous solution commonly exist as zwitter ions, where the amino group ( ) is protonated to and the carboxyl group ( ) is deprotonated to . (ii) Peptide bond:
A peptide bond is a covalent bond formed between the group of one amino acid and the group of another with the elimination of a water molecule (condensation reaction). It has partial double bond character due to resonance, making it rigid and planar. (iii) Primary structure of protein:
The primary structure of a protein is the linear sequence of amino acids in a polypeptide chain, linked together by peptide bonds. This sequence determines the higher levels of protein structure and ultimately governs its biological function. Example: the exact order of amino acids in insulin. (iv) Polysaccharides:
Polysaccharides are long-chain carbohydrates formed by the condensation of numerous monosaccharide units joined by glycosidic bonds. They can be storage polysaccharides (e.g., starch, glycogen) or structural polysaccharides (e.g., cellulose, chitin). They are usually insoluble in water and serve as energy reserves or structural materials in living organisms.