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CHEMISTRY
2024
CHEMISTRY OLD QUESTION PAPER SOLVED
1. What is the molarity of a solution containing 5g of NaOH in 450ml solution?
Ans: To calculate the molarity (M) of a solution, we use the formula:
M = Number of moles of solute/Volume of solution in liters
The molar mass of NaOH is:
NaOH = 23(Na) + 16(O) + 1(H) = 40g/mol
Moles of NaOH:
Moles = Mass of NaOH/Molar mass of NaOH = 5/40 = 0.125 mol
Convert the volume of solution to liters
450ml = 450/1000 = 0.45L
Calculate the molarity
M= Moles of NaOH/ Volume in liters = 0.125/0.45
= 0.278M.
2. How much H₂ in grams will be liberated if 1F electricity is passed through acidified water?
Ans: 1 gram of H2 will be liberated when 1 Faraday of electricity is passed through acidified water.
3. Identify the reaction order from the following rate constants:
k = 2.3×10-51 mol-1s-1
Ans: The reaction order is 2, as the unit of the rate constant (k = 2.3×10−5 L mol−1s−1k) corresponds to a second-order reaction.
4. Which of the 3d transition metals exhibits the largest number of oxidation states?
Ans: Manganese (Mn) exhibits the largest number of oxidation states, ranging from +2 to +7, among the 3d transition metals.
5. Write the chemical formula for the following coordination compounds:
Mercury (1) tetra thiocyanato-s-cobaltate (III)
Ans: The chemical formula for Mercury (I) tetra thiocyanato-s-cobaltate (III) is: Hg2[Co(SCN)4]
6. In the following pairs of halogen compounds, which would undergo SN2 reaction faster?
(a)
(b)
Ans: Cyclohexylmethyl chloride (CH₂Cl) will undergo an SN2 reaction faster than cyclohexyl chloride.
1-Chloropropane will undergo an SN2 reaction faster than 2-chloropropane.
7. Write the products that are formed by heating of the following ether with HI:
Ans: When ethers react with HI on heating, they undergo cleavage to form an alcohol and an alkyl iodide. For example, R-O-R’ + HI → R-I + R’-OH.
8. What is the basic structural difference between starch and cellulose.
Ans: The basic structural difference between starch and cellulose is:
Starch: Composed of α-D-glucose units linked by α-1,4-glycosidic bonds (with occasional α-1,6 bonds in amylopectin), forming a helical structure.
Cellulose: Composed of β-D-glucose units linked by β-1,4-glycosidic bonds, forming straight chains that align to form strong fibers.
9. Diazonium salts of aromatic amines are more stable than those of aliphatic amines. Why?
Ans: Diazonium salts of aromatic amines are more stable than those of aliphatic amines because the aromatic ring can stabilize the positive charge on the diazonium ion through resonance. On other hand aliphatic amines lack this resonance stabilization, making their diazonium salts more prone to decomposition.
OR
Why do primary amines have higher boiling point than tertiary amines ?
Ans: Primary amines have higher boiling points than tertiary amines because primary amines can form hydrogen bonds with other molecules, while tertiary amines cannot. Intermolecular hydrogen bonding is present in primary amines but not in tertiary amines (H-atom absent in amino group) so primary amines have higher boiling point than tertiary amines.
10. Calculate the equilibrium constant of the reaction:
Cu(s)+2Ag+ (aq) →Cu²+ (aq)+2Ag(s)
Given, Eocell = 0.46V
Ans: In K = nFEocell/RT
n=2, F = 96485 C/mol,Eocell = 0.46V,
R = 8.314J/mol
K,T = 298K
In K = nFEocell/RT = 2.96485.0.46/8.314.298 = 35.85
K = e35.85 = 2.8 х 1015
OR
Write the Nernst equation and emf of the following cells at 298K:
(i) Mg(s) | Mg2+ (0.001M) || Cu2+ (0.0001M) Cu(s)
Ans: Standard EMF:
E∘cell = E∘Cu2+/Cu− E∘Mg2+/Mg = 0.34−(−2.37) = 2.71V
Nernst EquationEMF:
Ecell = E∘cell – 0.0591/n log [Mg2+]/[Cu2+]
EMF Calculation:
Ecell = 2.71 – (0.0591/2) log (0.001/0.000)
Ecell = 2.71-0.02955.1 = 2.680V.
(ii) Sn(s) | Sn2+ (0.050M) H+ (0.020M) H2(g)(1 bar) | Pt(s)
Ans: Standard EMF:
E∘cell = E∘H+/H2− E∘Sn2+/Mg = 0.00-(-0.14) = 0.14V
Nernst EquationEMF:
Ecell = E∘cell – 0.0591/n log [Sn2+]/[H2+].PH2
EMF Calculation:
Ecell = 0.14-(0.0591/2) log (0.050/(0.020)2.1)
Ecell = 0.14- 0.02955 log (0.050/(0.0004)
Ecell = 0.14−0.02955⋅2.096 = 0.14−0.0619 = 0.078V
11. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration
Ans: Conductivity: Ability of a solution to conduct electricity.
Molar conductivity: Conductivity per mole of solute in a solution.
Variation with concentration:
Conductivity increases with concentration, while molar conductivity decreases.
12. Show that the half-life period of a first order reaction is independent of the initial concentration of the reactant.
Ans: Half-life equation for a first-order reaction does not contain the initial concentration [A]₀. This means that the half-life is solely dependent on the rate constant (k), which is a characteristic of the reaction itself.
OR
The initial concentration of N₂Os in the following first order reaction N₂Os (g) → 2NO2(g)+1/20₂(g) was 1.24 x 10-2mol L-¹ at 318K. The concentration of N₂Os after 60 minutes was 0.20 × 10-2 mol L. Calculate the rate constant of the reaction at 318K.
Ans: To calculation the rate constant k for a first-order reaction,
In [(A)t/(A)o] = -kt
The rate constant kkk for the reaction at 318 K is 0.0304 min⁻¹.
13. A reaction is first order in A and second order in B.
(a) Write the differential rate equation.
Ans: Rate = k[A]1[B]2
Where:
Rate is the rate of the reaction,
k is the rate constant,
[A] is the concentration of reactant A,
[B] is the concentration of reactant B.
Thus, the differential rate equation is: d[A]/dt =−k[A][B]2
This equation describes the rate of change of concentration of reactant A with respect to time in a reaction that is first-order with respect to A and second-order with respect to B.
(b) How is the rate affected on increasing the concentration of three times?
Ans: If the concentration of A is increased three times, the rate will increase by a factor of 3 because the reaction is first-order in A.
If the concentration of B is increased three times, the rate will increase by a factor of 32 = 9 because the reaction is second-order in B.
Thus, the overall rate will increase by a factor of 3×9 = 27.
14. Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.
Ans: For a divalent ion with atomic number 25 (which is Manganese, Mn²⁺), the electron configuration is:
Mn: [Ar]3d54s2
For Mn²⁺, two electrons are removed from the 4s2
shell, leaving 3d5.
The number of unpaired electrons in the 3d orbital is 5, so the magnetic moment can be calculated using the formula:
μ = √n(n = 2) BM
Where n is the number of unpaired electrons. For Mn²⁺, n = 5.
μ = √5(5+2)
= √35 = 5.92 BM
Thus, the magnetic moment of Mn²⁺ is approximately 5.92 Bohr Magnetons (BM).
OR
What are transition elements? Give the general electronic configuration of transition elements.
Ans: Transition elements (also known as transition metals) are elements that have partially filled d orbitals. IUPAC defines transition elements as an element having a d subshell that is partially filled with electrons, or an element that has the ability to form stable cations with an incompletely filled d orbital.
General electronic configuration of inner transition elements is (n−2)f1−14(n−1)d0−1ns2.
15. After having completely filled d orbitals (4d10) in silver atom in its ground state, how can you say that silver is a transition element?
Ans: The general electronic configuration of transition elements is (n−1)d1−10ns0−2, where n is the principal quantum number. Silver (Ag) has the configuration [Kr]4d105s1, with a completely filled 4d10 orbital. However, it is a transition element because its ion (Ag⁺) has an incomplete d-subshell (4d10).
Or
Zn2+ salt are colourless, while Cu2+ salts are coloured. Give reason.
Ans: The difference in the color of Zn²⁺ and Cu²⁺ salts can be attributed to the electronic configuration and d-orbital activity of these ions. Zn²⁺ has a 3d¹⁰ configuration, meaning all its d-orbitals are fully occupied, so no electronic transitions occur within the d-orbitals when light is absorbed. As a result, Zn²⁺ salts appear colorless. On the other hand, Cu²⁺ has a 3d⁹ configuration, with an incomplete d-orbital. This allows electronic transitions between d-orbitals upon absorption of visible light, leading to the absorption of specific wavelengths and the emission of colored light. Zn2+ has 3d10 configuration and no unpaired electrons, hence it is colourless. Whereas in Cu2+ state which has 3d9 configuration there is one unpaired electron and it undergoes d-d transitions emitting colour so Cu2+ salts are blue in colour.
16. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Ans: The steady decrease in the atomic size of lanthanides on increasing the atomic number is due to the increasing the nuclear charge and electrons entering the inner ( n – 2 ) f orbital or anti penultimate shell.
Lanthanoid contraction leads to similar ionic radii among successive lanthanides, affecting their chemical behavior, making them hard to separate. It also influences the chemistry of elements in higher periods.
17. Explain why [Ti(H2O)6]3+ is violet in colour.
Ans: Any compound or ion showing colour is due to presence of unpaired electron. [Ti(H2O)6]+3 has d1 configuration and have one unpaired electron and because of this it shows d-d transition and has purple colour. The energy difference between the d-orbitals causes the absorption of light in the visible region, specifically in the red-yellow part of the spectrum, resulting in the complementary violet color being observed. This absorption occurs due to electronic transitions from lower to higher energy d-orbitals in the presence of the ligand field.
Or
[NiCl4]2- is paramagnetic while (Ni(CO)4) is diamagnetic though both are tetrahedral. Explain why.
Ans: Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl4]2− is paramagnetic. In [NiCl₄]²⁻, nickel is in the +2 oxidation state (Ni²⁺), with a 3d⁸ configuration. Chloride ions are weak field ligands, so they don’t cause significant splitting of the d-orbitals. This results in unpaired electrons and makes [NiCl₄]²⁻ paramagnetic. In Ni(CO)₄, nickel is in the zero oxidation state (Ni⁰), with a 3d⁸4s² configuration.
But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, causing pairing of electrons in the d-orbitals, making Ni(CO)₄ diamagnetic.
18. (a) Draw structures of geometrical isomers of Fe(NH3)2 (CN)4]–.
Ans: Structures of geometrical isomers of Fe(NH3)2 (CN)4]–.
(b) Out of the following two coordination entities which is chiral (optically active)? Explain.
Ans: To determine which coordination entity is chiral, we analyze their structures for symmetry. A chiral coordination compound lacks a plane of symmetry and cannot be superimposed on its mirror image. A simple example of a chiral coordination compound is [Co(en)₃]³⁺, where “en” is ethylenediamine, a bidentate ligand. Its three ligands create a non-superimposable mirror image, making it optically active.
19. Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Ans: A chemical test that is most commonly used for the identification of primary, secondary and tertiary amines is called the Hinsberg test. An amine in the presence of an aqueous alkali interacts with a Hinsberg reagent. Thus, this is what is meant as the Hinsberg test.
Primary amine reaction: RNH₂+C6H5SO2Cl→RNH-SO₂C₆H₅ (soluble)
Secondary amine reaction:
R₂NH+C6H5SO2Cl→R₂N-SO₂C₆H₅(insoluble)
Tertiary amine reaction: R₃N+C6H5SO2Cl→No reaction
20. The time required for 10% completion of a first order reaction at 298K is equal to that required for 25% completion at 308K. Calculate Ea.
Ans: To calculate the activation energy (Ea):
Given: Time for 10% completion at 298K equals time for 25% completion at 308K for a first-order reaction.
The rate constants k1 are related as k2 = k2/k1 = ⅖.
Apply the Arrhenius equation:
In (k2/k1) = Ea/R(1/T1-1/T2)
Substitute k2 = k2/k1 = ⅖
T1 = 298K ,
T2 = 308K.
1/298−1/308 = 0.000112
In(⅖) = In(0.4) = – 0.916
− 0.916 = (Ea/ 8.314 × 0.000112)
Ea = (−0.916×8.314/ 0.000112)
= 9-7.625/ 0.000112) = 68,090J/mol
Thus, the activation energy Ea is approximately 68.1 kJ/mol.
21. Write the products of the following reactions:
(a)
Ans: Reaction: C₆H₅CH = CH₂ + HBr → C₆H₅CHBrCH₃
The resulting product is 1-phenylethyl bromide.
(b)
Ans: Reaction: CH₃-CH₂-CH=CH₂ + HCl → CH₃-CH₂-CHCl-CH₃
The resulting product is 2-chlorobutane.
(c)
Ans: Reaction: C₆H₅CH=CH₂ + HBr (peroxide) → C₆H₅CH₂CH₂Br
The resulting product is 1-phenylethyl bromide.
(d)
Ans: Reaction: C₆H₁₁CH = CH₂ + HCl → C₆H₁₁CHClCH₃
The resulting product is 1-chloro-1-methylcyclohexane.
22. How will you bring about the following conversions?
(a) Ethane to bromoethene.
Ans: Ethane is converted to bromoethene by first reacting ethane with bromine in presence of U.V light to form bromoethane. Then bromoethane is reacted with alcoholic potassium hydroxide to form ethene.
(b) Bromomethane to propanone Antid.
Ans: Bromomethane is converted to bromoethene by first React bromomethane with sodium cyanide (NaCN) to form acetonitrile (CH₃CN). Then Bromomethane Hydrolyze acetonitrile with dilute acid (H₂O/H⁺) to yield propanone (CH₃COCH₃), through nucleophilic attack by water on the nitrile group.
(c) Benzene to biphenyl.
Ans: To convert benzene to biphenyl, use the Wurtz-Fittig reaction. React benzene (C₆H₆) with bromobenzene (C₆H₅Br) in the presence of sodium (Na) in dry ether. This reaction promotes the coupling of two benzene molecules, forming biphenyl (C₆H₅–C₆H₅) through a radical mechanism.
(d) Ethyl magnesium chloride to Propan-1-o1.
Ans: Ethyl magnesium chloride (C₂H₅MgCl), a Grignard reagent, reacts with formaldehyde (H₂C = O) to form an intermediate adduct, which, upon hydrolysis, yields propan-1-ol (C₃H₇OH). The reaction is as follows:
C₂H₅MgCl + H₂C = O → (C₂H₅)₂C(OMgCl)₂.
The ethyl group (C₂H₅⁻) acts as a nucleophile and attacks the electrophilic carbonyl carbon of formaldehyde, resulting in the formation of a tetrahedral alkoxide intermediate. In the second step, hydrolysis of this intermediate produces propan-1-ol.
23. Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
Ans: The structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names
OR
Give the structures and IUPAC names of the products expected from the following reactions:
(a) Catalytic reduction of butanal
Ans: IUPAC names: Butan-1-ol
Structure: CH₃-CH₂-CH₂-CH₂OH
(b) Hydration of propene in the presence of dilute sulphuric acid
Ans: IUPAC names: Propan-2-ol
Structure: CH₃-CH(OH)-CH₃
(c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis
Ans: IUPAC names: Hydrolysis → 2-Methylpropan-2-ol
Structure: CH₃-C(OH)(CH₃)-CH₃
24. Explain why:
(a) Ortho nitrophenols are more acidic than phenol.
Ans: Ortho-nitrophenols are more acidic than phenol due to the presence of the nitro group at the ortho position. The nitro-group is an electron-withdrawing group. The presence of this group in the ortho or para position decreases the electron density in the OH bond. As a result, it is easier to lose a proton. This hydrogen bonding stabilizes the negative charge on the oxygen atom of the phenoxide ion (after deprotonation), making it easier for the proton to dissociate, and thereby increasing the acidity of ortho-nitrophenol compared to phenol. Hence, ortho and para nitrophenols are stronger acids than phenol.
(b) Ethanol has higher boiling point than that of the methoxymethane.
Ans: Ethanol has a higher boiling point than methoxymethane because ethanol has hydrogen bonding, which requires more energy to break than the weaker forces in methoxymethane.
(c) Ortho nitrophenol is more acidic than ortho methoxyphenol.
Ans: Ortho nitrophenol is more acidic than ortho methoxyphenol because nitro group is an electron withdrawing and it will increase + ve charge on the oxygen atom to make it more acidic.
(d) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
Ans: Cyclohexanone forms cyanohydrin in good yield but 2,2,6 trimethylcyclohexanone does not. ii. There are two -NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
25. An organic compound with the molecular formula CoH100 forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Ans: The compound with the molecular formula C₆H₁₀₀ forms a 2,4-dinitrophenylhydrazone (2,4-DNP) derivative, indicating the presence of an aldehyde or ketone group. It reduces Tollens’ reagent, confirming the presence of an aldehyde group. The compound undergoes the Cannizzaro reaction, which occurs with aldehydes that do not have an α-hydrogen, suggesting it is a non-α-hydrogen aldehyde. Upon vigorous oxidation, it gives 1,2-benzenedicarboxylic acid (phthalic acid), indicating that the aldehyde group is attached to a benzene ring. Therefore, the compound is benzaldehyde (C₆H₅CHO).
OR
Give simple chemical tests to distinguish between the following pairs of compounds
(a) Acetophenone and Benzophenone.
Ans: Benzophenone: Benzophenone does not have any methyl group at the alpha position, so does not give an iodoform test.
Acetophenone: Acetophenone does have a methyl group at the alpha position, so give a positive iodoform test.
(b) Phenol and Benzoic acid.
Ans: Phenol: Phenol reacts with neutral FeCl3 to form an iron-phenol complex giving violet colouration.
Benzoic acid: Benzoic acid reacts with neutral FeCl3 to give a buff coloured ppt.
(c) Ethanal and Propanal.
Ans: Ethanal: Methyl ketones give a positive iodoform test. Ethanal has one methyl group linked to the carbonyl carbon atom and therefore, responds to this test. Propanal does not respond to this test.
Propanal: Propanal being an aldehyde reduces Fehling’s solution to form a red-brown precipitate of cuprous oxide.
26. Complete the following reactions:
(a)
Ans: Reaction: COOH + SOCl2 → COCl + SO2 + HCl
(b)
Ans: Reaction: C₆H₅CHO + NH₂CONHNH₂ → C₆H₅CH=NNHCONH₂ + H₂O
(c)
Ans: Reaction: CH₃COCH₂COOC₂H₅ + NaBH₄ → CH₃CH(OH)CH₂COOC₂H₅
(d)
Ans: Reaction: Cyclohexanol + CrO₃ → Cyclohexanone + CrO₂ + H₂O
27. Give reasons for the following:
(a) pK of aniline is more than that of methylamine.
Ans: The pK of aniline is higher than methylamine because the amino group in aniline is less basic. The lone pair of electrons on the nitrogen of aniline is delocalized into the benzene ring, reducing its availability for protonation, making it less basic than methylamine, where the lone pair on nitrogen is more readily available. On the other hand in case of methylamine due to the +I effect of methyl group the electron density on the N-atom is increased. As a result aniline is less basic than methylamine. Thus pKb of aniline is more than that of methylamine.
(b) Aniline does not undergo Friedel-Crafts reaction.
Ans: Friedel-Crafts reaction is carried out in the presence of AlCl3. But AlCl3 is acidic in nature while aniline is a strong base. Thus aniline reacts with AlCl3 to form a salt as shown in the following equation.Due to the positive charge on the N-atom electrophilic substitution in the benzene ring is deactivated, but it also makes the nitrogen more basic. The nitrogen’s lone pair coordinates with the Lewis acid catalyst (like AlCl₃), deactivating the ring and preventing the reaction.
(c) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Ans: Gabriel phthalimide reaction gives pure primary amines without any contamination of secondary and tertiary amines. Therefore, it is preferred for synthesising primary amines. The reaction involves the nucleophilic substitution of phthalimide with an alkyl halide, followed by hydrolysis to give a primary amine.
(d) Aliphatic amines are stronger bases than aromatic amines.
Ans: In aromatic amines, the −NH2 group is attached to a −C6H5 group, which is an electron withdrawing group. So, the availability of a lone pair of electrons on N is decreased. Therefore aliphatic amines are more basic than aromatic amines.In aromatic amines, the lone pair is delocalized into the aromatic ring, making it less available for protonation, thus reducing the basicity.
28. Complete the following reactions:
(i) C6H5NH2 + CHCl3 + ale. KОН →
Ans: C6H5NH2 + CHCl3 + ale. KОН → C₆H₅NC + 3KCl + 3H₂O
(ii) C6H5OH + HNO3 (conc.)→
Ans: C6H5OH + HNO3 (conc.)→C₆H₄(NO₂)OH + H₂O
(iii) C6H5N₂C1+CH₂OH →
Ans: C6H5N₂C1+CH₂OH →C₆H₅OCH₃ + N₂ + HCl
(iv) CH3CONH₂ + Br₂ +4NaOH→
Ans: CH3CONH₂ + Br₂ +4NaOH→CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
(v) C6H5CHO+HNO3/H2SO4+ Δ→
Ans: C6H5CHO+HNO3/H2SO4+ Δ→C₆H₄(NO₂)CHO + H₂O
(vi) C6H5ONa+C₂H₃Cl→
Ans: C6H5ONa+C₂H₃Cl→C₆H₅OCH₂CH₃ + NaCl
29. (a) What are essential and non-essential amino acids? Give two examples of each type.
Ans: Essential amino acid: Essential amino acids are those that the body cannot synthesize on its own and must be obtained from food sources.
Examples: include leucine and valine.
Non-essential amino acids: Non-essential amino acids, on the other hand, can be synthesized by the body.
Examples: include alanine and glutamine.
(b) How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Ans: Vitamins are classified into two groups based on their solubility: the fat-soluble vitamins (A, D, E and K) and the water-soluble vitamins (B vitamins, vitamin C and folate, to name a few). Fat-soluble vitamins are usually absorbed passively and must be transported with dietary lipids.
Vitamin K is responsible for the coagulation of blood.
30. (a) What are nucleic acids? Mention their two important functions.
Ans: Nucleic acids are large biomolecules that play essential roles in all cells and viruses. A major function of nucleic acids involves the storage and expression of genomic information. The two main types of nucleic acids are DNA (Deoxyribonucleic acid), RNA (Ribonucleic acid).
Here are two functions of nucleic acids:
(i) Responsible for protein synthesis in cell.
(ii) Responsible for transfer of characters from one generation to another.
OR
Write the important structural and functional differences between DNA and RNA.
Ans: Structural and Functional Differences Between DNA and RNA:
| Basic of difference | DNA (Deoxyribonucleic Acid) | RNA (Ribonucleic Acid) |
| Sugar | Deoxyribose (lacks one oxygen atom at the 2′ carbon) | Ribose (has a hydroxyl group (-OH) at the 2′ carbon) |
| Strands | Double-stranded (double helix structure) | Single-stranded |
| Types | Only one type (DNA) | Several types (mRNA, tRNA, rRNA, etc.) |
(b) Write a way to determine the A value of water.
Ans: The A value of water, which represents its affinity for moisture, can be determined using a hygroscopicity experiment. By placing a known quantity of dry water in a sealed container with a controlled humidity level, the increase in mass due to absorbed water vapor is measured. The A value is calculated based on the mass change relative to the dry weight.
OR
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Ans: Werner’s postulates explain the bonding in coordination compounds as follows:i A metal exhibits two types of valencies namely primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valencies are satisfied by both negative and neutral ions. The metal ion forms primary bonds with ligands in fixed geometries (octahedral, tetrahedral, square planar), and secondary bonds determine the compound’s overall stability and color. The metal exhibits specific coordination numbers.
