CBSE Class 12 Chemistry Coordination Compounds MCQs Set 11

Question 1. How many ions will be produced by the complex compound \( [\text{Cr(en)}_3]\text{Cl}_3 \), when it is dissolved in water?
(a) 2
(b) 4
(c) 7
(d) 10
Answer: (b) 4
When \( [\text{Cr(en)}_3]\text{Cl}_3 \) dissolves in water, it yields three \( \text{Cl}^- \) ions and one \( [\text{Cr(en)}_3]^{3+} \) cationic complex. Thus, the total quantity of ions produced is four.
In simple words: The complex splits into one big positive ion and three small negative chloride ions, making four ions in total.

Exam Tip: Remember that the species inside the coordination sphere (square brackets) remains intact as a single entity when dissolved in water.

Question 2. On adding \( \text{AgNO}_3 \) solution to 1 mole of \( \text{PdCl}_2 \cdot 4\text{NH}_3 \), two moles of \( \text{AgCl} \) are formed. The secondary valency of Pd in the complex will be
(a) 0
(b) 2
(c) 4
(d) 1
Answer: (c) 4
The reaction occurs as: \( \text{PdCl}_2 \cdot 4\text{NH}_3 + 2\text{AgNO}_3 \rightarrow [\text{Pd(NH}_3)_4]^{2+} + 2\text{AgCl} + 2\text{NO}_3^- \). Because four \( \text{NH}_3 \) ligands are bound directly to the palladium metal ion, its coordination number or secondary valency is four.
In simple words: Since two chlorides form a precipitate, they must be outside the bracket. This leaves four ammonias inside the bracket directly attached to Pd, which gives a coordination number of four.

Exam Tip: Secondary valency corresponds directly to the coordination number of the central metal ion in the complex.

Question 3. In the complex compound \( \text{Fe}_4[\text{Fe(CN)}_6]_3 \), oxidation states of counter ion Fe and central metal ion Fe respectively are
(a) II, III
(b) III, II
(c) IV, III
(d) II, II
Answer: (b) III, II
The compound dissociates as follows: \( \text{Fe}_4[\text{Fe(CN)}_6]_3 \rightarrow 4\text{Fe}^{3+} + 3[\text{Fe(CN)}_6]^{4-} \). Hence, the oxidation state of the counter ion is \( +3 \). For the coordination sphere \( [\text{Fe(CN)}_6]^{4-} \), let the oxidation state of the central iron be \( x \): \( x + 6(-1) = -4 \implies x = +2 \). Therefore, the respective oxidation states are III and II.
In simple words: When the compound splits, the iron outside the brackets has a charge of \( +3 \), while the iron inside the brackets has a charge of \( +2 \).

Exam Tip: Cross-multiply the subscripts to easily find the charges on the counter ion and the complex coordination sphere before solving for the central atom.

Question 4. In which of the following does the central atom exhibit an oxidation state of +3 ?
(a) \( \text{K}_2[\text{Ni(CN)}_4] \)
(b) \( \text{K}_4[\text{Fe(CN)}_6] \)
(c) \( [\text{Fe(C}_2\text{O}_4)_3]^{3-} \)
(d) \( [\text{Cu(NH}_3)_4]^{2+} \)
Answer: (c) \( [\text{Fe(C}_2\text{O}_4)_3]^{3-} \)
In \( [\text{Fe(C}_2\text{O}_4)_3]^{3-} \), let the oxidation state of iron be \( x \). Since oxalate is a bidentate ligand carrying a \( -2 \) charge, we get: \( x + 3(-2) = -3 \implies x = +3 \).
In simple words: The three oxalate ligands have a total charge of \( -6 \). Since the entire complex has a charge of \( -3 \), the iron must have a charge of \( +3 \) to balance it.

Exam Tip: Always remember to check the correct charge on common ligands like oxalate (\( \text{C}_2\text{O}_4^{2-} \)) to prevent arithmetic errors.

Question 5. A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent?
(a) Thiosulphate
(b) Oxalato
(c) Glycinato
(d) Ethane-1, 2-diamine
Answer: (a) Thiosulphate
Thiosulphate (\( \text{S}_2\text{O}_3^{2-} \joinrel \)) is not a chelating ligand because its geometric structure makes it difficult to bind to the same metal ion through two different atoms simultaneously.
In simple words: Even though thiosulphate has multiple potential donor atoms, they are not arranged in a way that allows them to grab the same metal ion together to form a ring.

Exam Tip: Chelating ligands must be able to form stable 5- or 6-membered ring structures with the central metal ion.

Question 6. Which of the following complexes is a 'Chelate' complex?
(a) \( [\text{Co(NH}_3)_6]^{3+} \)
(b) \( [\text{Co(en)}_3]^{3+} \)
(c) \( [\text{Co(NH}_3)_4\text{Cl}_2]^+ \)
(d) \( [\text{CoF}_6]^{3-} \)
Answer: (b) \( [\text{Co(en)}_3]^{3+} \)
The complex \( [\text{Co(en)}_3]^{3+} \) is a chelate complex. In this species, 'en' (ethane-1,2-diamine) acts as a bidentate chelating ligand that forms stable ring structures with the central cobalt ion.
In simple words: Ethane-1,2-diamine has two donor atoms that bind to the cobalt ion, forming a stable ring structure known as a chelate ring.

Exam Tip: Look for polydentate ligands like 'en' or 'ox' to easily identify chelate complexes.

Question 7. Amongst the following, the most stable complex is
(a) \( [\text{Fe(H}_2\text{O})_6]^{3+} \)
(b) \( [\text{Fe(NH}_3)_6]^{3+} \)
(c) \( [\text{Fe(C}_2\text{O}_4)_3]^{3-} \)
(d) \( [\text{FeCl}_6]^{3-} \)
Answer: (c) \( [\text{Fe(C}_2\text{O}_4)_3]^{3-} \)
In all the given options, iron exists in the \( +3 \) oxidation state. However, the complex \( [\text{Fe(C}_2\text{O}_4)_3]^{3-} \) is exceptionally stable because the oxalate ions act as bidentate chelating ligands, creating stable five-membered rings.
In simple words: Chelating ligands form rings with the metal, which makes the entire structure much more stable than complexes containing only simple, non-ring-forming ligands.

Exam Tip: The chelate effect always enhances the thermodynamic stability of coordination complexes compared to similar complexes with monodentate ligands.

Question 8. The formula of the complex dichloridobis(ethane-1, 2-diamine)platinum(IV) nitrate is
(a) \( [\text{PtCl}_2(\text{en})_2(\text{NO}_3)_2] \)
(b) \( [\text{PtCl}_2(\text{en})_2](\text{NO}_3)_2 \)
(c) \( [\text{PtCl}_2(\text{en})_2(\text{NO}_3)]\text{NO}_3 \)
(d) \( [\text{Pt(en)}_2(\text{NO}_3)_2]\text{Cl}_2 \)
Answer: (b) \( [\text{PtCl}_2(\text{en})_2](\text{NO}_3)_2 \)
The correct chemical representation for dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate is \( [\text{PtCl}_2(\text{en})_2](\text{NO}_3)_2 \). The nitrate ions balance the \( +2 \) charge of the coordination sphere.
In simple words: The name tells us that platinum, two chlorides, and two 'en' molecules are inside the bracket, while the nitrate ions are outside to balance the overall charge.

Exam Tip: Work out the charge of the complex cation by summing the oxidation states of the metal and inner ligands to find the correct number of counter ions outside.

Question 9. The correct IUPAC name of \( [\text{Pt(NH}_3)_2\text{Cl}_2]^{2+} \) is
(a) Diamminedichloridoplatinum (II)
(b) Diamminedichloridoplatinum (IV)
(c) Diamminedichloridoplatinum (0)
(d) Diamminedichloridoplatinate (IV)
Answer: (b) Diamminedichloridoplatinum (IV)
Since the complex carries a \( +2 \) charge, the oxidation state of platinum is calculated as: \( x + 2(0) + 2(-1) = +2 \implies x = +4 \). Thus, the correct name is diamminedichloridoplatinum (IV).
In simple words: The platinum ion has an oxidation state of \( +4 \), and the ligands are named alphabetically: 'diammine' before 'dichlorido'.

Exam Tip: Remember that if the coordination sphere has a positive charge, the metal name ends with its standard name (platinum), not with '-ate' (platinate).

Question 10. Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type \( [\text{Pd(C}_6\text{H}_5)_2(\text{SCN})_2] \) and \( [\text{Pd(C}_6\text{H}_5)_2(\text{NCS})_2] \) are
(a) linkage isomers
(b) coordination isomers
(c) ionisation isomers
(d) geometrical isomers
Answer: (a) linkage isomers
Because the thiocyanate ligand (\( \text{SCN}^- \)) is ambidentate, it can bind to palladium through either the sulfur atom or the nitrogen atom. This leads to linkage isomerism.
In simple words: Linkage isomerism happens when a ligand can attach itself to the central metal in two different ways, like using either its nitrogen or sulfur atom.

Exam Tip: Look for classic ambidentate ligands like \( \text{SCN}^- \), \( \text{NO}_2^- \), or \( \text{CN}^- \) to quickly identify linkage isomers.

Question 11. \( [\text{Co(NH}_3)_5\text{NO}_3]\text{SO}_4 \) and \( [\text{Co(NH}_3)_5\text{SO}_4]\text{NO}_3 \) exhibit
(a) linkage isomerism
(b) ionisation isomerism
(c) optical isomerism
(d) coordination isomerism
Answer: (b) ionisation isomerism
These two coordination compounds yield different ions when dissolved in water, even though they share the same overall chemical formula. This behavior is called ionisation isomerism.
In simple words: Ionisation isomers swap their counter ions (the ones outside the brackets) with the ligands inside the brackets, producing different ions in solution.

Exam Tip: Test for ionisation isomerism by checking if the compounds release different anions upon dissolving in water.

Question 12. A co-ordination compound pentaamminechloridocobalt(III) sulphate is dissolved in water. When a few drops of chemical 'A' is added to the solution, it gives white precipitate. Identify chemical 'A'.
(a) \( \text{AgCl} \)
(b) \( \text{AgNO}_3 \)
(c) \( \text{BaSO}_4 \)
(d) \( \text{BaCl}_2 \)
Answer: (d) \( \text{BaCl}_2 \)
When pentaamminechloridocobalt(III) sulphate, \( [\text{Co(NH}_3)_5\text{Cl]SO}_4 \), dissolves in water, it releases free sulphate (\( \text{SO}_4^{2-} \)) ions. Adding barium chloride (\( \text{BaCl}_2 \)) results in the formation of a white precipitate of barium sulphate (\( \text{BaSO}_4 \)).
In simple words: The sulphate group is outside the coordination sphere, so it reacts with barium ions from barium chloride to form a white solid that doesn't dissolve.

Exam Tip: Identify the counter ion (in this case, sulphate) to predict which analytical reagent will form a characteristic precipitate.

Question 13. The complexes \( [\text{Co(NH}_3)_6][\text{Cr(CN)}_6] \) and \( [\text{Cr(NH}_3)_6][\text{Co(CN)}_6] \) are the examples of which type of isomerism?
(a) Linkage isomerism
(b) Ionisation isomerism
(c) Coordination isomerism
(d) Geometrical isomerism
Answer: (c) Coordination isomerism
These compounds exhibit coordination isomerism. This phenomenon is observed in systems containing both complex cations and complex anions, where the ligands are exchanged between the two metal centers.
In simple words: Coordination isomerism is like a partner swap where the ligands switch places between the positive and negative complex parts of the compounds.

Exam Tip: Coordination isomerism is easy to recognize because both the cation and the anion in the formula are enclosed in square brackets.

Question 14. Indicate the complex ion which shows geometrical isomerism.
(a) \( [\text{Cr(H}_2\text{O})_4\text{Cl}_2]^+ \)
(b) \( [\text{Pt(NH}_3)_3\text{Cl}]^+ \)
(c) \( [\text{Co(NH}_3)_6]^{3+} \)
(d) \( [\text{Co(CN)}_5(\text{NC})]^{3-} \)
Answer: (a) \( [\text{Cr(H}_2\text{O})_4\text{Cl}_2]^+ \)
The complex ion \( [\text{Cr(H}_2\text{O})_4\text{Cl}_2]^+ \) exhibits geometrical isomerism. As a coordination compound of the \( \text{MA}_4\text{B}_2 \) type, it can exist in cis and trans arrangements depending on the positions of the two chloride ligands.
In simple words: Since it has four of one ligand and two of another, the two identical chloride groups can either be right next to each other (cis) or opposite each other (trans).

Exam Tip: For octahedral complexes, the minimum requirement for geometrical isomerism is typically a formula like \( \text{MA}_4\text{B}_2 \) or \( \text{MA}_3\text{B}_3 \).

Question 15. \( [\text{M(AA)X}_2\text{Y}_2] \) is a type of a co-ordinate compound in which M = metal ion, AA = didentate ligand, X = monodentate ligand, and Y = monodentate ligand. Which of the following isomerisms does this compound exhibit?
(a) Co-ordination isomerism
(b) Linkage isomerism
(c) Geometrical isomerism
(d) Optical isomerism
Answer: (d) Optical isomerism
The complex \( [\text{M(AA)X}_2\text{Y}_2] \) has a configuration that lacks a plane of symmetry in its active forms. Consequently, it exhibits optical isomerism due to the presence of non-superimposable mirror images.
In simple words: Since this molecule is asymmetrical, it can exist as left-handed and right-handed forms that are mirror images of each other.

Exam Tip: Remember that octahedral complexes containing didentate ligands often lack a plane of symmetry in their cis or unsymmetrical configurations, leading to optical activity.

Question 16. Optical isomerism is not shown by the complex
(a) \( [\text{Co(en)}_2\text{Cl}_2]^+ \) trans-form
(b) \( [\text{Co(en)}_2\text{Cl}_2]^+ \) cis-form
(c) \( [\text{Cr(ox)}_2]^{2-} \)
(d) \( [\text{Cr(en)}_3]^{3+} \)
Answer: (a) \( [\text{Co(en)}_2\text{Cl}_2]^+ \) trans-form
The trans-isomer of \( [\text{Co(en)}_2\text{Cl}_2]^+ \) does not show optical isomerism. This is because the trans-form possesses a plane of symmetry, making its mirror image completely superimposable.
In simple words: The trans form is perfectly symmetrical, so its mirror image is identical to itself. This prevents it from being optically active.

Exam Tip: Always remember that trans-isomers of type \( [\text{M(AA)}_2\text{X}_2] \) are symmetrical and optically inactive, whereas their cis-isomers are chiral and optically active.

Question 17. The magnetic moment of \( [\text{NiCl}_4]^{2-} \) is
(a) 1.82 BM
(b) 2.82 BM
(c) 4.42 BM
(d) 5.46 BM
Answer: (b) 2.82 BM
In \( [\text{NiCl}_4]^{2-} \), nickel has an oxidation state of \( +2 \), corresponding to a \( 3d^8 \) electronic configuration. Because chloride is a weak field ligand, no pairing occurs, leaving two unpaired electrons (\( n = 2 \)). The magnetic moment is calculated as: \[ \mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \text{ BM} \]
In simple words: Since the weak field chloride ligands do not pair up the electrons, nickel has two unpaired electrons, giving a magnetic moment of 2.82 Bohr Magnetons.

Exam Tip: You can quickly estimate the magnetic moment: if the number of unpaired electrons is \( n \), the magnetic moment is always slightly above \( n \text{ BM} \) (e.g., for \( n=2 \), it is \( \approx 2.8 \text{ BM} \)).

Question 18. Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?
(a) \( [\text{Ni(NH}_3)_6]^{2+} \)
(b) \( [\text{Zn(NH}_3)_6]^{2+} \)
(c) \( [\text{Cr(NH}_3)_6]^{3+} \)
(d) \( [\text{Co(NH}_3)_6]^{3+} \)
Answer: (a) \( [\text{Ni(NH}_3)_6]^{2+} \)
The complex \( [\text{Ni(NH}_3)_6]^{2+} \) uses \( 4d \)-orbitals for bonding, making it an outer orbital complex. The central nickel ion has a \( 3d^8 \) configuration with two unpaired electrons, which accounts for its paramagnetic behavior.
In simple words: Because nickel has eight d-electrons, it cannot clear up two inner d-orbitals for bonding, so it must use outer orbitals, leaving two unpaired electrons that make it magnetic.

Exam Tip: Complexes of metal ions with \( d^8, d^9, d^{10} \) configurations always form outer orbital octahedral complexes because the inner \( 3d \) orbitals cannot be made vacant.

Question 19. Which one of these statements about \( [\text{Co(CN)}_6]^{3-} \) is true?
(a) \( [\text{Co(CN)}_6]^{3-} \) has no unpaired electrons and will be in a low-spin configuration.
(b) \( [\text{Co(CN)}_6]^{3-} \) has four unpaired electrons and will be in a low-spin configuration.
(c) \( [\text{Co(CN)}_6]^{3-} \) has four unpaired electrons and will be in a high-spin configuration.
(d) \( [\text{Co(CN)}_6]^{3-} \) has no unpaired electrons and will be in a high-spin configuration.
Answer: (a) \( [\text{Co(CN)}_6]^{3-} \) has no unpaired electrons and will be in a low-spin configuration.
In \( [\text{Co(CN)}_6]^{3-} \), cobalt is in the \( +3 \) oxidation state with a \( 3d^6 \) configuration. Because cyanide is a strong field ligand, it causes complete pairing of the six d-electrons in the \( t_{2g} \) level, resulting in a low-spin configuration with zero unpaired electrons.
In simple words: The cyanide ligands are strong enough to force all of cobalt's six d-electrons to pair up nicely, leaving no unpaired electrons behind in a low-spin state.

Exam Tip: Cobalt(III) almost always forms low-spin, diamagnetic octahedral complexes with strong field ligands like \( \text{CN}^- \).

Question 20. Which of the following facts about the complex \( [\text{Cr(NH}_3)_6]\text{Cl}_3 \) is incorrect?
(a) The complex involves \( d^2sp^3 \) hybridisation and is octahedral in shape.
(b) The complex is paramagnetic.
(c) The complex is in outer orbital complex.
(d) The complex gives white precipitate with \( \text{AgNO}_3 \) solution.
Answer: (c) The complex is in outer orbital complex.
The complex \( [\text{Cr(NH}_3)_6]\text{Cl}_3 \) has a central \( \text{Cr}^{3+} \) ion with a \( 3d^3 \) electronic configuration. It undergoes \( d^2sp^3 \) hybridisation to form an inner orbital complex, which makes statement (c) incorrect.
In simple words: Since chromium(III) has only three electrons in its 3d subshell, the two inner d-orbitals are naturally empty and ready for bonding, making this an inner orbital complex, not an outer one.

Exam Tip: Remember that metal ions with \( d^1, d^2, d^3 \) configurations always form inner orbital complexes regardless of whether the ligand is weak or strong.

Question 21. Atomic number of Mn, Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?
I. \( [\text{Co(NH}_3)_6]^{3+} \)
II. \( [\text{Mn(CN)}_6]^{3-} \)
III. \( [\text{Fe(CN)}_6]^{4-} \)
IV. \( [\text{Fe(CN)}_6]^{3-} \)

(a) II and III
(b) I and IV
(c) I and III
(d) II and IV
Answer: (c) I and III
Both \( [\text{Co(NH}_3)_6]^{3+} \) (with \( \text{Co}^{3+}: 3d^6 \)) and \( [\text{Fe(CN)}_6]^{4-} \) (with \( \text{Fe}^{2+}: 3d^6 \)) contain six d-electrons. Under the influence of strong field ligands, all six electrons pair up completely in the \( t_{2g} \) orbitals, making these complexes diamagnetic.
In simple words: In both of these complexes, the metal has six d-electrons, and the strong ligands force them all to pair up in pairs of two, leaving zero unpaired electrons.

Exam Tip: A \( d^6 \) configuration in a strong ligand field results in a completely filled \( t_{2g} \) subshell, which is highly stable and diamagnetic.

Question 22. The CFSE for octahedral \( [\text{CoCl}_6]^{4-} \) is 18,000 \( \text{cm}^{-1} \). The CFSE for tetrahedral \( [\text{CoCl}_4]^{2-} \) will be
(a) 18,000 \( \text{cm}^{-1} \)
(b) 16,000 \( \text{cm}^{-1} \)
(c) 8,000 \( \text{cm}^{-1} \)
(d) 20,000 \( \text{cm}^{-1} \)
Answer: (c) 8,000 \( \text{cm}^{-1} \)
The crystal field splitting energy for a tetrahedral complex (\( \Delta_t \)) is related to that of an octahedral complex (\( \Delta_o \)) by the formula: \[ \Delta_t = \frac{4}{9}\Delta_o \] Substituting the given value: \[ \Delta_t = \frac{4}{9} \times 18,000 \text{ cm}^{-1} = 8,000 \text{ cm}^{-1} \]
In simple words: The tetrahedral splitting energy is always smaller than the octahedral splitting energy, being exactly four-ninths of its value.

Exam Tip: Always use the relationship \( \Delta_t = \frac{4}{9} \Delta_o \) to quickly convert crystal field splitting parameters between octahedral and tetrahedral geometries.

Question 23. The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, \( [\text{Co(NH}_3)_6]^{3+} \), \( [\text{Co(CN)}_6]^{3-} \), \( [\text{Co(H}_2\text{O})_6]^{3+} \)?
(a) \( [\text{Co(CN)}_6]^{3-} > [\text{Co(NH}_3)_6]^{3+} > [\text{Co(H}_2\text{O})_6]^{3+} \)
(b) \( [\text{Co(NH}_3)_6]^{3+} > [\text{Co(H}_2\text{O})_6]^{3+} > [\text{Co(CN)}_6]^{3-} \)
(c) \( [\text{Co(H}_2\text{O})_6]^{3+} > [\text{Co(NH}_3)_6]^{3+} > [\text{Co(CN)}_6]^{3-} \)
(d) \( [\text{Co(CN)}_6]^{3-} > [\text{Co(NH}_3)_6]^{3+} > [\text{Co(H}_2\text{O})_6]^{3+} \)
Answer: (c) \( [\text{Co(H}_2\text{O})_6]^{3+} > [\text{Co(NH}_3)_6]^{3+} > [\text{Co(CN)}_6]^{3-} \)
The energy absorbed is inversely proportional to the wavelength of light absorbed (\( \Delta E = \frac{hc}{\lambda} \)). Since ligand strength increases in the order \( \text{H}_2\text{O} < \text{NH}_3 < \text{CN}^- \), the crystal field splitting energy (\( \Delta E \)) also increases in this order. Thus, the wavelength of absorbed light (\( \lambda \)) follows the opposite order: \( [\text{Co(H}_2\text{O})_6]^{3+} > [\text{Co(NH}_3)_6]^{3+} > [\text{Co(CN)}_6]^{3-} \).
In simple words: Stronger ligands cause larger energy splits, which means they absorb light with higher energy (shorter wavelength). So, the weakest ligand absorbs the longest wavelength.

Exam Tip: Remember that ligand field strength is directly proportional to splitting energy and inversely proportional to the wavelength of absorption.

Question 24. EDTA is used for the estimation of
(a) \( \text{Na}^+ \) and \( \text{K}^+ \) ions
(b) \( \text{Cu}^{2+} \) and \( \text{Cs}^+ \) ions
(c) \( \text{Cl}^- \) and \( \text{Br}^- \) ions
(d) \( \text{Ca}^{2+} \) and \( \text{Mg}^{2+} \) ions
Answer: (d) \( \text{Ca}^{2+} \) and \( \text{Mg}^{2+} \) ions
Ethylenediaminetetraacetic acid (EDTA) forms very stable chelate complexes with divalent alkaline earth metal ions. Therefore, it is standardly used to estimate \( \text{Ca}^{2+} \) and \( \text{Mg}^{2+} \) ions, which help determine the hardness of water.
In simple words: EDTA is a powerful chelating agent that grabs calcium and magnesium ions in water, helping scientists measure how hard the water is.

Exam Tip: EDTA titrations are a fundamental laboratory technique for determining water hardness due to calcium and magnesium content.

Question 25. For lead-poisoning, the antidote used is
(a) white of an egg
(b) cis-platin
(c) nickel
(d) EDTA
Answer: (d) EDTA
Calcium-disodium salt of EDTA is used as an effective antidote for treating lead poisoning. The chelating ligand binds tightly to lead ions, forming a soluble complex that is easily excreted from the body.
In simple words: EDTA acts like a claw that traps toxic lead ions in the bloodstream, allowing the body to safely flush them out through urine.

Exam Tip: Remember that chelation therapy uses ligands like EDTA to treat heavy metal poisoning by forming non-toxic, water-soluble complexes.

Question 26. In photography, the use of \( \text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O} \) is
(a) for converting \( \text{AgBr} \) into \( \text{Ag}_2\text{SO}_4 \)
(b) for converting \( \text{AgBr} \) into soluble thiosulphate complex
(c) for converting \( \text{AgBr} \) into silver thiosulphate
(d) in reduction of Ag metal from \( \text{AgBr} \)
Answer: (b) for converting \( \text{AgBr} \) into soluble thiosulphate complex
Sodium thiosulphate acts as a fixing agent in photographic development. It converts the unexposed silver bromide (\( \text{AgBr} \)) on the film into a soluble coordination complex, \( \text{Na}_3[\text{Ag(S}_2\text{O}_3)_2] \), which can be easily washed away.
In simple words: Thiosulphate reacts with the leftover silver bromide on film, turning it into a dissolved complex that washes right off.

Exam Tip: The chemistry of photographic "fixing" relies on the formation of the highly soluble sodium dithiosulphatoargentate(I) complex.

Question 27. Identify X and Y in the following table regarding Cobalt (III) Chloride-Ammonia Complexes:

(a) \( \text{X} = [\text{Co(NH}_3)_6]^{3+} 3\text{Cl}^- \), \( \text{Y} = 1:3 \)
(b) \( \text{X} = [\text{Co(NH}_3)_4\text{Cl}_2]^+ \text{Cl}^- \), \( \text{Y} = 1:3 \)
(c) \( \text{X} = [\text{Co(NH}_3)_4\text{Cl}_2]^+ \text{Cl}^- \), \( \text{Y} = 1:1 \)
(d) \( \text{X} = [\text{Co(NH}_3)_4\text{Cl}_2]^{2+} 3\text{Cl}^- \), \( \text{Y} = 1:1 \)
Answer: (b) \( \text{X} = [\text{Co(NH}_3)_4\text{Cl}_2]^+ \text{Cl}^- \), \( \text{Y} = 1:3 \)
For the yellow complex \( [\text{Co(NH}_3)_6]\text{Cl}_3 \), there are a total of four ions (one complex cation and three chloride anions), which corresponds to a \( 1:3 \) electrolyte, so \( \text{Y} = 1:3 \). For the green complex showing a \( 1:1 \) electrolyte conductivity, the formula must have only one chloride ion outside the coordination sphere, giving \( \text{X} = [\text{Co(NH}_3)_4\text{Cl}_2]^+\text{Cl}^- \).
In simple words: The yellow compound releases four ions in total, making it a 1:3 electrolyte. The green compound must release only two ions (one positive, one negative) to be a 1:1 electrolyte, so its formula is \( [\text{Co(NH}_3)_4\text{Cl}_2]\text{Cl} \).

Exam Tip: The number of ions produced outside the coordination sphere determines the electrolytic conductivity and ratio of the complex solution.

Question 28. Which of the following is the most stable complex species?
(a) \( [\text{Fe(CN)}_6]^{3-} \)
(b) \( [\text{Fe(CO)}_5]^{3+} \)
(c) \( [\text{Fe(C}_2\text{O}_4)_3]^{3-} \)
(d) \( [\text{Fe(H}_2\text{O})_6]^{3+} \)
Answer: (c) \( [\text{Fe(C}_2\text{O}_4)_3]^{3-} \)
The complex \( [\text{Fe(C}_2\text{O}_4)_3]^{3-} \) is highly stable because the oxalate group is a bidentate ligand that undergoes chelation. Chelate complexes are significantly more stable than non-chelated complexes.
In simple words: The oxalate ligand forms strong ring-like bonds with iron, making this complex much harder to break apart compared to the others.

Exam Tip: Chelate rings lead to high thermodynamic stability because breaking one bond still leaves the ligand attached, allowing it to easily re-coordinate.

Question 29. Assertion (A) In the coordination compound \( [\text{Co(H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_3]_2(\text{SO}_4)_3 \), ethane-1,2-diamine is a neutral ligand.
Reason (R) Oxidation number of Co in the complex ion is +3.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Ethane-1,2-diamine is a neutral ligand since it does not carry any electrical charge. In the complex ion \( [\text{Co(en)}_3]^{3+} \), the oxidation state of cobalt is indeed \( +3 \). Although both statements are correct, the oxidation state of the metal does not explain why the ligand is neutral.
In simple words: Both facts are true, but the metal's charge being \( +3 \) has nothing to do with whether the 'en' ligand is neutral.

Exam Tip: Carefully check if the Reason logically explains the statement in the Assertion before choosing option (a) or (b).

Question 30. Assertion (A) The stability of \( [\text{Ni(en)}_3]\text{Cl}_2 \) is more than that of \( [\text{Ni(NH}_3)_6]\text{Cl}_2 \).
Reason (R) The geometry of Ni is trigonal bipyramidal in \( [\text{Ni(en)}_3]\text{Cl}_2 \).
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
The Assertion is true because \( [\text{Ni(en)}_3]\text{Cl}_2 \) is stabilized by the chelate effect. However, the Reason is incorrect because nickel has a coordination number of six in this complex, giving it an octahedral geometry rather than trigonal bipyramidal.
In simple words: The first statement is true because ring-forming complexes are stronger. The second statement is false because the nickel ion is surrounded by six donor atoms, making it octahedral, not trigonal bipyramidal.

Exam Tip: Any complex with a coordination number of six adopts an octahedral geometry, not a five-coordinate geometry like trigonal bipyramidal.

Question 31. Assertion (A) \( [\text{Cr(H}_2\text{O})_6]\text{Cl}_2 \) and \( [\text{Fe(H}_2\text{O})_6]\text{Cl}_2 \) are examples of homoleptic complexes.
Reason (R) All the ligands attached to the metal are the same.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Complexes in which the central metal atom is bound to only one type of ligand are called homoleptic complexes. Since both \( [\text{Cr(H}_2\text{O})_6]\text{Cl}_2 \) and \( [\text{Fe(H}_2\text{O})_6]\text{Cl}_2 \) contain only water molecules as ligands, they are homoleptic, making the Reason the correct explanation.
In simple words: These complexes are homoleptic because they contain only one single kind of ligand (water) inside their brackets.

Exam Tip: Homoleptic complexes have identical ligands, whereas heteroleptic complexes contain more than one kind of ligand.

Question 32. Assertion (A) The ligands of nitro and nitrito are called ambidentate ligands.
Reason (R) These ligands give linkage isomers.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Nitro (\( -\text{NO}_2 \)) and nitrito (\( -\text{ONO} \)) are ambidentate ligands because they possess two distinct donor atoms through which they can coordinate. While they do form linkage isomers, this isomeric property is a consequence of their ambidentate nature, not the definition or explanation of it.
In simple words: Both statements are correct, but saying they form linkage isomers doesn't explain *why* they are called ambidentate ligands (which is because they have two different bonding sites).

Exam Tip: The definition of an ambidentate ligand relies on its dual bonding sites, which is the underlying cause for the occurrence of linkage isomerism.

Question 33. Assertion (A) trans-\( [\text{CrCl}_2(\text{ox})_2]^{3-} \) shows optical isomerism.
Reason (R) Optical isomerism is common in octahedral complexes involving didentate ligands.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
The trans-isomer of \( [\text{CrCl}_2(\text{ox})_2]^{3-} \) is symmetric and has a plane of symmetry, meaning its mirror images are superimposable and it cannot show optical isomerism. The Reason is a true general statement, but the Assertion is false.
In simple words: The first statement is false because the trans version of this complex is highly symmetrical and cannot rotate light. The second statement is true as octahedral complexes with double-bonding ligands often show optical activity.

Exam Tip: Keep in mind that only the cis-isomer of \( [\text{M(AA)}_2\text{X}_2] \) shows optical isomerism, while the trans-isomer is always optically inactive.

Question 34. Assertion (A) Complexes of \( \text{MX}_6 \) and \( \text{MX}_5\text{L} \) type (X and L are unidentate) do not show geometrical isomerism.
Reason (R) Geometrical isomerism is not shown by complexes of coordination number 6.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
Highly symmetric complexes like \( \text{MX}_6 \) and \( \text{MX}_5\text{L} \) cannot show geometrical isomerism because any arrangement of the single ligand \( \text{L} \) remains equivalent. However, the Reason is false because many other complexes with coordination number 6 (like \( \text{MX}_4\text{Y}_2 \)) do show geometrical isomerism.
In simple words: The first statement is true because you can't have different spatial arrangements when all or almost all ligands are the same. The second statement is false because six-coordinate complexes can show cis-trans isomerism.

Exam Tip: Geometrical isomerism in octahedral complexes requires a minimum distribution of ligands, such as \( \text{MA}_4\text{B}_2 \) or \( \text{MA}_3\text{B}_3 \).

Question 35. Assertion (A) The total number of isomers shown by \( [\text{Co(en)}_2\text{Cl}_2]^+ \) complex ion is three.
Reason (R) \( [\text{Co(en)}_2\text{Cl}_2]^+ \) complex ion has an octahedral geometry.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
The complex \( [\text{Co(en)}_2\text{Cl}_2]^+ \) exists as three isomers: a trans-isomer (optically inactive) and two enantiomeric forms of the cis-isomer (cis-dextro and cis-laevo). While it does have an octahedral geometry, this shape alone does not explain why there are exactly three isomers.
In simple words: There are three isomers because the cis form has a pair of mirror-image twins (optical isomers) while the trans form does not. This is not directly explained by just having an octahedral shape.

Exam Tip: Remember that \( [\text{M(en)}_2\text{Cl}_2]^+ \) type complexes have 1 trans-isomer and 2 optical cis-isomers, making a total of 3 stereoisomers.

Question 36. Assertion (A) Tetrahedral complexes do not show geometrical isomerism.
Reason (R) The relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In a tetrahedral geometry, all four coordination positions are adjacent and equidistant from each other. As a result, no cis-trans configurations can be created, making the Reason the correct explanation for why geometrical isomerism is impossible.
In simple words: In a tetrahedron, every corner is equally close to every other corner, so you cannot put two groups "opposite" each other to make a trans-isomer.

Exam Tip: Geometrical isomerism is never found in tetrahedral complexes, though they can show optical isomerism if all four ligands are different.

Question 37. Assertion (A) \( [\text{Mn(CN)}_6]^{3-} \) has a magnetic moment of two unpaired electrons while \( [\text{MnCl}_6]^{3-} \) has a magnetic moment of four unpaired electrons.
Reason (R) \( [\text{Mn(CN)}_6]^{3-} \) is inner orbital complex involving \( d^2sp^3 \)-hybridisation, on the other hand, \( [\text{MnCl}_6]^{3-} \) is outer orbital complex involving \( sp^3d^2 \)-hybridisation.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
For both complexes, manganese is in the \( +3 \) oxidation state (\( 3d^4 \)). Cyanide is a strong field ligand that forces pairing, yielding a low-spin \( d^2sp^3 \) inner orbital complex with two unpaired electrons. Chloride is a weak field ligand that does not cause pairing, resulting in a high-spin \( sp^3d^2 \) outer orbital complex with four unpaired electrons. Hence, the Reason perfectly explains the Assertion.
In simple words: The strong cyanide ligand packs the d-electrons together to leave only two unpaired, while the weak chloride ligand leaves all four d-electrons unpacked and unpaired.

Exam Tip: Strong-field ligands create inner-orbital low-spin complexes, whereas weak-field ligands create outer-orbital high-spin complexes.

Question 38. Assertion (A) \( [\text{Fe(CN)}_6]^{3-} \) ion shows magnetic moment corresponding to two unpaired electrons.
Reason (R) Because it has \( d^2sp^3 \) type hybridisation.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In the complex ion \( [\text{Fe(CN)}_6]^{3-} \), iron is in the \( +3 \) oxidation state with a \( 3d^5 \) configuration. Under the influence of the strong cyanide ligand, the five d-electrons pair up, leaving exactly one unpaired electron. Thus, the Assertion is false, while the Reason is true.
In simple words: The first statement is false because the paired-up \( 3d^5 \) configuration has only one single unpaired electron, not two. The second statement is true since it uses inner d-orbitals for bonding.

Exam Tip: A \( d^5 \) strong field octahedral complex always has exactly 1 unpaired electron, resulting in a magnetic moment of \( \approx 1.73 \text{ BM} \).

Question 39. Assertion (A) \( [\text{Ni(CN)}_4]^{2-} \) is square planar and diamagnetic.
Reason (R) It has no unpaired electrons due to presence of strong field ligand.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
The central nickel ion in \( [\text{Ni(CN)}_4]^{2-} \) has a \( 3d^8 \) configuration. Because cyanide is a very strong field ligand, it forces the pairing of the two d-electrons, clearing a \( 3d \) orbital for \( dsp^2 \) hybridisation. This results in a square planar, diamagnetic geometry, making the Reason the correct explanation.
In simple words: Cyanide forces the electrons to pair up completely, which leaves no unpaired electrons (making it diamagnetic) and allows it to form a flat, square planar shape.

Exam Tip: Remember that \( d^8 \) metal ions with strong-field ligands coordinate to four groups to form square planar complexes, whereas with weak-field ligands they form tetrahedral complexes.

Question 40. Assertion (A) Low spin tetrahedral complexes are rarely observed.
Reason (R) Crystal field splitting energy is less than pairing energy for tetrahedral complexes.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Because the crystal field splitting energy for tetrahedral complexes (\( \Delta_t \)) is quite small, it is almost always less than the pairing energy (\( P \)) of the electrons. Consequently, electrons prefer to occupy higher energy levels rather than pair up, explaining why low-spin tetrahedral complexes are extremely rare.
In simple words: The energy gap in tetrahedral shapes is too small to force electrons to pair up, so they always choose to jump to the next level instead, keeping the complex high-spin.

Exam Tip: Since \( \Delta_t < P \joinrel \), virtually all tetrahedral complexes are high-spin.

Question 41. Assertion (A) Toxic metal ions are removed by the chelating ligands.
Reason (R) Chelate complexes tend to be more stable.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Chelating ligands can wrap around toxic metal ions to form highly stable ring-structured complexes. Because these chelated complexes are so thermodynamically stable, they lock the metal ions away safely, allowing them to be easily eliminated from the body.
In simple words: Chelators act like molecular claws that hold onto toxic metals very tightly, making a stable complex that the body can safely flush away.

Exam Tip: The exceptional stability of chelates (the chelate effect) is the chemical basis for heavy metal detox treatments in medicine.

42. Read the given case and answer the following questions.
In 1893, Werner was the first to propose correct structures for coordination compounds containing complex ions, in which a central transition metal atom is surrounded by neutral or anionic ligands.
For example, it was known that cobalt forms a “complex” hexamminecobalt(III) chloride, with formula CoCl₃ -6NH_3, but the nature of the association indicated by the dot was mysterious. Werner proposed the structure [Co(NH₃)g]Cl₃, with the Co ion surrounded by six NH 3, at the vertices of an octahedron.
The three Cl“ are dissociated as free ions, which Werner confirmed by measuring the conductivity of the compound in aqueous solution, and also by chloride anion analysis using precipitation with silver nitrate. Later, magnetic susceptibility analysis was also used to confirm Werner’s proposal for the chemical nature of CoCl₃ ● 6NH₃.
For complexes with more than one type of ligand,
Werner succeeded in explaining the number of isomers observed. For example, he explained the existence of two tetraammine isomers, for (Co(NH₃)4Cl₃, one green and one purple. Werner proposed that these are two geometrical isomers of formula [Co(NH₃)₄Cl₂]Cl, with one Cl“ ion dissociated as confirmed by conductivity measurements.

The Co atom is surrounded by four NH₃ and two Cl ligands at the vertices of an octahedron. The green isomer is “trans” with the two Cl ligands at opposite vertices and the purple is “cis” with the two Cl at adjacent vertices.
Werner also prepared complexes with optical isomers, and in 1914 he reported the first synthetic chiral compound lacking carbon, known as hexol with formula [Co(Co(NH₃)₄ (OH)₂}₃ jBr⁶.

Question 42. Based on the passage, answer the following questions:
(i) Why is CO a stronger ligand than \( \text{NH}_3 \) in complexes?
(ii) Write the IUPAC name of the following complex: \( [\text{CoBr}_2(\text{en})_2]^+ \)
Or What type of isomerism is shown by the complex \( [\text{Co(NH}_3)_5\text{SO}_4]\text{Br} \)?
(iii) For the complex ion \( [\text{Ni(NH}_3)_6]^{2+} \), state the geometry of the ion.
Answer:
(i) Carbon monoxide (CO) possesses empty \( \pi^* \) antibonding orbitals that can overlap with the filled \( d \)-orbitals (\( t_{2g} \)) of the transition metal, resulting in \( \pi \)-backbonding. This synergistic back-bonding significantly increases the crystal field splitting energy (\( \Delta_o \)), making CO a strong ligand. In contrast, ammonia (\( \text{NH}_3 \)) acts as a simple \( \sigma \)-donor and cannot form \( \pi \)-bonds.
(ii) The IUPAC name of \( [\text{CoBr}_2(\text{en})_2]^+ \) is dibromidobis(ethane-1,2-diamine)cobalt(III) ion.
Or The complex \( [\text{Co(NH}_3)_5\text{SO}_4]\text{Br} \) exhibits ionisation isomerism. It can exchange its counter bromide ion with the inner sulphate ligand to yield \( [\text{Co(Br)(NH}_3)_5]\text{SO}_4 \).
(iii) The geometry of the complex ion \( [\text{Ni(NH}_3)_6]^{2+} \) is octahedral, corresponding to \( sp^3d^2 \) hybridisation.
In simple words: (i) CO does back-bonding to share extra electrons with the metal, which makes its bond much stronger than ammonia's. (ii) The name is dibromidobis(ethane-1,2-diamine)cobalt(III) ion. Or It shows ionisation isomerism because the bromine outside and sulphate inside can swap places. (iii) The geometry is octahedral.

Exam Tip: When writing IUPAC names, always put the ligands in alphabetical order and state the oxidation state of the metal in Roman numerals inside parentheses.

43. Read the given case and answer the following questions.
Complex compounds play an important role in our daily life. These compounds contain central metal and ligands. Ligands can be negatively charged or neutral molecules. Ligands can be differentiated on the basis of the number of donor sites which is simply called as denticity. It can be monodentate, didentate, polydentate, ambidentate ligands. Polydentate ligands are more stable than monodentate ligands.
Complex compounds are named according the lUPAC system. Many approaches have been put forth to explain the nature of bonding in complex compounds.
These are valence bond theory (VBT), crystal field theory (CFT), ligand field theory (LFT) and Molecular orbital theory (MOT).

Question 43. Based on the passage, answer the following questions:
(i) (a) What is the oxidation state of Cu in \( [\text{Cu(NH}_3)_4\text{Cl}_2] \)?
(b) What is the IUPAC name of the complex \( [\text{Ni(NH}_3)_6]\text{Cl}_2 \)?
(ii) Which one of the following forms a low spin complex?
(a) \( [\text{Co(CN)}_6]^{3-} \)
(b) \( [\text{NiCl}_4]^{2-} \)
(c) \( [\text{Ni(H}_2\text{O})_6]^{2+} \)
(d) \( [\text{MnF}_6]^{3-} \)
(iii) The magnetic property and the number of unpaired electrons in the complex \( [\text{Co(NH}_3)_6]^{3+} \) will be:
(a) Paramagnetic, 2
(b) Paramagnetic, 1
(c) Diamagnetic, 0
(d) Diamagnetic, 2
Or The hybridisation and geometry for the complex \( [\text{Co(NH}_3)_6]^{3+} \) is:
(a) \( d^2sp^3 \), octahedral
(b) \( sp^3d^2 \), trigonal bipyramidal
(c) \( sp^3 \), tetrahedral
(d) \( dsp^2 \), tetrahedral
Answer:
(i) (a) Let the oxidation state of Cu be \( x \). Since ammonia is neutral and chloride carries a \( -1 \) charge: \( x + 4(0) + 2(-1) = 0 \implies x = +2 \). Thus, the oxidation state of Cu is \( +2 \).
(b) The IUPAC name is hexaamminenickel(II) chloride.
(ii) (a) \( [\text{Co(CN)}_6]^{3-} \) forms a low spin complex because cyanide is a strong field ligand.
(iii) (c) Diamagnetic, 0. In \( [\text{Co(NH}_3)_6]^{3+} \), \( \text{Co}^{3+} \) has a \( 3d^6 \) configuration. Since \( \text{NH}_3 \) acts as a strong field ligand here, it causes complete pairing of electrons, leaving zero unpaired electrons.
Or (a) \( d^2sp^3 \), octahedral. The complex is an inner orbital complex involving \( d^2sp^3 \) hybridisation and has an octahedral geometry.
In simple words: (i) (a) Copper has a charge of \( +2 \). (b) The name is hexaamminenickel(II) chloride. (ii) Option (a) is correct because cyanide is a very strong ligand that forces electrons to pair up. (iii) Option (c) is correct because all electrons pair up, making it non-magnetic. Or Option (a) is correct because it uses inner d-orbitals for an octahedral shape.

Exam Tip: Remember that ammonia behaves as a strong field ligand for \( \text{Co}^{3+} \), forcing all six d-electrons to pair up to form a diamagnetic \( d^2sp^3 \) complex.

44. Read the given case and answer the following questions.
Crystal field theory was proposed by H. Bethe and John van Vleck in 1930. The Crystal Field Theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. In an octahedral coordination entity with six ligands surrounding the metal atom/ion, there will be repulsion between the electrons in metal d-orbitals and the electrons of the ligands. The degeneracy of the d-orbitals has been removed due to ligand-metal electron repulsions in the octahedral complex to yield three orbitals of lower energy, \( t_{2g} \) set and two orbitals of higher energy, \( e_g \) set.

Question 44. Based on the passage, answer the following questions:
(i) What will be the correct order for the wavelengths of absorption in the visible region for the following: \( [\text{Ni(NO}_2)_6]^{4-} \), \( [\text{Ni(NH}_3)_6]^{2+} \), \( [\text{Ni(H}_2\text{O})_6]^{2+} \)?
(ii) Explain the violet colour of the complex \( [\text{Ti(H}_2\text{O})_6]^{3+} \) on the basis of crystal field theory.
Or How does the colour of the coordination complex \( [\text{Ti(H}_2\text{O})_6]^{3+} \) ion change on heating?
(iii) What is meant by spectrochemical series?
Answer:
(i) The energy separation \( \Delta_o \) increases with ligand strength: \( \text{H}_2\text{O} < \text{NH}_3 < \text{NO}_2^- \). Since wavelength is inversely proportional to energy (\( \lambda \propto \frac{1}{\Delta_o} \)), the correct order of absorbed wavelengths is: \( [\text{Ni(H}_2\text{O})_6]^{2+} > [\text{Ni(NH}_3)_6]^{2+} > [\text{Ni(NO}_2)_6]^{4-} \).
(ii) In \( [\text{Ti(H}_2\text{O})_6]^{3+} \), titanium is in the \( +3 \) oxidation state with a \( 3d^1 \) configuration. This single electron occupies the lower-energy \( t_{2g} \) level. When it absorbs yellow-green light, the electron is excited to the higher-energy \( e_g \) level (d-d transition), and the complex transmits violet light, which is the complementary colour.
Or Upon heating, \( [\text{Ti(H}_2\text{O})_6]^{3+} \) loses its coordinated water molecules. Without ligands, crystal field splitting does not occur, making the complex colourless.
(iii) The spectrochemical series is a systematic arrangement of ligands in the order of their increasing crystal field splitting energy or field strength.
In simple words: (i) Stronger ligands cause bigger splits, meaning they absorb higher energy (shorter wavelength). So, the order of wavelength is water > ammonia > nitro. (ii) The single electron absorbs yellow-green light to jump up a level, leaving behind violet light that we see. Or Heating drives away the water ligands, so the energy levels don't split anymore and the colour disappears. (iii) It is a list that ranks ligands from weakest to strongest.

Exam Tip: For explaining colours in complexes, always mention the key term "d-d transition" as this is what examiners look for to award full marks.

Question 1. \( \text{FeSO}_4 \) solution mixed with \( (\text{NH}_4)_2\text{SO}_4 \) solution in 1:1 molar ratio gives the test of \( \text{Fe}^{2+} \) ion but \( \text{CuSO}_4 \) solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of \( \text{Cu}^{2+} \) ion. Why?
Answer: Mixing \( \text{FeSO}_4 \) and \( (\text{NH}_4)_2\text{SO}_4 \) in a \( 1:1 \) ratio forms Mohr's salt, which is a double salt. In water, it fully dissociates into simple ions, including free \( \text{Fe}^{2+} \) ions, thus giving a positive test for iron. In contrast, mixing \( \text{CuSO}_4 \) and ammonia in a \( 1:4 \) ratio yields a stable coordination complex, \( [\text{Cu(NH}_3)_4]\text{SO}_4 \). Because this complex does not dissociate to release free \( \text{Cu}^{2+} \) ions, the solution does not give a positive test for copper.
In simple words: Mohr's salt is a double salt that breaks down completely into individual ions in water. The copper-ammonia mixture forms a complex compound where the copper is locked inside a stable bracket and cannot be detected as free ions.

Exam Tip: Clearly distinguish between double salts (which fully dissociate in solution) and coordination complexes (which do not dissociate into their constituent metal and ligand ions).

Question 2. What is meant by the unidentate, bidentate and ambidentate ligands? Give two examples of each.
Answer:
- Unidentate Ligands: These ligands possess only a single donor atom capable of forming a coordinate bond with the metal. Examples: \( \text{H}_2\text{O} \), \( \text{NH}_3 \).
- Bidentate Ligands: These ligands contain two donor atoms that can simultaneously form two coordinate bonds with the central metal ion. Examples: ethylene diamine (en), oxalate ion (\( \text{C}_2\text{O}_4^{2-} \)).
- Ambidentate Ligands: These ligands have two different donor atoms but can only coordinate through one of them at any given time. Examples: \( \text{SCN}^- \) (thiocyanate), \( \text{NO}_2^- \) (nitrite).
In simple words: Unidentate ligands attach by one atom, bidentate ligands attach using two atoms at once like a clamp, and ambidentate ligands have two choices of atoms but can only use one at a time.

Exam Tip: When giving examples, write down both the name and chemical formula of the ligands to show a thorough understanding.

Question 3. (i) What is a chelate complex? Give one example.
(ii) What are heteroleptic complexes? Give one example.
Answer:
(i) A chelate complex is a coordination compound in which a polydentate ligand binds to the central metal ion through multiple donor atoms to form one or more stable ring structures. Example: \( [\text{Co(en)}_3]^{3+} \).
(ii) Heteroleptic complexes are coordination complexes where the central metal ion is coordinated to more than one kind of ligand. Example: \( [\text{Co(NH}_3)_4\text{Cl}_2]^+ \).
In simple words: A chelate complex is a ring-shaped compound made when a ligand grabs a metal with two or more arms. A heteroleptic complex is one that contains a mix of different types of ligands.

Exam Tip: Remember that chelate complexes have high stability, and heteroleptic complexes contain different kinds of ligands (e.g., both amine and chloride).