CBSE Class 12 Chemistry Coordination Compounds NEET JEE Questions Set 04

Single Correct Answer

Question 1. Coordination number of Cr is six. A complex ion of Cr with C₂O₄²⁻, en and superoxide ion O₂⁻ has the formula [Cr(C₂O₄)_x(en)_y(O₂)_z]⁻. The ratio x : y : z will be
(a) 1 : 1 : 1
(b) 1 : 1 : 2
(c) 1 : 2 : 2
(d) 2 : 1 : 1
Answer:
Ans. (b)
Oxalate and en are bidentate, superoxide is monodentate: 2x + 2y + z = 6.
Taking Cr as +3, charge: 3 − 2x − z = −1, or 2x + z = 4.
x = 1, y = 1, z = 2 satisfies both: 2 + 2 + 2 = 6 and 2 + 2 = 4. Hence x : y : z = 1 : 1 : 2.

Teacher's Note:
1. Use two equations: one for the coordination number (count denticity) and one for the overall charge.

Question 2. [Cr(H₂O)₆] Cl₃ (at. no. of Cr = 24) has a magnetic moment of 3.83 B.M. The correct distribution of 3d electrons in the chromium of the complex is
(a) 3d_xy¹, 3d_yz¹, 3d_xz¹
(b) 3d_xy¹, 3d_yz¹, 3d_z²¹
(c) 3d_(x²−y²)¹, 3d_z²¹, 3d_xz¹
(d) 3d_xy¹, 3d_(x²−y²)¹, 3d_yz¹
Answer:
Ans. (a)
Cr³⁺ = 3d³; μ = 3.83 B.M. confirms 3 unpaired electrons.
In an octahedral field the three electrons occupy the lower t_2g orbitals singly: 3d_xy¹, 3d_yz¹, 3d_xz¹.

Teacher's Note:
1. t_2g = d_xy, d_yz, d_xz; the e_g orbitals (d_z², d_x²−y²) stay empty for d³.

Question 3. The ratio of the uncomplexed to the complexed Zn²⁺ ion in a solution of 10 M NH₃ is (Given that the stability constant of [Zn(NH₃)₄]²⁺ is 3 × 10⁹)
(a) 3.3 × 10⁻⁹
(b) 3.3 × 10⁻¹¹
(c) 3.3 × 10⁻¹³
(d) 3.3 × 10⁻¹⁴
Answer:
Ans. (d)
Zn²⁺ + 4NH₃ ⇄ [Zn(NH₃)₄]²⁺, \( K = \frac{[\mathrm{Zn(NH_3)_4^{2+}}]}{[\mathrm{Zn^{2+}}][\mathrm{NH_3}]^4} \)
\( \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Zn(NH_3)_4^{2+}}]} = \frac{1}{K[\mathrm{NH_3}]^4} = \frac{1}{3\times10^{9}\times(10)^4} \) = 3.3 × 10⁻¹⁴

Teacher's Note:
1. Do not forget to raise [NH₃] to the fourth power, matching the four ligands in the complex.

Question 4. The octahedral complex of a metal ion M³⁺ with four monodentate ligands L₁, L₂, L₃ and L₄ absorbs wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is (JEE Main 2014)
(a) L₁ < L₂ < L₄ < L₃
(b) L₄ < L₃ < L₂ < L₁
(c) L₁ < L₃ < L₂ < L₄
(d) L₃ < L₂ < L₄ < L₁
Answer:
Ans. (c)
VIBGYOR: wavelength (λ) increases from violet to red.
According to spectrochemical series, greater the adsorption frequency, stronger is the ligand or CFSE (Δ₀) ∝ Strength of ligand ∝ Frequency absorbed (ν) or \( \propto \frac{1}{\lambda} \)
Hence, increasing order of ligand strength will be L₁ (Red) < L₃ (Yellow) < L₂ (Green) < L₄ (Blue).

Teacher's Note:
1. Arrange the absorbed colours by wavelength (red longest, blue shortest); the shortest wavelength belongs to the strongest ligand.

Question 5. Among the following complexes, the one which shows zero crystal field stabilization energy (CFSE) is (AIPMT 2014)
(a) [Mn(H₂O)₆]³⁺
(b) [Fe(H₂O)₆]³⁺
(c) [Co(H₂O)₆]²⁺
(d) [Co(H₂O)₆]³⁺
Answer:
Ans. (b)
H₂O is a weak field ligand. Hence, Δ₀ < pairing energy
CFSE = (− 0.4 x + 0.6 y) Δ₀, where x and y are no. of electrons occupying t_2g and e_g orbitals respectively.
For [Fe(H₂O)₆]³⁺ complex ion, Fe³⁺ (3d⁵) = t_2g³ e_g²
∴ CFSE = − 0.4 × 3 + 0.6 × 2 = 0 Dq

Teacher's Note:
1. High spin d⁵ (and d⁰, d¹⁰) always has zero CFSE because the gains and losses cancel.

Question 6. Consider the following complexes :
1. Fe(CO)₅ 2. Ni(CO)₄ 3. K₄[Fe(CN)₆] 4. Na₃[Cu(SCN)₄]
Which of the above complexes obey EAN rule ? (IAS Prelim 2010)
(a) 1 and 2 only
(b) 2, 3 and 4 only
(c) 1, 2, 3 and 4
(d) none of the above
Answer:
Ans. (c)
All the given complexes follow EAN rule:
Fe(CO)₅ : 26 − 0 + 2 × 5 = 36; Ni(CO)₄ : 28 − 0 + 2 × 4 = 36;
K₄[Fe(CN)₆] : 26 − 2 + 2 × 6 = 36; Na₃[Cu(SCN)₄] : 29 − 1 + 2 × 4 = 36.

Teacher's Note:
1. EAN = Z − oxidation state + 2 × coordination number; 36 is the atomic number of Kr.

Question 7. EAN of cobalt is 36 in [Co(NH₃)₂O₂(en) Cl]. Knowing that atomic number of Co is 27, O₂ is present as
(a) peroxide ion
(b) dioxide ion
(c) superoxide ion
(d) oxide ion
Answer:
Ans. (a)
Peroxide = O₂²⁻, dioxide = 2 O²⁻, superoxide = O₂⁻, oxide = O²⁻
EAN = Z − Oxidation no. + 2 × coordination no.
If O₂ is present as O₂⁻, oxidation no. of Co = + 2; EAN = 27 − 2 + 2 × 6 = 37
If O₂ is present as O₂²⁻, oxidation no. of Co = + 3; EAN = 27 − 3 + 2 × 6 = 36.

Teacher's Note:
1. Coordination number here is 6 (2 NH₃ + O₂ + en (2) + Cl); then try each charge of O₂ until EAN = 36.

Question 8. Complex A has a composition of H₁₂O₆Cl₃Cr. If the complex on treatment with conc. H₂SO₄ loses 13.5% of its original mass, the correct molecular formula of A is (Given : atomic mass of Cr = 52 amu and Cl = 35 amu) (JEE Main 2020)
(a) [Cr(H₂O)₆]Cl₃
(b) [Cr(H₂O)₄Cl₂]Cl·2 H₂O
(c) [Cr(H₂O)₃Cl₃]·3 H₂O
(d) [Cr(H₂O)₅Cl]Cl₂·H₂O
Answer:
Ans. (b)
13.5% of H₁₂O₆Cl₃Cr (265) = 36 approximately
∴ About 2 moles of water are lost during treatment with conc. H₂SO₄.
Formula of the complex H₁₂O₆Cl₃Cr will be [Cr(H₂O)₄Cl₂]Cl·2H₂O

Teacher's Note:
1. Conc. H₂SO₄ removes only the water of crystallisation; coordinated water stays bound to Cr.

Question 9. An example of a sigma bonded organometallic compound is (NEET 2017)
(a) ruthenocene
(b) Grignard's reagent
(c) ferrocene
(d) cobaltocene
Answer:
Ans. (b)
Grignard's reagent, R–Mg–X is formed through a normal 2 electron covalent bond (σ bond) between R group and Mg atom.

Teacher's Note:
1. Metallocenes (ferrocene, ruthenocene, cobaltocene) are π-bonded sandwich compounds.

Question 10. Fe₂(CO)₉ is diamagnetic. Which of the following is the correct reason ?
(a) One CO is present as bridge group
(b) CO is a π-acceptor ligand
(c) CO can form π-bond with Fe by back bonding
(d) Metal-metal (Fe–Fe) bonding takes place.
Answer:
Ans. (d)
Metal-metal bonding pairs up the unpaired electron.

Teacher's Note:
1. Each Fe in Fe₂(CO)₉ has one odd electron; the Fe–Fe bond pairs them and also completes 18 electrons on each Fe.

Question 11. Among the following metal carbonyls, the C–O bond order is lowest in (IIT 2007)
(a) [Mn(CO)₆]⁺
(b) [Fe(CO)₅]
(c) [Cr(CO)₆]
(d) [V(CO)₆]⁻
Answer:
Ans. (d)
An anionic complex can delocalise more electron density to antibonding π-orbital (dπ – pπ back bonding) of CO and thus lowers the bond order.

Teacher's Note:
1. More negative charge on the metal means more back-donation into CO π*, which lowers C–O bond order and stretching frequency.

Question 12. In Fe(CO)₅, the Fe ← CO σ bond results by the overlap between filled sp hybrid orbital of C-atom of CO molecule and vacant (DPMT 2010)
(a) d²sp³
(b) sp³
(c) dsp³
(d) dsp² hybrid orbital of Fe
Answer:
Ans. (c)
₂₆Fe = 3d⁶4s². CO is a strong field ligand. Hence, electrons pair up to give 3d⁸ with one 3d orbital empty.
One 3d, one 4s and three 4p orbitals hybridise: dsp³ (trigonal bipyramidal).

Teacher's Note:
1. Five ligands need five hybrid orbitals: dsp³ for Fe(CO)₅, compared with sp³ for Ni(CO)₄ and d²sp³ for Cr(CO)₆.

Question 13. Which of the following carbonyls will have the strongest C–O bond ?
(a) Mn (CO)₆⁺
(b) Cr (CO)₆
(c) V (CO)₆⁻
(d) Fe (CO)₅
Answer:
Ans. (a)
As positive charge on the central metal atom increases, the metal shows lesser tendency to donate its electron density to the antibonding π orbitals of CO ligand. This causes strengthening of the C–O bond. Hence, CO bond would be strongest in Mn(CO)₆⁺.

Teacher's Note:
1. A cationic carbonyl has the least back-bonding, so its C–O bond is closest to that of free CO.

Question 14. If the length of CO bond in carbon monoxide is 1.128 Å, then what is the value of CO bond length in Fe(CO)₅ ?
(a) 1.15 Å
(b) 1.128 Å
(c) 1.72 Å
(d) 1.118 Å
Answer:
Ans. (a)
The metal-carbon bond in carbonyls possesses both σ and π character—weak σ bond being formed by donation of electron pair from carbon to metal (M ← C ≡ O) and a stronger π-bond being formed by back donation of electrons from filled d_xy orbital of metal into empty antibonding π*_2py orbital of carbon (π M → C bond). The total bonding is thus M = C = O. Thus, bond order of C–O bond is reduced from triple bond to double bond. As a result, C–O bond length of 1.128 Å in CO increases to about 1.15 Å in many carbonyls.

Teacher's Note:
1. Back-bonding always lengthens C–O slightly, so the answer must be a little more than 1.128 Å, not a large jump like 1.72 Å.

Question 15. Which of the following has longest C–O bond length ? (Free bond length in CO is 1.128 Å) (NEET Phase I 2016)
(a) [Fe(CO)₄]²⁻
(b) [Mn(CO)₆]⁺
(c) Ni(CO)₄
(d) [Co(CO)₄]⁻
Answer:
Ans. (a)
The greater the negative charge on the carbonyl complex, the easier it would be for the metal to allow its electrons to participate in the back bonding. As a result, higher would be the M–C bond order and simultaneously, there would be greater reduction in the C–O bond order and hence larger bond length. Thus, [Fe(CO)₄]²⁻ has the lowest C–O bond order and, therefore, the longest bond length.

Teacher's Note:
1. Order of C–O length here: [Fe(CO)₄]²⁻ > [Co(CO)₄]⁻ > Ni(CO)₄ > [Mn(CO)₆]⁺.

Question 16. Low oxidation state of metals in their complexes are common when ligands : (JEE Main 2022)
(a) have good π-accepting character
(b) have good σ-donor character
(c) are having good π-donating ability
(d) are having poor σ-donating ability
Answer:
Ans. (a)
Ligands like : CO are sigma donor and π-acceptor and they make stronger bond with lower oxidation state metal ion. In this case, back-bonding is more effective.

Teacher's Note:
1. A π-acceptor removes excess electron density from an electron-rich (low oxidation state) metal, which stabilises states like Ni(0) and Fe(0).

Question 17. When EDTA solution is added to Mg²⁺ ion solution, then which of the following statements is not true ? (DCE 2009)
(a) Four coordinate sites of Mg²⁺ are occupied by EDTA and remaining two sites are occupied by water molecules
(b) All six coordinate sites of Mg²⁺ are occupied
(c) pH of the solution is decreased
(d) Colourless [Mg-EDTA]²⁻ chelate is formed.
Answer:
Ans. (a)
Mg²⁺ forms an octahedral complex with EDTA where all six coordinate sites of Mg²⁺ are occupied.

Teacher's Note:
1. When H₄EDTA/H₂EDTA²⁻ binds a metal ion, H⁺ ions are released, which is why the pH falls.

Question 18. The oxidation state of Fe in the brown ring complex [Fe(H₂O)₅NO] SO₄ is (WB JEE 2008 ; Karnataka CET 2009)
(a) +1
(b) +2
(c) +3
(d) +4
Answer:
Ans. (a)
[Fe (H₂O)₅ NO]²⁺, with H₂O = 0 and NO = +1 : x + 0 + 1 = + 2 or x = +1.

Teacher's Note:
1. In the brown ring complex NO is present as NO⁺, so iron is +1, not +2.

Question 19. Dipole moment will be zero in the complexes
I. [Ni(CN)₄]²⁻ II. cis-Pt[(NH₃)₂Cl₂] III. trans-[Pt(NH₃)₂Cl₂]
(a) I and II
(b) I and III
(c) II and III
(d) I, II and III
Answer:
Ans. (b)
[Ni(CN)₄]²⁻ is square planar with four identical CN groups: the bond dipoles cancel, μ = 0.
In trans-[Pt(NH₃)₂Cl₂], identical groups are opposite each other, so μ = 0. The cis-isomer has a net dipole moment.

Teacher's Note:
1. In square planar complexes, bond dipoles cancel only when identical ligands are trans to each other.

Question 20. In nitroprusside ion, iron and NO exist as Fe (II) and NO⁺ rather than Fe (III) and NO. These forms can be differentiated by
(a) estimating the concentration of iron
(b) measuring the concentration of CN⁻
(c) measuring the solid state magnetic moment
(d) thermally decomposing the compound.
Answer:
Ans. (c)
In nitroprusside ion, Fe (II) has d⁶ configuration, i.e., 4 unpaired electrons which pair up for d² sp³-hybridisation and NO⁺ has also no unpaired electron. Hence, its magnetic moment is zero, i.e., it is diamagnetic. On the other hand, Fe (III) has d⁵ configuration and NO has unpaired electron. Even after pairing up, unpaired electrons are present and it would have net magnetic moment.

Teacher's Note:
1. Magnetic measurements are a direct way to find the number of unpaired electrons and hence the oxidation state.

One or More Correct Answers

Question 21. The correct statement(s) regarding the binary transition metal carbonyl compounds is (are) (Atomic numbers : Fe = 26, Ni = 28) (JEE Advanced 2018)
(a) Total number of valence shell electrons at metal centre in Fe(CO)₅ or Ni(CO)₄ is 16
(b) These are predominantly low spin in nature
(c) Metal-carbon bond strengthens when the oxidation state of the metal is lowered
(d) The carbonyl C–O bond weakens when the oxidation state of the metal is increased.
Answer:
Ans. (b, c)
Fe = [Ar] 3d⁶4s², Ni = [Ar] 3d⁸4s², CO is strong field ligand.
Valence electrons : Fe(CO)₅ = 8 + 2 × 5 = 18; Ni(CO)₄ = 10 + 4 × 2 = 18 (Low spin complexes)
If oxidation state of M is increased, M–C bond strength decreases and C–O bond strength increases due to less extent of back bonding. If oxidation state of M is decreased, M–C bond strength increases and C–O bond strength decreases due to greater extent of back bonding.

Teacher's Note:
1. M–C and C–O bond strengths always change in opposite directions, because both depend on back-donation into CO π*.

Question 22. Choose the correct statement(s) among the following : (JEE Advanced 2020)
(a) [FeCl₄]⁻ has tetrahedral geometry
(b) [Co(en)(NH₃)₂ Cl₂]⁺ has 2 geometrical isomers
(c) [FeCl₄]⁻ has higher spin only magnetic moment than [Co(en)(NH₃)₂ Cl₂]⁺
(d) Cobalt in [Co(en)(NH₃)₂ Cl₂]⁺ has sp³d² hybridisation
Answer:
Ans. (a, c)
(a) Fe³⁺ (3d⁵) with weak field Cl⁻: sp³, tetrahedral. Correct.
(b) [Co(en)(NH₃)₂Cl₂]⁺ is of the type [M(AA)B₂C₂] and has 3 geometrical isomers (both Cl trans; both NH₃ trans; all cis). Wrong.
(c) [FeCl₄]⁻ has 5 unpaired electrons (μ = 5.92 BM), while Co³⁺ (3d⁶) with en/NH₃ is low spin with 0 unpaired electrons. Correct.
(d) Co³⁺ in this complex is d²sp³ (inner orbital), not sp³d². Wrong.

Teacher's Note:
1. For [M(AA)B₂C₂], list which pair is trans: B–B trans, C–C trans, or neither; that gives three geometrical isomers.

Question 23. The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are)
Note : py = pyridine
Given : Atomic nos. of Fe, Co, Ni and Cu are 26, 27, 28 and 29 respectively (JEE Advanced 2021)
(a) [FeCl₄]⁻ and [Fe(CO)₄]²⁻
(b) [Co(CO)₄]⁻ and [CoCl₄]²⁻
(c) [Ni(CO)₄] and [Ni(CN)₄]²⁻
(d) [Cu(py)₄]⁺ and [Cu(CN)₄]³⁻
Answer:
Ans. (a, b, d)
(a) [FeCl₄]⁻, Fe⁺³ = 3d⁵, Cl⁻ = weak field ligand, sp³
[Fe(CO)₄]²⁻, Fe⁻² = 3d⁸4s², CO = strong field ligand, pairing occurs = 3d¹⁰. Hence, sp³.
(b) [Co(CO)₄]⁻, Co⁻ = 3d⁸4s², CO = strong field ligand, pairing occurs = 3d¹⁰ = sp³
[CoCl₄]²⁻, Co²⁺ = 3d⁷, Cl⁻ = weak field ligand = sp³
(c) [Ni(CN)₄]²⁻ is square planar (dsp²), so this pair is not both tetrahedral.
(d) [Cu(py)₄]⁺, Cu⁺ = 3d¹⁰ = sp³; [Cu(CN)₄]³⁻, Cu⁺ = 3d¹⁰ = sp³

Teacher's Note:
1. A d¹⁰ centre with four ligands is always tetrahedral, because no inner d orbital is free for dsp².

Passage-Based Questions

Question 24. Comprehension 3: The coordination number of Ni²⁺ is 4
NiCl₂ + KCN (excess) → A (cyano complex)
A + conc. HCl (excess) → B (chloro complex) (IIT 2006)
Predict the magnetic nature of A and B
(a) Both are diamagnetic
(b) A is diamagnetic and B is paramagnetic with one unpaired electron.
(c) A is diamagnetic and B is paramagnetic with two unpaired electrons
(d) Both are paramagnetic
Answer:
Ans. (c)
A = [Ni(CN)₄]²⁻: all eight 3d electrons paired, dsp² (diamagnetic)
B = [NiCl₄]²⁻: 3d⁸ with two unpaired electrons, sp³ (paramagnetic with 2 unpaired electrons)

Teacher's Note:
1. Strong-field CN⁻ pairs the Ni²⁺ electrons; weak-field Cl⁻ does not.

Question 25. Comprehension 3: The coordination number of Ni²⁺ is 4
NiCl₂ + KCN (excess) → A (cyano complex)
A + conc. HCl (excess) → B (chloro complex) (IIT 2006)
The hybridisation of A and B are
(a) dsp², sp³
(b) sp³, sp³
(c) dsp², dsp²
(d) sp³d², d²sp³
Answer:
Ans. (a)
A = dsp², B = sp³.

Teacher's Note:
1. Square planar [Ni(CN)₄]²⁻ (dsp²) and tetrahedral [NiCl₄]²⁻ (sp³) are the classic pair for Ni(II).

Matrix Match Questions

Teacher's Note:
1. Check for a change of oxidation state before calling a reaction redox: chromate to dichromate is only an acid–base condensation.

Teacher's Note:
1. d¹–d³ ions always have two empty inner d orbitals, so even weak ligands give d²sp³.

Teacher's Note:
1. Option-elimination works well here: settle one row firmly (I → R, T) and strike out every option that contradicts it.

Integer and Numerical Answer Questions

Question 29. [Ti(H₂O)₆]³⁺ absorbs light of wavelength 498 nm during d-d transition. The octahedral splitting energy for the above complex is ........... × 10⁻¹⁹ J. (Round off to the nearest integer) h = 6.626 × 10⁻³⁴ Js ; c = 3 × 10⁸ m s⁻¹ (JEE Main 2021)
Answer:
Ans. 4
\( \Delta_0 = \frac{hc}{\lambda_{abs}} = \frac{(6.626\times10^{-34})\times(3\times10^{8})}{498\times10^{-9}} \) = 3.99 × 10⁻¹⁹ J
Nearest integer = 4 × 10⁻¹⁹ J

Teacher's Note:
1. Convert nm to m (× 10⁻⁹) before dividing; for a d¹ ion the absorbed photon energy equals Δ₀.

Question 30. In the cobalt-carbonyl complex, [(Co)₂(CO)₈], number of Co–Co bonds is “X” and the terminal CO ligands is “Y”. X + Y = ......... . (JEE Main 2022)
Answer:
Ans. 7
In Co₂(CO)₈, the two Co atoms are joined by one Co–Co bond and two bridging CO groups; each Co carries three terminal CO.
No. of Co–Co bonds (X) = 1
No. of terminal CO ligands (Y) = 6
∴ X + Y = 1 + 6 = 7

Teacher's Note:
1. Of the eight CO in Co₂(CO)₈, two are bridging; do not count them as terminal.

Question 31. If [Cu(H₂O)₄]²⁺ absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field splitting energy for [Cu(H₂O)₆]²⁺ will be ............ × 10⁻²¹ J (Nearest integer) (Given : h = 6.63 × 10⁻³⁴ Js and c = 3.08 × 10⁸ ms⁻¹) (JEE Main 2022)
Answer:
Ans. 766
[Cu(H₂O)₄]²⁺ is tetrahedral and [Cu(H₂O)₆]²⁺ is octahedral.
\( \Delta_t = \frac{4}{9}\times\Delta_0 \)
or \( \Delta_0 = \frac{9}{4}\times\Delta_t = \frac{9}{4}\times\frac{hc}{\lambda} \)
\( = \frac{9}{4}\times\frac{(6.63\times10^{-34})(3.08\times10^{8})}{600\times10^{-9}} \)
= 765.8 × 10⁻²¹ J ≈ 766 × 10⁻²¹ J

Teacher's Note:
1. The absorbed photon gives Δ_t of the tetrahedral ion; multiply by 9/4 to get Δ₀.

Question 32. The spin only magnetic moment value of an octahedral complex among CoCl₃ . 4 NH₃, NiCl₂ . 6 H₂O and PtCl₄ . 2 HCl which upon reaction with excess of AgNO₃ gives 2 moles of AgCl is .......... B.M. (Nearest integer) (JEE Main 2022)
Answer:
Ans. 3
[Co(NH₃)₄Cl₂] Cl, [Ni(H₂O)₆] Cl₂, H₂[PtCl₆]
Only [Ni(H₂O)₆] Cl₂ will give two moles of AgCl
[Ni(H₂O)₆]²⁺ : Ni²⁺ = d⁸ = t_2g⁶ e_g² = 2 unpaired electrons.
Magnetic moment \( = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \) ≈ 3 BM

Teacher's Note:
1. PtCl₄·2HCl is H₂[PtCl₆]: all six Cl are coordinated, so it gives no AgCl.

Question 33. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl₂.6 H₂O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of NiCl₂.6H₂O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is.......... . (Atomic weights in g mol⁻¹ : H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)
Answer:
Ans. 2992
(NH₄)₂SO₄ + Ca(OH)₂ → 2 NH₃ + CaSO₄.2 H₂O (gypsum); \( \frac{1584}{132} \) = 12 mol → 24 mol NH₃ and 12 mol gypsum
NiCl₂.6H₂O + 6 NH₃ → [Ni(NH₃)₆]Cl₂ + 6 H₂O; \( \frac{952}{238} \) = 4 mol → 4 mol
Gypsum (CaSO₄.2 H₂O) formed = 12 mol = 12 × 172 g = 2064 g
[Ni(NH₃)₆]Cl₂ coordination compound formed = 4 mol = 4 × 232 g = 928 g
Combined weight (in g) = 2064 + 928 = 2992 g

Teacher's Note:
1. Check the limiting reagent: 4 mol Ni needs exactly 24 mol NH₃, which is what 12 mol ammonium sulphate supplies.

Question 34. Cu²⁺ + NH₃ ⇄ [Cu(NH₃)]²⁺, K₁ = 10⁴
[Cu(NH₃)]²⁺ + NH₃ ⇄ [Cu(NH₃)₂]²⁺, K₂ = 1.58 × 10³
[Cu(NH₃)₂]²⁺ + NH₃ ⇄ [Cu(NH₃)₃]²⁺, K₃ = 5 × 10²
[Cu(NH₃)₃]²⁺ + NH₃ ⇄ [Cu(NH₃)₄]²⁺, K₄ = 10²
If the dissociation constant of [Cu(NH₃)₄]²⁺ is x × 10⁻¹², determine x. (JEE Main 2021)
Answer:
Ans. 1.26
Overall stability constant (β) = K₁ × K₂ × K₃ × K₄ = (10⁴) × (1.58 × 10³) (5 × 10²) (10²) = 7.9 × 10¹¹
∴ Dissociation constant \( = \frac{1}{\beta} = \frac{1}{7.9\times10^{11}} \) = 1.26 × 10⁻¹²

Teacher's Note:
1. Stepwise constants multiply to give β₄; the overall dissociation (instability) constant is its reciprocal.

Assertion–Reason Questions

Question 35. Assertion: The [Ni(en)₃]Cl₂ (en = ethylenediamine) has lower stability than [Ni(NH₃)₆]Cl₂.
Reason: In [Ni(en)₃]Cl₂, the geometry of Ni is trigonal bipyramidal.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. [Ni(en)₃]Cl₂ has greater stability than [Ni(NH₃)₆]Cl₂.
Correct R. [Ni(en)₃]Cl₂ has bidentate ligand, en, forming chelates (and is octahedral).

Teacher's Note:
1. Three didentate en give coordination number 6, so the geometry is octahedral; chelation makes it more stable.

Question 36. Assertion: [Al(H₂O)₆]³⁺ is a stronger acid than [Mg(H₂O)₆]²⁺.
Reason: Size of [Al(H₂O)₆]³⁺ is smaller than [Mg(H₂O)₆]²⁺ and possesses more effective nuclear charge. (AIIMS 2008)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A. This is because [Al(H₂O)₆]³⁺ has greater effective nuclear charge than [Mg(H₂O)₆]²⁺ and thus attracts electron pairs from donor groups more effectively. This also makes it relatively more acidic.

Teacher's Note:
1. A small, highly charged metal ion polarises the O–H bonds of coordinated water, making it easier to release H⁺.

Question 37. Assertion: If β₄ for [Cu(NH₃)₄]²⁺ is 2.1 × 10¹³, its instability constant is 4.76 × 10⁻¹⁴.
Reason: Overall dissociation equilibrium constant varies inversely with formation constant. (AIIMS 2009)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A.
Dissociation constant (instability constant) \( = \frac{1}{\text{Formation constant } \beta_4} = \frac{1}{2.1\times10^{13}} \) = 4.76 × 10⁻¹⁴.

Teacher's Note:
1. Instability constant = 1/β; a larger β means a more stable complex.

Question 38. Assertion: Oxidation state of Fe in Fe(CO)₅ is zero.
Reason: EAN of Fe in this complex is 36. (AIIMS 2010)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
A is true because CO is neutral ligand.
R is true because EAN of Fe = At. No. of Fe + Electrons gained by coordination = 26 + 2 × 5 = 36
But R is not the correct explanation.
Correct explanation. Oxidation state of Fe in Fe(CO)₅ is zero because CO is a neutral ligand.

Teacher's Note:
1. EAN follows from the oxidation state, not the other way round, so R cannot explain A.

Question 39. Assertion: When NO reacts with FeSO₄, a brown coloured complex is formed.
Reason: In the complex, the coordination number of Fe is + 6. (AIIMS 2009)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Both A and R are true: the brown complex is [Fe(H₂O)₅NO]²⁺, in which Fe has coordination number 6.
Correct explanation. The colour of the complex formed, [Fe(H₂O)₅NO]²⁺ is due to charge transfer (transfer of odd e⁻ on N to Fe, changing NO to NO⁺ and Fe²⁺ to Fe⁺).

Teacher's Note:
1. Coordination number is a count of donor atoms and carries no sign; colour here arises from charge transfer.