CBSE Class 12 Chemistry Coordination Compounds Case Study Questions Set 01

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Case Study 1

The extent to which the set of d-orbitals is split in the electrostatic field produced by the ligands depends upon several factors. Two of the most important factors are the nature of the ligands and the nature of the metal ion. In order to see this effect, consider the complex ion [Ti(H2O)6]3+. The Ti3+ ion has a single electron in the 3d-orbital, and we refer to it as d1 ion. In the octahedral field generated by six H2O molecules, the single electron will reside in one of the three degenerate t2g orbitals. Under spectral excitation, the electron is promoted to an eg orbital giving rise to an absorption spectrum consisting of a single peak that can be represented as shown :20,300Wavenumber (cm−1)AbsorbanceThe maximum absorption in the spectrum for [Ti(H2O)6]3+ occurs at 20,300 cm−1 which is equal to 243 kJ mol−1. This gives the value of Δo directly, but only in case of simple d1 ions. Other complexes containing the Ti3+ ion (e.g., [Ti(NH3)6]3+, [TiF6]3−, etc.) could also be prepared and spectra obtained for these complexes. If this was done, it would be observed that the absorption maximum occurs at a different energy for each complex. Because the maximum corresponds to the splitting of d-orbitals, the ligands could be ranked in terms of their ability to cause the splitting of orbital energies. Such a ranking is known as the spectrochemical series and for several common ligands the following order of decreasing energy is observed, CO > CN− > NO2− > en > NH3 > H2O > OH− > F−, Cl− > Br−. In general, the splitting in tetrahedral fields is only about half as large as that in octahedral fields.
Question 1: Which of the following ligands has lowest Δo value?
(a) CN−
(b) CO
(c) F−
(d) NH3
Show Answer & Explanation

Answer: (c) F−

Explanation:
1. Because F− is a weak field ligand.
2. In the series given, F− comes after CO, CN− and NH3, i.e., it produces the smallest splitting.

Teacher's Note:
1. Read the spectrochemical series from the passage – the ligand nearest the weak end gives the smallest Δo.

Question 2: The visible spectra of salts of the following complexes are measured in aqueous solution for which complex would the spectrum contain absorption with highest Emax values?
(a) [Co(H2O)6]2+
(b) [Co(H2O)6]3+
(c) [Co(NH3)6]3+
(d) [Co(CN)6]3−
Show Answer & Explanation

Answer: (d) [Co(CN)6]3−

Explanation:
1. [Co(CN)6]3− complex would contain absorption with highest Emax value because according to spectrochemical series, the crystal field splitting energy of CN− ion is very high.

Teacher's Note:
1. Higher oxidation state and a stronger ligand both increase Δo; Co(III) with CN− combines both.

Question 3: Which of the following statements is incorrect for complex [Ti(H2O)6]3+ ?
(a) [Ti(H2O)6]3+ is violet in colour.
(b) [Ti(H2O)6]3+ is an octahedral complex.
(c) Excitation of electron in [Ti(H2O)6]3+ occurs as, t2g1eg0 → t2g0eg1
(d) The colour of the complex [Ti(H2O)6]3+ arises due to d-d and f-f transition of the electron.
Show Answer & Explanation

Answer: (d) The colour of the complex [Ti(H2O)6]3+ arises due to d-d and f-f transition of the electron.

Explanation:
1. Ti3+ (3d1) has no f electrons; its violet colour arises only from the d-d transition t2g1eg0 → t2g0eg1.

Teacher's Note:
1. f-f transitions occur only in lanthanoid/actinoid ions; 3d complexes owe their colour to d-d transitions.

Case Study 2

Werner, a Swiss chemist in 1892 prepared and characterised a large number of coordination compounds and studied their physical and chemical behaviour. He proposed that, in coordination compounds, metals possess two types of valencies, viz. primary valencies, which are normally ionisable and secondary valencies which are non-ionisable. In a series of compounds of cobalt (III) chloride with ammonia, it was found that some of the chloride ions could be precipitated as AgCl on adding excess of AgNO3 solution in cold, but some remained in solution. The number of ions furnished by a complex in a solution can be determined by precipitation reactions. The measurement of molar conductance of solutions of coordination compounds helps to estimate the number of ions furnished by the compound in solution.

Question 4:
Assertion (A): The complex [Co(NH3)3Cl3] does not give precipitate with silver nitrate solution.
Reason (R): The given complex is non-ionisable.
(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.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. All three Cl− are inside the coordination sphere, so the complex gives no free Cl− ions and no AgCl is precipitated.

Teacher's Note:
1. A neutral complex with no counter-ions is a non-electrolyte – zero ions, zero precipitate.

Question 5:
Assertion (A): The complex [Co(NH3)4Cl2]Cl gives precipitate corresponding to 2 mol of AgCl with AgNO3 solution.
Reason (R): It ionises as [Co(NH3)4Cl2]+ + Cl−.
(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.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. [Co(NH3)4Cl2]Cl + AgNO3 → [Co(NH3)4Cl2]+ + AgCl↓
2. Thus it gives precipitate of 1 mol of AgCl.

Teacher's Note:
1. Count only the Cl− outside the square brackets – here just one.

Question 6:
Assertion (A): CoCl3·4NH3 gives 1 mol of AgCl on reacting with AgNO3, its secondary valency is 6.
Reason (R): Secondary valency corresponds to coordination number.
(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.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. CoCl3·4NH3 gives 1 mol AgCl on reaction with AgNO3, hence the complex can be represented as [CoCl2(NH3)4]Cl.

Teacher's Note:
1. Secondary valency 6 = 4 NH3 + 2 Cl inside the bracket; the one ionisable Cl− satisfies primary valency only.

Question 7:
Assertion (A): 1 mol of [CrCl2(H2O)4]Cl·2H2O will give 1 mol of AgCl on treating with AgNO3.
Reason (R): Cl− ions satisfying secondary valencies will not be precipitated.
(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.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The Cl− ions outside the coordination sphere can only be precipitated.

Teacher's Note:
1. Water of crystallisation (2H2O) is outside the complex and does not affect the AgCl count.

Question 8:
Assertion (A): CoCl3·3NH3 is not conducting while CoCl3·5NH3 is conducting.
Reason (R): The complex of CoCl3·3NH3 is [CoCl3(NH3)3] while that of CoCl3·5NH3 is [CoCl(NH3)5]Cl2.
(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.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. [CoCl3(NH3)3] gives no ions in solution, while [CoCl(NH3)5]Cl2 → [CoCl(NH3)5]2+ + 2Cl−, so only the second conducts.

Teacher's Note:
1. Conductivity is a quick test of how many ions a complex furnishes in solution.

Case Study 3

Valence bond theory considers the bonding between the metal ion and the ligands as purely covalent. On the other hand, crystal field theory considers the metal-ligand bond to be ionic arising from electrostatic interaction between the metal ion and the ligands. In coordination compounds, the interaction between the ligand and the metal ion causes the five d-orbitals to split-up. This is called crystal field splitting and the energy difference between the two sets of energy level is called crystal field splitting energy. The crystal field splitting energy (Δo) depends upon the nature of the ligand. The actual configuration of complexes is decided by the relative values of Δo and P (pairing energy).
If Δo < P, then complex will be high spin.
If Δo > P, then complex will be low spin.

Question 9: The crystal field splitting energy for octahedral (Δo) and tetrahedral (Δt) complex is related as
(a) Δt = \( \frac{1}{2} \)Δo
(b) Δt = \( \frac{4}{9} \)Δo
(c) Δt = \( \frac{3}{5} \)Δo
(d) Δt = \( \frac{2}{5} \)Δo
Show Answer & Explanation

Answer: (b) Δt = \( \frac{4}{9} \)Δo

Explanation:
1. Δt = \( \frac{4}{9} \) Δo, because a tetrahedral complex has only four ligands and none of them points directly at the d-orbitals.

Teacher's Note:
1. 3/5 and 2/5 are the energy shifts of eg and t2g in an octahedral field – do not confuse them with the Δt/Δo ratio.

Question 10: On the basis of crystal field theory, the electronic configuration of d4 in two situations :
(a) Δo > P and (b) Δo < P are
(a) (a) t2g4eg0, (b) t2g3eg1
(b) (a) t2g3eg1, (b) t2g4eg0
(c) (a) t2g3eg1, (b) t2g3eg1
(d) (a) t2g4eg0, (b) t2g4eg0
Show Answer & Explanation

Answer: (a) (a) t2g4eg0, (b) t2g3eg1

Explanation:
1. When Δo > P, the electrons paired up in the t2g level rather than going to the eg level, so
2. when Δo > P : t2g4eg0
3. and Δo < P : t2g3eg1

Teacher's Note:
1. Only d4 to d7 have two possible arrangements, so these are the configurations where Δo vs P matters.

Question 11: Using crystal field theory, calculate magnetic moment of central metal ion of [FeF6]4−.
(a) 1.79 B.M.
(b) 2.83 B.M.
(c) 3.85 B.M.
(d) 4.9 B.M.
Show Answer & Explanation

Answer: (d) 4.9 B.M.

Explanation:
1. Fe2+ : 3d6 => t2g4eg2 (Since, F− is a weak field ligand)
2. Hence four unpaired electrons are present.
3. Magnetic moment (μ) = \( \sqrt{n(n+2)} = \sqrt{4(4+2)} \) = 4.9 B.M.

Teacher's Note:
1. Learn the spin-only values: 1, 2, 3, 4, 5 unpaired electrons give 1.73, 2.83, 3.87, 4.90 and 5.92 B.M.

Question 12: Electronic configuration of d-orbitals in [Ti(H2O)6]3+ ion in an octahedral crystal field is
(a) t2g1eg0
(b) t2g2eg0
(c) t2g0eg1
(d) t2g1eg1
Show Answer & Explanation

Answer: (a) t2g1eg0

Explanation:
1. In [Ti(H2O)6]3+, Ti is in +3 oxidation state and there is only one electron in d-orbital.

Teacher's Note:
1. A single electron always goes into the lower-energy t2g set in the ground state.

Case Study 4

Coordination compounds are formulated and named according to the IUPAC system.
Few rules for naming coordination compounds are :
(I) In ionic complex, the cation is named first and then the anion.
(II) In the coordination entity, the ligands are named first and then the central metal ion.
(III) When more than one type of ligands are present, they are named in alphabetical order of preference without any consideration of charge.

Question 13: The IUPAC name of the complex [Pt(NH3)3Br(NO2)Cl]Cl is
(a) triamminechlorobromonitroplatinum(IV) chloride
(b) triamminebromonitrochloroplatinum(IV) chloride
(c) triamminebromidochloridonitroplatinum(IV) chloride
(d) triamminenitrochlorobromoplatinum(IV) chloride.
Show Answer & Explanation

Answer: (c) triamminebromidochloridonitroplatinum(IV) chloride

Explanation:
1. Ligands are named in alphabetical order irrespective of their charge.
2. Pt + 3(0) + (−1) + (−1) + (−1) = +1, so Pt = +4: triamminebromidochloridonitroplatinum(IV) chloride.

Teacher's Note:
1. Alphabetise by the ligand name, ignoring multiplying prefixes: ammine, bromido, chlorido, nitro.

Question 14: The IUPAC name of [Ni(CO)4] is
(a) tetracarbonylnickel(II)
(b) tetracarbonylnickel(0)
(c) tetracarbonylnickelate(II)
(d) tetracarbonylnickelate(0).
Show Answer & Explanation

Answer: (b) tetracarbonylnickel(0)

Explanation:
1. CO is a neutral ligand and the complex is neutral, so Ni is in 0 oxidation state; a neutral complex keeps the plain metal name: tetracarbonylnickel(0).

Teacher's Note:
1. The suffix ‘ate’ is used only when the complex itself is an anion.

Question 15: As per IUPAC nomenclature, the name of the complex [Co(H2O)4(NH3)2]Cl3 is
(a) tetraaquadiamminecobalt(II) chloride
(b) tetraaquadiamminecobalt(III) chloride
(c) diamminetetraaquacobalt(II) chloride
(d) diamminetetraaquacobalt(III) chloride.
Show Answer & Explanation

Answer: (d) diamminetetraaquacobalt(III) chloride.

Explanation:
1. Ammine comes before aqua in alphabetical order (prefixes di, tetra are ignored), and Co + 0 = +3 gives cobalt(III).
2. Hence the name is diamminetetraaquacobalt(III) chloride.

Teacher's Note:
1. Never alphabetise by the prefix: ‘tetraaqua’ is sorted under ‘a’ for aqua, after ‘ammine’.

Question 16: Which of the following represents correct formula of dichloridobis(ethane-1, 2-diamine)cobalt(III) ion?
(a) [CoCl2(en)]2+
(b) [CoCl2(en)2]2+
(c) [CoCl2(en)]+
(d) [CoCl2(en)2]+
Show Answer & Explanation

Answer: (d) [CoCl2(en)2]+

Explanation:
1. ‘bis’ means two en ligands; charge = +3 + 2(−1) + 0 = +1, so the formula is [CoCl2(en)2]+.

Teacher's Note:
1. ‘bis’, ‘tris’ are used instead of di, tri for ligands like en whose names already contain a number.

Question 17: Correct formula of pentaamminenitrito-O-cobalt(III) sulphate is
(a) [Co(NO2)(NH3)5]SO4
(b) [Co(ONO)(NH3)5]SO4
(c) [Co(NO2)(NH3)4](SO4)2
(d) [Co(ONO)(NH3)4](SO4)2
Show Answer & Explanation

Answer: (b) [Co(ONO)(NH3)5]SO4

Explanation:
1. Ligand NO2− is ambidentate ligand as it can donate electrons through either nitrogen (NO2) or oxygen (ONO).
2. Nitrito-O means bonding through oxygen, so the formula is [Co(ONO)(NH3)5]SO4.

Teacher's Note:
1. Write ONO for O-bonded nitrite and NO2 for N-bonded nitrite – the formula must show the donor atom.

Case Study 5

To explain bonding in coordination compounds various theories were proposed. One of the important theory was valence bond theory. According to that, the central metal ion in the complex makes available a number of empty orbitals for the formation of coordination bonds with suitable ligands. The appropriate atomic orbitals of the metal hybridise to give a set of equivalent orbitals of definite geometry.
The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e., (n − 1)d or outer d-orbitals i.e., nd.
For example, Co3+ forms both inner orbital and outer orbital complexes, with ammonia it forms [Co(NH3)6]3+ and with fluorine it forms [CoF6]3− complex ion.

Question 18: Which of the following is not true for [CoF6]3−?
(a) It is paramagnetic.
(b) It has coordination number of 6.
(c) It is outer orbital complex.
(d) It involves d2sp3 hybridisation.
Show Answer & Explanation

Answer: (d) It involves d2sp3 hybridisation.

Explanation:
1. It involves sp3d2 hybridisation and not d2sp3.

Teacher's Note:
1. F− is weak field, so the four unpaired 3d electrons of Co3+ stay unpaired and outer 4d orbitals are used.

Question 19: [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 central metal of the complex is
(a) 3dxy1, 3dx2−y21, 3dyz1
(b) 3dxy1, 3dyz1, 3dzx1
(c) 3dxy1, 3dzy1, 3dz21
(d) 3dx2−y21, 3dz21, 3dxz1
Show Answer & Explanation

Answer: (b) 3dxy1, 3dyz1, 3dzx1

Explanation:
1. Magnetic moment of 3.83 B.M. suggests that it has 3 unpaired electrons,
2. ∴ n = 3 i.e., Cr3+ : 3d3
3. It involves d2sp3 hybridisation so correct distribution of electrons is 3dxy1, 3dyz1, 3dzx1.

Teacher's Note:
1. In d2sp3 the two eg orbitals (dx²−y², dz²) are used for bonding, so the three electrons must sit in the t2g set.

Question 20: Which of the following is true for [Co(NH3)6]3+ ?
(a) It is an octahedral, diamagnetic and outer orbital complex.
(b) It is an octahedral, paramagnetic and outer orbital complex.
(c) It is an octahedral, paramagnetic and inner orbital complex.
(d) It is an octahedral, diamagnetic and inner orbital complex.
Show Answer & Explanation

Answer: (d) It is an octahedral, diamagnetic and inner orbital complex.

Explanation:
1. [Co(NH3)6]3+ is d2sp3 hybridised with all electrons paired hence, it is diamagnetic and inner orbital complex.

Teacher's Note:
1. NH3 behaves as a strong field ligand towards Co3+, so the 3d6 electrons pair up.

Question 21: The paramagnetism of [CoF6]3− is due to
(a) 3 electrons
(b) 4 electrons
(c) 2 electrons
(d) 1 electron.
Show Answer & Explanation

Answer: (b) 4 electrons

Explanation:
1. Co3+ :

3d4s4p
↑↓↑↑↑↑    

2. [CoF6]3− :
3d4s4p4d
↑↓↑↑↑↑××××××××××××   
sp3d2 hybridisation (six F−)

Teacher's Note:
1. Co3+ is 3d6: one pair plus four single electrons when the ligand is weak field.

Question 22: Which of the following is an inner orbital or low spin complex?
(a) [Ni(H2O)6]3+
(b) [FeF6]3−
(c) [Co(CN)6]3−
(d) [NiCl4]2−
Show Answer & Explanation

Answer: (c) [Co(CN)6]3−

Explanation:
1. Inner orbital complexes are formed with strong ligands as they force electrons to pair up and hence the complex will be either diamagnetic or will have less number of unpaired electrons.
2. CN− is a strong field ligand, so [Co(CN)6]3− is a d2sp3 inner orbital, low spin complex.

Teacher's Note:
1. Look for the strong field ligand (CN−, CO) first; F−, Cl− and H2O usually give outer orbital/high spin complexes.

Case Study 6

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. Ligands are treated as point charges in case of anions or dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands (either anions or the negative ends of dipolar molecules like NH3 and H2O) in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals.

Question 23: Consider the coordination compound, Na2[Pt(CN)4], identify the lewis acid.
(a) [Pt(CN)4]2−
(b) Na+
(c) Pt2+
(d) CN−
Show Answer & Explanation

Answer: (c) Pt2+

Explanation:
1. Lewis acids are electron deficient and Pt has vacant orbital to accept electrons whereas lewis bases are electron donors i.e., CN−.

Teacher's Note:
1. In every complex the central metal ion is the Lewis acid (electron-pair acceptor) and the ligands are Lewis bases.

Question 24: The CFSE for octahedral [CoCl6]4− is 18,000 cm−1. The CFSE for tetrahedral [CoCl4]2− will be
(a) 18,000 cm−1
(b) 16,000 cm−1
(c) 8,000 cm−1
(d) 20,000 cm−1
Show Answer & Explanation

Answer: (c) 8,000 cm−1

Explanation:
1. CFSE for tetrahedral complex is Δt = \( \frac{4}{9} \) Δo
2. Δt = \( \frac{4}{9} \) × 18,000 = 8,000 cm−1

Teacher's Note:
1. The 4/9 relation holds only for the same metal ion, the same ligands and the same metal–ligand distance.

Question 25: An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is because _____________.
(a) [Co(H2O)6]2+ is transformed into [CoCl6]4−.
(b) [Co(H2O)6]2+ is transformed into [CoCl4]2−.
(c) tetrahedral complexes have larger crystal field splitting than octahedral complex.
(d) None of the above
Show Answer & Explanation

Answer: (b) [Co(H2O)6]2+ is transformed into [CoCl4]2−.

Explanation:
1. Due to d–d transition in [Co(H2O)6]2+, where electron from t2g shifts into eg energy level, it absorbs light and shows complementary colour such that pink coloured aqueous CoCl2 is obtained. Thus, on excess addition, HCl transforms [Co(H2O)6]2+ into [CoCl4]2−.

Teacher's Note:
1. The colour change shows a change of geometry: octahedral [Co(H2O)6]2+ (pink) to tetrahedral [CoCl4]2− (blue), whose smaller splitting shifts the absorption.

Question 26: 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) thiosulphato
(b) oxalato
(c) glycinato
(d) ethane-1,2-diamine
Show Answer & Explanation

Answer: (a) thiosulphato

Explanation:
1. Thiosulphato is a monodentate ligand whereas oxalato, glycinato and ethylene diamine are bidentate ligands and can form rings with the central metal ion. So, they are also chelating ligands. Thiosulphato is a monodentate ligand and hence, cannot form chelate rings. Hence, it is not a chelating ligand.

Teacher's Note:
1. Learn the common bidentate ligands: en, ox2− (oxalato), gly− (glycinato) and acac; EDTA4− is hexadentate.

Case Study 7

According to Valence Bond Theory, the metal atom or ion under the influence of ligands can use its (n − 1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with the ligand orbitals that can donate electron pairs for bonding.

Question 27:
Assertion (A): In the diamagnetic octahedral complex, [Co(NH3)6]3+, the cobalt ion is in +3 oxidation state.
Reason (R): Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals.
(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.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. In the octahedral complex, [Co(NH3)6]3+, the cobalt ion is in +3 oxidation state and has the electronic configuration 3d6. NH3 being a strong field ligand uses the d, s, p orbitals to hybridise and pair up 4e− and form inner orbital complex. Unavailability of unpaired electrons in the complex makes it diamagnetic.

Teacher's Note:
1. The oxidation state comes from charge balance (x + 0 = +3); the reason describes bonding, which does not explain the oxidation state.

Question 28:
Assertion (A): [NiCl4]2− is an inner orbital complex.
Reason (R): An inner orbital or low spin or spin paired complex uses inner d orbitals of the metal ion for 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.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. [NiCl4]2− is high spin complex because Cl− is a weak field ligand.
2. Hence, Ni2+ uses outer 4s and 4p orbitals (sp3), so it is an outer orbital complex.

Teacher's Note:
1. Tetrahedral sp3 complexes never use d orbitals, so they cannot be inner orbital complexes.

Question 29:
Assertion (A): [Ni(CN)4]2− is square planar complex, the hybridisation involved is dsp2.
Reason (R): In [Ni(CN)4]2−, nickel is in +2 oxidation state and has the electronic configuration 3d8.
(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.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. In [Ni(CN)4]2−, CN− being strong field ligand makes involvement of inner-d-orbitals and form inner orbital complex.
2.

3d4s4p
Ni2+↑↓↑↓↑↓↑↑    
Ni in [Ni(CN)4]2−↑↓↑↓↑↓↑↓×××××××× 

3. (×× = electron pair from CN−)
4. Hybridisation: dsp2; Geometry: Square planar

Teacher's Note:
1. A 3d8 configuration alone does not decide the geometry: it is the strong field CN− that pairs the electrons and frees one 3d orbital for dsp2.

Question 30:
Assertion (A): The paramagnetic octahedral complex, [CoF6]3− uses outer orbital (4d) in hybridisation (sp3d2).
Reason (R): It is a high spin complex.
(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.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. In [CoF6]3−, cobalt (Co+3) ion uses outer 4d orbital in hybridisation due to weak field ligand (F−).
2.

3d4s4p4d
Co3+ ion↑↓↑↑↑↑         
Co in [CoF6]3−↑↓↑↑↑↑××××××××××××   

3. (×× = electron pair from F−)
4. Hybridisation: sp3d2
5. Hence, complex is outer orbital or high spin complex.

Teacher's Note:
1. Outer orbital and high spin go together in VBT, but the cause of both is the weak field F− ligand, not one statement causing the other.

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Chemistry Class 12 Exam Resources: Unit 5 Coordination Compounds

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