CBSE Class 12 Chemistry Coordination Compounds NEET JEE Questions Set 02

Single Correct Answer

Question 1. Which one of the following complexes shows optical isomerism ? (JEE Main 2016)
(a) [Co(NH₃)₃Cl₃]
(b) [Co(en)₂Cl₂] Cl
(c) trans [Co(en)₂Cl₂] Cl
(d) [Co(NH₃)₄Cl₂] Cl (en = ethylenediamine)
Answer:
Ans. (b)
(a) [Co(NH₃)₃Cl₃] has two geometrical isomers but both are optically inactive due to plane of symmetry.
(b) cis [Co(en)₂Cl₂] Cl is optically active.
(c) trans [Co(en)₂Cl₂] Cl is optically inactive due to plane of symmetry.
(d) [Co(NH₃)₄Cl₂] Cl has two geometrical isomers but both are optically inactive due to plane of symmetry.

Teacher's Note:
1. Option (b) has no prefix, so it includes the cis form, which is chiral; option (c) names only the trans form.

Question 2. Consider the following reaction and statements :
[Co(NH₃)₄Br₂]⁺ + Br⁻ → [Co(NH₃)₃Br₃] + NH₃
(I) Two isomers are produced if the reactant complex ion is a cis-isomer
(II) Two isomers are produced if the reactant complex ion is a trans-isomer
(III) Only one isomer is produced if the reactant complex ion is a trans-isomer
(IV) Only one isomer is produced if the reactant complex ion is a cis-isomer
The correct statements are : (JEE Main 2018)
(a) (I) and (II)
(b) (I) and (III)
(c) (III) and (IV)
(d) (II) and (IV)
Answer:
Ans. (b)
Case (i) : In the cis-isomer, the two Br are adjacent. If the incoming Br⁻ replaces an NH₃ adjacent to both Br, the fac-isomer is formed; if it replaces an NH₃ trans to one Br, the mer-isomer is formed. (Two isomers)
Case (ii) : In the trans-isomer, all four NH₃ lie in one plane between the two Br. Replacing any one of them gives three Br along a meridian, i.e., the mer-isomer only. (One isomer)
Hence, statements (I) and (III) are correct.

Teacher's Note:
1. Picture the octahedron: in the trans reactant all four NH₃ are equivalent, so substitution can give only one product.

Question 3. The type of isomerism shown by the complex [CoCl₂(en)₂] is (NEET 2018)
(a) ionization isomerism
(b) coordination isomerism
(c) geometrical isomerism
(d) linkage isomerism
Answer:
Ans. (c)
The complex [CoCl₂(en)₂] is a disubstituted octahedral with C.N. = 6. It is of [M(AA)₂B₂] type and shows geometrical isomerism having cis- and trans-isomers: the cis-form is optically active and the trans-form is optically inactive.

Teacher's Note:
1. There is no counter ion to exchange and no ambidentate ligand, so ionisation and linkage isomerism are ruled out at once.

Question 4. The number of possible optical isomers for the complexes MA₂B₂ with sp³ and dsp² hybridized metal atom, respectively, is
Note : A and B are unidentate neutral and unidentate monoanionic ligands, respectively. (JEE Main 2020)
(a) 0 and 0
(b) 2 and 2
(c) 0 and 2
(d) 0 and 1
Answer:
Ans. (a)
Tetrahedral complexes with unidentate neutral ligands of the type MA₂B₂ will not show optical isomerism. Square planar complexes of the type MA₂B₂ where A and B are unidentate monoanionic ligands will have a plane of symmetry and therefore, will not show optical isomerism.

Teacher's Note:
1. A tetrahedral centre is chiral only with four different groups; a square planar complex always has its molecular plane as a mirror plane.

Question 5. An excess of AgNO₃ is added to 100 mL of 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitated would be : (AIPMT 2013)
(a) 0.001
(b) 0.002
(c) 0.003
(d) 0.01
Answer:
Ans. (a)
The complex is [Cr(H₂O)₄Cl₂] Cl. It ionizes as
[Cr(H₂O)₄Cl₂] Cl ⇄ [Cr(H₂O)₄Cl₂]⁺ + Cl⁻ (1 mole → 1 mole)
Thus, 1 mole of the complex produces 1 mole of Cl⁻ ions and hence will precipitate out 1 mole of AgCl.
100 mL of 0.01 M complex solution contains moles of the complex \( = \frac{0.01}{1000}\times 100 \) = 0.001 mole
and will precipitate out 0.001 mole of AgCl.

Teacher's Note:
1. Write the formula from the name first: “dichloro” Cl are inside the sphere, so only one Cl⁻ per formula unit is ionisable.

Question 6. The correct order of the stoichiometries of AgCl formed when AgNO₃ in excess is treated with the complexes CoCl₃.6 NH₃, CoCl₃.5 NH₃, CoCl₃.4 NH₃ respectively is (NEET 2017)
(a) 1 AgCl, 3 AgCl, 2 AgCl
(b) 3 AgCl, 1 AgCl, 2 AgCl
(c) 3 AgCl, 2 AgCl, 1 AgCl
(d) 2 AgCl, 3 AgCl, 1 AgCl
Answer:
Ans. (c)
Remembering that coordination number of Co is 6, the complex CoCl₃.6 NH₃ can be written as [Co(NH₃)₆] Cl₃ which provides 3 Cl⁻ ions to be precipitated with excess of AgNO₃ solution giving 3 moles of AgCl precipitate. Similarly, CoCl₃.5 NH₃ is [Co(NH₃)₅Cl] Cl₂ which gives 2 moles of AgCl precipitate and CoCl₃.4 NH₃ is [Co(NH₃)₄Cl₂] Cl which gives 1 mole of AgCl precipitate with AgNO₃ solution. Thus, the correct order of stoichiometries of AgCl formed is 3 AgCl, 2 AgCl and 1 AgCl.

Teacher's Note:
1. Fill the six coordination sites with NH₃ first, then with Cl; whatever Cl is left over is ionisable.

Question 7. A solution containing 0.319 g of CrCl₃.6 H₂O was passed through a cation exchange resin and acid coming out of the cation exchange resin required 28.5 ml of 0.125 M NaOH. Determine correct formula of the complex [Mol. wt. of the complex = 266.5] (AIIMS 2014)
(a) [Cr(H₂O)₆] Cl₃
(b) [Cr(H₂O)₅Cl] H₂O.Cl₂
(c) [Cr(H₂O)₄Cl₂] Cl.2 H₂O
(d) [Cr(H₂O)₃Cl₃].3 H₂O
Answer:
Ans. (a)
Suppose the number of ionizable Cl⁻ ions (present outside the coordination sphere) = n. When the solution of the complex passes through the cation exchanger, these Cl⁻ ions will combine with H⁺ ions of the cation exchanger to form HCl
n Cl⁻ + n H⁺ → n HCl
Thus, 1 mole of the complex will form n moles of HCl
1 mole of the complex ≡ n moles of HCl ≡ n moles of NaOH
Moles of the complex \( = \frac{0.319}{266.5} \) = 0.0012 mole
Moles of NaOH used \( = \frac{28.5\times 0.125}{1000} \) = 0.0036 mole
0.0012 mole of the complex = 0.0036 mole NaOH ≡ 0.0036 mole HCl
∴ 1 mole of the complex \( = \frac{0.0036}{0.0012} \) = 3 moles of HCl = 3 moles of Cl⁻ ions
Thus, all the Cl⁻ ions are outside the coordination sphere. Hence, the complex is [Cr(H₂O)₆] Cl₃.

Teacher's Note:
1. Each ionisable Cl⁻ releases one H⁺ from the resin, so moles of NaOH used per mole of complex equal the number of ionisable Cl⁻.

Question 8. The two isomers X and Y with the formula Cr(H₂O)₅ClBr₂ were taken for experiment on depression in freezing point. It was found that one mole of X gave depression corresponding to 2 moles of particles and one mole of Y gave depression due to 3 moles of particles. The structural formulae of X and Y respectively are (Kerala PET 2008)
(a) [Cr(H₂O)₅Cl]Br₂, [Cr(H₂O)₄Br₂]Cl . H₂O
(b) [Cr(H₂O)₅Cl]Br₂, [Cr(H₂O)₃ClBr₂] . 2 H₂O
(c) [Cr(H₂O)₅Br]BrCl, [Cr(H₂O)₄ClBr]Br . H₂O
(d) [Cr(H₂O)₅Cl]Br₂, [Cr(H₂O)₄ClBr]Br . H₂O
(e) [Cr(H₂O)₄Br₂]Cl . H₂O, [Cr(H₂O)₅Cl]Br₂
Answer:
Ans. (e)
1 mole of complex X giving 2 moles of particles will be [Cr(H₂O)₄Br₂] Cl . H₂O, i.e.,
[Cr(H₂O)₄Br₂]⁺ + Cl⁻
1 mole of complex Y giving 3 moles of particles will be [Cr(H₂O)₅Cl]Br₂, i.e.,
[Cr(H₂O)₅Cl]²⁺ + 2 Br⁻.

Teacher's Note:
1. Water of crystallisation does not ionise; count only the complex ion plus the counter ions to get the number of particles.

Question 9. The correct statement with respect to the complexes Ni(CO)₄ and [Ni(CN)₄]²⁻ is (Kerala PET 2011)
(a) nickel is in the same oxidation state in both
(b) both have tetrahedral geometry
(c) both have square planar geometry
(d) have square planar and tetrahedral geometry respectively
(e) have tetrahedral and square planar geometry respectively
Answer:
Ans. (e)
Ni(CO)₄ → sp³ hybridisation → tetrahedral
[Ni(CN)₄]²⁻ → dsp² hybridisation → square planar.

Teacher's Note:
1. Ni(0) in Ni(CO)₄ becomes 3d¹⁰ (no free d orbital), while Ni(II) with strong-field CN⁻ pairs its 3d⁸ electrons and frees one 3d orbital for dsp².

Question 10. The complex which has the highest magnetic moment among the following is (Kerala PET 2011)
(a) [CoF₆]³⁻
(b) [Co(NH₃)₆]³⁺
(c) [Ni(NH₃)₄]²⁺
(d) [Ni(CN)₄]²⁻
(e) [Fe(CN)₆]⁴⁻
Answer:
Ans. (a)
No. of unpaired electrons are 4 in (a), 0 in (b), (c), (d) and (e). Hence, [CoF₆]³⁻ has highest magnetic moment.

Teacher's Note:
1. F⁻ is a weak-field ligand, so Co³⁺ (3d⁶) stays high spin with four unpaired electrons.

Question 11. Among [Ni(CO)₄], [Ni(CN)₄]²⁻, [NiCl₄]²⁻ species, the hybridisation states at the Ni atom are, respectively (Atomic number of Ni = 28) (IIT 2008, AMU Engg. 2011)
(a) sp³, dsp², dsp²
(b) sp³, dsp², sp³
(c) sp³, sp³, dsp²
(d) dsp², sp³, sp³
Answer:
Ans. (b)
[Ni(CO)₄]: Ni(0), 3d¹⁰ after pairing by CO, so 4s and three 4p orbitals are used: sp³ (tetrahedral).
[Ni(CN)₄]²⁻: Ni²⁺, 3d⁸; strong-field CN⁻ pairs the electrons and one 3d orbital becomes free: dsp² (square planar).
[NiCl₄]²⁻: Ni²⁺, 3d⁸; weak-field Cl⁻ does not cause pairing: sp³ (tetrahedral, 2 unpaired electrons).

Teacher's Note:
1. For four-coordinate Ni(II), the ligand decides: strong field gives square planar dsp², weak field gives tetrahedral sp³.

Question 12. The hybridization involved in the complex [Ni(CN)₄]²⁻ is (At. No. of Ni = 28) (RE-AIPMT 2015)
(a) d²sp²
(b) d²sp³
(c) dsp²
(d) sp³
Answer:
Ans. (c)
[Ni(CN)₄]²⁻ has Ni in +2 oxidation state.
Ni²⁺ = [Ar]¹⁸ 3d⁸4s⁰ (two unpaired electrons in 3d)
In presence of strong field ligand CN⁻, electrons in d-orbitals pair up, leaving one 3d orbital empty. One 3d, one 4s and two 4p orbitals accept the four CN⁻ electron pairs: dsp² hybridization.

Teacher's Note:
1. dsp² uses the inner 3d_x²−y² orbital, which is why the complex is square planar and diamagnetic.

Question 13. Which of the following complexes are not correctly matched with the hybridisation of their central metal ion ?
(i) [Ni(CO)₄] sp³ (ii) [Ni(CN)₄]²⁻ sp³
(iii) [CoF₆]³⁻ d²sp³ (iv) [Fe(CN)₆]³⁻ sp³d²
Select the correct answer using the codes given below :
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (i), (iii) and (iv)
(e) (ii), (iii) and (iv)
Answer:
Ans. (e)
Ni(CO)₄ = sp³, [Ni(CN)₄]²⁻ = dsp², [CoF₆]³⁻ = sp³d², [Fe(CN)₆]³⁻ = d²sp³.
(F⁻ is a weak ligand and cannot force d electrons to pair up forming outer orbital complex.)
Thus (ii), (iii) and (iv) are wrongly matched.

Teacher's Note:
1. Inner orbital (d²sp³) needs a strong-field ligand like CN⁻; weak F⁻ gives outer orbital sp³d².

Question 14. Pick out the correct statement with respect to [Mn(CN)₆]³⁻ (NEET 2017)
(a) It is sp³d²-hybridized and octahedral
(b) It is sp³d²-hybridized and tetrahedral
(c) It is d²sp³-hybridized and octahedral
(d) It is dsp²-hybridized and square planar
Answer:
Ans. (c)
In [Mn(CN)₆]³⁻, Mn³⁺ = 3d⁴ (four unpaired electrons).
In presence of strong field ligand, CN⁻, two electrons pair up, so 3d has one paired and two unpaired electrons and two 3d orbitals become empty.
Two 3d, one 4s and three 4p orbitals accept six CN⁻ electron pairs: d²sp³-hybridization.
Thus, the complex is d²sp³-hybridized having octahedral shape.

Teacher's Note:
1. Six ligands always mean octahedral; only the choice between d²sp³ and sp³d² depends on the ligand strength.

Question 15. Which of the following complex exhibits the highest paramagnetic behaviour ?
where gly = glycine, en = ethylene diamine and bpy = bipyridyl moities (At Nos. Ti = 22, V = 23, Fe = 26, Co = 27) (AIPMT 2008)
(a) [V(gly)₂ (OH)₂(NH₃)₂]⁺
(b) [Fe(en)(bpy) (NH₃)₂]²⁺
(c) [Co(OX)₂(OH)₂]²⁻
(d) [Ti(NH₃)₆]³⁺
Answer:
Ans. (c)
Ox. state of V in (a) : x + 2(−1) + 2(−1) + 0 = +1 or x = +5
∴ Valence shell config. = 3d⁰ (i.e., no unpaired electron)
Ox. state of Fe in (b) : x + 0 + 0 + 0 = +2 or x = +2
∴ Valence shell config. = 3d⁶ (i.e., 4 unpaired electrons)
Ox. state of Co in (c) : x + 2(−2) + 2(−1) = −2 or x = +4
∴ Valence shell config. = 3d⁵ (i.e., 5 unpaired electrons)
Ox. state of Ti in (d) : x + 6(0) = +3 or x = +3
∴ Valence shell config. = 3d¹ (i.e., 1 unpaired electron)
Thus, complex of cobalt has maximum unpaired electrons (viz. 5) and hence has highest paramagnetism.

Teacher's Note:
1. Find the oxidation state first: glycinate and OH⁻ carry −1 each and oxalate −2, then count the d electrons.

Question 16. The pair having the same magnetic moment is [At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27] (JEE Main 2016)
(a) [Cr(H₂O)₆]²⁺ and [CoCl₄]²⁻
(b) [Cr(H₂O)₆]²⁺ and [Fe(H₂O)₆]²⁺
(c) [Mn(H₂O)₆]²⁺ and [Cr(H₂O)₆]²⁺
(d) [CoCl₄]²⁻ and [Fe(H₂O)₆]²⁺
Answer:
Ans. (b)
[Cr(H₂O)₆]²⁺ : Cr²⁺ = [Ar] 3d⁴ 4s⁰ → 4 unpaired electrons
[CoCl₄]²⁻ : Co²⁺ = [Ar] 3d⁷ 4s⁰ → 3 unpaired electrons
[Fe(H₂O)₆]²⁺ : Fe²⁺ = [Ar] 3d⁶ 4s⁰ → 4 unpaired electrons
[Mn(H₂O)₆]²⁺ : Mn²⁺ = [Ar] 3d⁵ 4s⁰ → 5 unpaired electrons
Thus, [Cr(H₂O)₆]²⁺ and [Fe(H₂O)₆]²⁺ have the same number of unpaired electrons and hence have same magnetic moment.

Teacher's Note:
1. With weak-field H₂O and Cl⁻ there is no pairing, so d⁴ and d⁶ both have four unpaired electrons.

Question 17. Complex X of composition of Cr(H₂O)₆Cl_n has a spin only magnetic moment of 3.83 BM. It reacts with AgNO₃ and shows geometrical isomerism. The IUPAC nomenclature of X is (JEE Main 2020)
(a) dichloridotetraaqua chromium (IV) chloride dihydrate
(b) hexaaqua chromium (III) chloride
(c) tetraaquadichloridochromium(III) chloride dihydrate
(d) tetraaquadichlorido chromium (IV) chloride dihydrate
Answer:
Ans. (c)
The complex Cr(H₂O)₆Cl_n has spin only μ = 3.83 BM
This shows n = 3 and the O.N. of Cr = + 3
Thus, the complex is Cr(H₂O)₆Cl₃ which shows geometrical isomerism.
Therefore, formula of the complex is [Cr(H₂O)₄Cl₂]Cl·2H₂O
IUPAC name is tetraaquadichloridochromium(III) chloride dihydrate.

Teacher's Note:
1. μ = 3.87 BM corresponds to three unpaired electrons, i.e. Cr³⁺ (3d³); ligands are named alphabetically (aqua before chlorido).

Question 18. The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 BM. The suitable ligand for this complex is (JEE Main 2019)
(a) CN⁻
(b) NCS⁻
(c) CO
(d) ethylenediamine
Answer:
Ans. (b)
\( \mu = \sqrt{n(n+2)} \) = 5.9 BM. This is so when n = 5
i.e., Mn²⁺ should have 5 unpaired electrons
Mn²⁺ = [Ar]¹⁸ 3d⁵
The 5 unpaired electrons will remain unpaired only in presence of weak field ligand viz. NCS⁻.

Teacher's Note:
1. 5.9 BM means high spin d⁵; of the options only thiocyanate is a weak-field ligand.

Question 19. In which of the following orders, the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment ?
(i) [FeF₆]³⁻ (ii) [Co(NH₃)₆]³⁺ (iii) [NiCl₄]²⁻ (iv) [Cu(NH₃)₄]²⁺ (JEE Main 2021)
(a) (ii) > (i) > (iii) > (iv)
(b) (iii) > (iv) > (ii) > (i)
(c) (ii) > (iii) > (i) > (iv)
(d) (i) > (iii) > (iv) > (ii)
Answer:
Ans. (d)
(i) [FeF₆]³⁻ : Fe³⁺ = 3d⁵ → 5 unpaired electrons as F⁻ is weak field ligand
(ii) [Co(NH₃)₆]³⁺ : Co³⁺ = 3d⁶ → No unpaired electrons as NH₃ is strong field ligand and causes pairing
(iii) [NiCl₄]²⁻ : Ni²⁺ = 3d⁸ → 2 unpaired electrons
(iv) [Cu(NH₃)₄]²⁺ : Cu²⁺ = 3d⁹ → 1 unpaired electron.

Teacher's Note:
1. Spin-only moment rises with the number of unpaired electrons, so just rank 5 > 2 > 1 > 0.

Question 20. Among [Ni(CO)₄], [NiCl₄]²⁻, [Co(NH₃)₄Cl₂] Cl, Na₃[CoF₆], Na₂O₂ and CsO₂, the total number of paramagnetic compounds is (JEE Advanced 2016)
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
Ans. (b)
Unpaired electrons :
[Ni(CO)₄] : 3d¹⁰ (sp³) → 0
[NiCl₄]²⁻ : 3d⁸ (sp³) → 2
[Co(NH₃)₄Cl₂] Cl : 3d⁶ (d²sp³) → 0
Na₃ [CoF₆] : 3d⁶ (sp³d²) → 4
Na₂O₂ : O₂²⁻ → 0
CsO₂ : O₂⁻ → 1
Thus, 3 compounds are paramagnetic.

Teacher's Note:
1. Peroxide O₂²⁻ has all electrons paired, but superoxide O₂⁻ has one unpaired electron.

One or More Correct Answers

Question 21. Which of the following statements are not correct ?
(a) If Δ_o < P, low spin state is more stable
(b) CO is a very weak ligand
(c) The colour of a complex depends only on the nature of metal ion
(d) Tetrahedral complexes have nearly 50% CFSE value than octahedral complexes
Answer:
Ans. (a, b, c)
If P < Δ_o, low spin state is more stable. Hence, (a) is wrong. CO is a very strong ligand. Hence, (b) is wrong. The colour of the complex depends not only on the metal ion but also on the nature of the ligand. Hence, (c) is wrong.
(d) is correct since Δ_t = 4/9 Δ_o, i.e. nearly half.

Teacher's Note:
1. Read “not correct” questions twice: here you must mark the false statements.

Question 22. Which of the following are outer orbital complexes ?
(a) [Co(NH₃)₆]³⁺
(b) [Mn(CN)₆]³⁻
(c) [Ni(NH₃)₆]²⁺
(d) [FeF₆]³⁻
Answer:
Ans. (c, d)
[Ni(NH₃)₆]²⁺: Ni²⁺ is 3d⁸; even after pairing no two 3d orbitals are empty, so 4s4p³4d² (sp³d²) is used.
[FeF₆]³⁻: F⁻ is weak, Fe³⁺ (3d⁵) remains high spin, sp³d².
[Co(NH₃)₆]³⁺ and [Mn(CN)₆]³⁻ are inner orbital (d²sp³) complexes.

Teacher's Note:
1. Octahedral Ni(II) complexes are always outer orbital, whatever the ligand, because 3d⁸ cannot free two inner d orbitals.

Question 23. Which of the following are π-bonded organometallic compounds ?
(a) Zeise's salt
(b) Furacene
(c) Dibenzene chromium
(d) Grignard reagent
Answer:
Ans. (a, c)
Zeise's salt, K[PtCl₃(η²-C₂H₄)], and dibenzene chromium, [Cr(η⁶-C₆H₆)₂], are π-bonded organometallic compounds.
Grignard reagent (R–Mg–X) is σ-bonded.

Teacher's Note:
1. In π-bonded organometallics the metal binds to the π-electron cloud of an alkene or arene, not to a single carbon.

Passage-Based Questions

Question 24. Comprehension 2: A metal complex having the composition Cr (NH₃)₄ Cl₂ Br has been isolated in two forms A and B. The form A reacts with AgNO₃ to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.
The formula of the complex A is
(a) [Cr(NH₃)₄ Br] Cl₂
(b) [Cr(NH₃)₄Br Cl] Cl
(c) [Cr (NH₃)₄ Cl₂] Br
(d) Both (a) and (b) are possible
Answer:
Ans. (b)
As form A is readily soluble in dilute aqueous ammonia, it must have Cl⁻ outside the coordination sphere, i.e., its formula is [Cr (NH₃)₄BrCl]⁺ Cl⁻.
(a) is ruled out because Cr³⁺ with four NH₃ and one Br would have coordination number 5, not 6.

Teacher's Note:
1. White AgCl dissolves in dilute NH₃; pale yellow AgBr needs concentrated NH₃; yellow AgI is insoluble.

Question 25. Comprehension 2: A metal complex having the composition Cr (NH₃)₄ Cl₂ Br has been isolated in two forms A and B. The form A reacts with AgNO₃ to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.
The hybridisation of Cr in the complexes A and B respectively is
(a) d²sp³ and sp³d²
(b) sp³d² and d²sp³
(c) sp³d and dsp³
(d) d²sp³ and d²sp³
Answer:
Ans. (d)
Both the above forms have Cr in +3 oxidation state. Hence, its electronic configuration is [Ar] 3d³, i.e., three unpaired electrons in 3d with two 3d orbitals empty.
Thus, the hybridisation of Cr in each case is d²sp³, i.e., both the complexes are octahedral.

Teacher's Note:
1. Cr³⁺ (3d³) is always d²sp³ in octahedral complexes, whatever the ligands.

Matrix Match Questions

Teacher's Note:
1. Both sp³ and dsp² give coordination number 4; the d orbital in dsp² is what makes it planar.

Teacher's Note:
1. Same metal, two oxidation states, two ligands: work out n for each and match with √n(n+2).

Teacher's Note:
1. A didentate ligand whose two donor atoms bind the same metal ion forms a ring, i.e. acts as a chelating ligand.

Integer and Numerical Answer Questions

Question 29. EDTA⁴⁻ is ethylenediaminetetraacetate ion. The total number of N–Co–O bond angles in [Co(EDTA)]¹⁻ complex ion is (JEE Advanced 2013)
Answer:
Ans. 8
In [Co(EDTA)]⁻ the two N atoms (N^I, N^II) and four O atoms (O^I–O^IV) are arranged octahedrally around Co. Each N is cis to all four O atoms.
The N–Co–O bond angles are :
(1) ∠N^ICoO^I (2) ∠N^ICoO^II (3) ∠N^ICoO^III (4) ∠N^ICoO^IV
(5) ∠N^IICoO^I (6) ∠N^IICoO^II (7) ∠N^IICoO^III (8) ∠N^II CoO^IV
Hence, total N–Co–O bond angles = 8.

Teacher's Note:
1. In the EDTA complex the two N atoms are cis to each other, so each one has all four O atoms at 90°.

Question 30. For the octahedral complex of Fe³⁺ in SCN⁻ (thiocyanato-S) and in CN⁻ ligand environments, the difference between the spin only magnetic moments in Bohr magnetons (when approximated to the nearest integer) is [Atomic number of Fe = 26] (JEE Advanced 2015)
Answer:
Ans. 4
Fe (26) = 3d⁶4s², Fe³⁺ = 3d⁵
In [Fe(CN)₆]³⁻, CN⁻ is a strong field ligand which causes pairing of electrons: one unpaired electron, d²sp³ hybridization
\( \mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \) = 1.732 BM
In [Fe(SCN)₆]³⁻, SCN⁻ being a weak field ligand does not cause pairing of electrons: five unpaired electrons, sp³d² hybridization
\( \mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \) = 5.916 BM
Difference = 5.916 − 1.732 = 4.184 ≈ 4 BM

Teacher's Note:
1. Round only at the end: rounding each moment first (2 and 6) also gives 4 here, but not in general.

Question 31. In dilute aqueous H₂SO₄, the complex diaquo-dioxalato ferrate (II) is oxidized by MnO₄⁻. For this reaction, the ratio of the rate of change of [H⁺] to the rate of change of [MnO₄⁻] is (JEE Advanced 2015)
Answer:
Ans. 8
The given complex, [Fe(C₂O₄)₂(H₂O)₂]²⁻ contains Fe in +2 oxidation state. In acidic medium, KMnO₄ oxidizes Fe²⁺ to Fe³⁺.
MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺
5 Fe²⁺ ≡ 5 [Fe(C₂O₄)₂(H₂O)₂]²⁻
∴ \( \frac{\text{Rate of change of [H}^+]}{\text{Rate of change of [MnO}_4^-]} = \frac{8}{1} \) = 8

Teacher's Note:
1. The ratio of rates of change equals the ratio of stoichiometric coefficients in the balanced equation.

Question 32. In the complex acetyl bromidodicarbonyl bis(triethyl phosphine) iron (II), the number of Fe–C bond (s) is (JEE Advanced 2015)
Answer:
Ans. 3
The given complex is [Fe(CH₃CO) (Br) (CO)₂ (PEt₃)₂]. In its octahedral structure Fe is bonded to two PEt₃, one Br, two CO and one acetyl group.
Thus, it has two Fe–CO bonds and one Fe ← C(=O)–CH₃ bond.
Total number of Fe–C bonds = 3.

Teacher's Note:
1. The acetyl ligand bonds to Fe through its carbonyl carbon, so it counts as one Fe–C bond; PEt₃ bonds through P.

Question 33. Among the complex ions
[Co(NH₂CH₂CH₂NH₂)₂Cl₂]⁺, [CrCl₂(C₂O₄)₂]³⁻, [Fe(H₂O)₄(OH)₂]⁺, [Fe(NH₃)₂(CN)₄]⁻, [Co(NH₂CH₂CH₂NH₂)₂(NH₃) Cl]²⁺ and [Co(NH₃)₄(H₂O) Cl]²⁺
the number of complex ion(s) that show (s) cis-trans isomerism is (JEE Advanced 2015)
Answer:
Ans. 6
[Co(NH₂CH₂CH₂NH₂)₂Cl₂]⁺ = [M(AA)₂B₂] → shows cis-trans isomerism
[CrCl₂(C₂O₄)₂]³⁻ = [MA₂(BB)₂] or [M(AA)₂B₂] → shows cis-trans isomerism
[Fe(H₂O)₄(OH)₂]⁺ = [MA₄B₂] → shows cis-trans isomerism
[Fe(NH₃)₂(CN)₄]⁻ = [MA₂B₄] → shows cis-trans isomerism
[Co(NH₂CH₂CH₂NH₂)₂ (NH₃) Cl]²⁺ = [M(AA)₂BC] → shows cis-trans isomerism
[Co(NH₃)₄(H₂O) Cl]²⁺ = [MA₄BC] → shows cis-trans isomerism (Cl and H₂O cis or trans to each other)

Teacher's Note:
1. Any octahedral complex with two different ligands that can sit either adjacent or opposite (MA₄B₂, MA₄BC, M(AA)₂B₂...) shows cis-trans isomerism.

Assertion–Reason Questions

Question 34. Statement-1: The geometrical isomers of the complex [M(NH₃)₄Cl₂] are optically inactive.
Statement-2: Both geometrical isomers of the complex [M(NH₃)₄Cl₂] possess axis of symmetry. (IIT 2008)
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Complexes of the type [MA₄B₂] show geometrical isomerism (cis- and trans-). Both the forms have axis of symmetry and hence optically inactive.

Teacher's Note:
1. Strictly, optical inactivity requires a mirror plane (or improper axis); both isomers of MA₄B₂ have mirror planes too.

Question 35. Statement-1: Co(NH₃)₅Cl₃ gives white precipitate with AgNO₃ solution.
Statement-2: Chlorine is not present in the ionising sphere. (Bihar Board 2011)
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Correct Statement-2. As the complex gives white precipitate with AgNO₃, one Cl must be present in the ionising sphere. (The complex is [Co(NH₃)₅Cl]Cl₂.)

Teacher's Note:
1. A white AgCl precipitate is direct evidence of Cl⁻ outside the coordination sphere.

Question 36. Statement 1: The identification of Ni²⁺ is carried out by dimethylglyoxime in the presence of NH₄OH.
Statement 2: Dimethylglyoxime is a bidentate neutral ligand. (JEE Main 2021)
(a) Both statement 1 and 2 are true.
(b) Both statements 1 and 2 are false.
(c) Statement 1 is true but statement 2 is false.
(d) Statement 1 is false but statement 2 is true.
Answer:
Ans. (c)
Correct Statement-2. Dimethylglyoxime is a negative bidentate ligand (it binds as the dimethylglyoximate ion, dmg⁻, forming red [Ni(dmg)₂]).

Teacher's Note:
1. NH₄OH removes a proton from each dmgH₂, which is why the test is done in ammoniacal medium.

Question 37. Statement 1: [Ni(CN)₄]²⁻ is square planar and diamagnetic complex with dsp² hybridisation for Ni but [Ni(CO)₄] is tetrahedral, paramagnetic with sp³ hybridisation for Ni.
Statement 2: [NiCl₄]²⁻ and [Ni(CO)₄] both have same d-electron configuration, have same geometry and are paramagnetic. (JEE Main 2022)
(a) Both statement 1 and 2 are true.
(b) Both statements 1 and 2 are false.
(c) Statement 1 is true but statement 2 is false.
(d) Statement 1 is false but statement 2 is true.
Answer:
Ans. (b)
Ni(CO)₄ is diamagnetic (Ni(0), 3d¹⁰), so Statement 1 is false. [NiCl₄]²⁻ is 3d⁸ while Ni(CO)₄ is 3d¹⁰, and Ni(CO)₄ is not paramagnetic, so Statement 2 is false. Hence, both statements are incorrect.

Teacher's Note:
1. [NiCl₄]²⁻ and Ni(CO)₄ share the tetrahedral shape, but only [NiCl₄]²⁻ is paramagnetic.