Coordination Compounds MCQ Quiz - Objective Question with Answer for Coordination Compounds - Download Free PDF

CONCEPT:

Electronic Configuration, Geometry & Magnetic Properties of Transition Metal Complexes

• The geometry and electronic configuration of a metal complex determine its magnetic behavior and structural distortions (like Jahn-Teller effect).
• Jahn-Teller (JT) distortion typically occurs in octahedral complexes with d⁹ or high-spin d⁷ configurations.
• Spin-only magnetic moment applies when orbital contribution is negligible (usually in high-spin octahedral fields).
• Tetrahedral complexes are often paramagnetic (due to high-spin configurations) unless completely filled or empty d-orbitals.
• Square planar d⁸ complexes (like Pd(0)) are usually diamagnetic.

EXPLANATION:

• a. [Co(H₂O)₆]²⁺ → i. Magnetic moment higher than spin-only value and weak JT distortion
  • High-spin d⁷ (octahedral Co²⁺) shows JT distortion and has both spin and orbital contributions, leading to magnetic moment higher than spin-only value.
• b. [Cr(H₂O)₆]³⁺ → ii. Spin-only magnetic moment and absence of JT distortion
  • Cr³⁺ is d³ (t_2g³), no eg electrons ⇒ no JT distortion, and purely spin-only moment.
• c. NiCl₂(PPh₃)₂ → iii. Paramagnetic and tetrahedral
  • Ni(II) is d⁸, and in a tetrahedral field, it is high-spin with two unpaired electrons ⇒ paramagnetic.
• d. Pd(PPh₃)₄ → iv. Diamagnetic and tetrahedral
  • Pd(0) is d¹⁰, fully filled configuration ⇒ diamagnetic, and forms tetrahedral complexes with bulky ligands.

Therefore, the correct answer is a-i, b-ii, c-iii, d-iv.

CONCEPT:

Term Symbol, Lande g-Factor and Magnetic Moment for f-block ions

• The ground state term for a given configuration is derived using Hund’s rules based on the number of electrons in f-orbitals.
• For Pr³⁺ (Z = 59), the configuration is [Xe] 4f² (i.e., 2 electrons in 4f orbitals).
• Term symbol is found using:
  • L (orbital angular momentum quantum number): For 4f², L = 5 → 'H'
  • S (total spin): 2 unpaired electrons → S = 1 → multiplicity = 2S + 1 = 3
  • J: For less than half-filled subshell → J = |L - S| = 5 - 1 = 4
• Therefore, ground state term = ³H₄

LANDE g-FACTOR:

The Lande g-factor is given by:

g = 1 + [ J(J + 1) + S(S + 1) - L(L + 1) ] / [ 2J(J + 1) ]

Substitute S = 1, L = 5, J = 4:

• g = 1 + [ 4(5) + 1(2) - 5(6) ] / [ 2 × 4(5) ]
• g = 1 + [20 + 2 - 30]/40 = 1 - 8/40 = 0.80

 MAGNETIC MOMENT:

μ_eff = g × √[J(J + 1)]

• μ = 0.80 × √[4 × 5] = 0.80 × √20 ≈ 0.80 × 4.47 ≈ 3.58 B.M.

Therefore, the correct values are:

• Term: ³H₄
• g-factor: 0.80
• μ_calc: 3.58 B.M.

Correct answer: Option 2

CONCEPT:

Geometry of Coordination Complexes

• The geometry of a coordination compound depends on:
  • The number and type of donor atoms
  • The oxidation state and electronic configuration of the metal
  • The steric and electronic properties of the ligands

EXPLANATION & MATCHING:

1. a. [Zn{N(CH₂CH₂NH₂)₃}Cl]⁺
   • Ligand: TREN (tris(2-aminoethyl)amine) → tetradentate + Cl⁻ → total of 5 donor atoms
   • Geometry: Trigonal bipyramidal → i
   • 
2. b. [Cu(2,2-bpy){NH(CH₂COO)₂}]
   • Ligands: bidentate 2,2'-bipyridine + tridentate amino diacetate-type ligand
   • Total coordination number = 5 → square pyramidal
   • Match: ii
   • 
3. c. [ZrF₇]³⁻
   • 7 fluorides around Zr(IV)
   • Known to adopt monocapped trigonal prismatic geometry
   • Match: iii
   • 
4. d. [AgTe₇]³⁻
   • Te atoms are large, soft ligands → may form planar structures with d¹⁰ Ag⁺
   • Match: iv (trigonal planar)
   • 

Correct Answer: Option 1 ✅

CONCEPT:

Overall Stability Constants (logβ) for Chelate Complexes

• logβ represents the overall formation constant of a complex; higher logβ means greater thermodynamic stability.
• In the series of divalent 3d metal ions (Mn²⁺ → Zn²⁺), stability generally increases and then decreases due to crystal field stabilization energy (CFSE) and other factors.
• Ligands:
  • en (ethylene diamine): bidentate, neutral, moderate chelating ability
  • EDTA: hexadentate, negatively charged, forms very stable chelates via multiple donor atoms

EXPLANATION OF STATEMENTS:

• A. The logβ is lowest for Mn²⁺ in both complexes → Correct
  • Mn²⁺ (d⁵, high spin) has no CFSE advantage and weak field stabilization → forms least stable complexes across the series.
• B. logβ of [Mn(EDTA)]²⁻ < [Mn(en)₃]²⁺ →  Incorrect
  • EDTA forms more stable complexes than en due to its hexadentate nature and negative charge → so [Mn(EDTA)]²⁻ has higher logβ.
• C. logβ increases monotonically Mn²⁺ → Zn²⁺ → Incorrect
  • There is a peak at Cu²⁺ (due to Jahn-Teller stabilization); logβ does not increase uniformly.
• D. ΔS° is nearly constant along the series → Correct
  • The entropy change mostly reflects ligand displacement and chelation; metal identity affects ΔH° more than ΔS° → ΔS° remains relatively constant.

Correct Statements: A and D only

CONCEPT:

Electronic Transitions Between Spin Multiplets (Term Symbols)

• In transition metal and lanthanide/actinide chemistry, electronic states are represented by term symbols, like ^2S+1L (e.g., ³D, ³F).
• Each term symbol is split into multiple levels due to spin-orbit coupling, which mixes spin and orbital angular momenta.
• J-values for a term ^2S+1L range from |L−S| to (L+S).

EXPLANATION:

• For ³D term:
  • S = 1, L = 2 (D → L = 2)
  • Possible J values = |L − S| to (L + S) = |2 − 1| to (2 + 1) = 1, 2, 3
  • So ³D splits into: ³D₁, ³D₂, ³D₃
• For ³F term:
  • S = 1, L = 3 (F → L = 3)
  • Possible J values = |3 − 1| to (3 + 1) = 2, 3, 4
  • So ³F splits into: ³F₂, ³F₃, ³F₄
• Allowed transitions (by selection rule ΔJ = 0, ±1; but not J = 0 ↔ 0):
• From 3 D-levels to 3 F-levels → maximum of 3 × 3 = 9 combinations
• However, applying selection rules, the allowed transitions are:
  • D₁ → F₂, F₁ →  (no F₁) → only F₂ → 1
  • D₂ → F₁, F₂, F₃ → F₂, F₃ → 2
  • D₃ → F₂, F₃, F₄ → all allowed → 3
• Total = 1 + 2 + 3 = 6 allowed transitions

Correct answer Total 6 electronic transitions.

Concept:

• Octahedral complexes usually follows dissociative pathway for substitution reaction. Dissociation step is the rate determining step.
• The mechanism is similar to SN1 reactions.
• the value of entropy change, \(\Delta S^0\), has positive value for dissociative mechanism.
• Mechanism:

​ \(ML_5X \ \rightleftharpoons ML_5 + X^-\)

 \(ML_5\;+\;Y^- \rightarrow \; ML_5Y\)​

Explanation:

(a) Correct.

The first step of substitution is dissociation of ligand. A sterically crowded system is unstable and thus breaking/ dissociation of ligand is easy. Hence, High steric hindrance between ligands in the metal complex favors fast dissociation of ligand.

(b) Incorrect.

Increase in positive charge on metal leads to increase in interaction between metal and ligand. This increases the M-L bond strength and dissociation is not favored. Hence, Increased charge on the metal atom/ion of the complex favors the acceptance of electron pair of the entering ligands.

(c) Correct.

The first step, that is, dissociation of ligand from octahedral complex leads to formation of penta-coordinated intermediate (which can be trigonal bipyramidal or square planar). 

(d) Incorrect.

Nature of the entering ligand significantly influences the reaction. -

ligand dissociation is the rate-determining step, the entering group should have no or very minor effect on the reaction rate. Changes in rate constants of less than a factor of 10 are generally considered sufficiently similar when making these judgments.

So in light of this more specific information, the 'd' option will be considered incorrect for a dissociative process, especially when the dissociation is the rate-determining step, so we should exclude it.

Thus the correct answer is a and c

Conclusion:

Hence, the correct statements are a and c

• A microstate is a specific way in which we can arrange the energy of the system.
• Many microstates are indistinguishable from each other.
• The more indistinguishable microstates, the higher the entropy.
• The formula for calculating the number of microstates is:

\(= {N! \over (n - r)! r!}\), Where N = Number of electron positions, r = number of electrons.

Calculation:

• As there are 5 orbitals in d, the number of orbitals × electrons = electron positions
• One electron can occupy two spin states, so positions 'N' = 5 × 2 = 10
• The number of electrons 'r' = 2

We know that the number of microstates is given by:

\(= {N! \over (n - r)! r!}\)

Substituting the values in the above equation, we get:
Microstates = \(x = {10! \over (10-2)!\times 2!} =45\)

hence, the number of microstates in d² is 45.

Concept:

Atomic term symbol can be given as:

^2s+1L

where: 2s + 1 → Spin multiplicity

L → orbital angular momentum

value of L can be given as:

       s    p     d     f      g ...

L→ 0    1    2     3     4

Now,

{No. of microstates for a given term = (2L + 1)(2s + 1)}

Explanation:

Given atomic Term: ⁴f

∴ L = 3, 2s + 1 = 4

∴ No. of microstates = (2 × 3 + 1) × 4

= 28

∴ option '3' is correct.

Conclusion:-

No. of microstates in atomic term symbol

⁴F is 28.

Concept: 

• A metallocene is a compound typically consisting of two cyclopentadienyl anions (Cp^- or C₅H₅^-) bound to a metal center M in the oxidation state II, with the compound of general formula \({\left( {Cp} \right)_2}M \) or \({\left( {{C_5}{H_5}} \right)_2}M \).
• Metallocenes are a broader class of compounds called sandwich compounds.
• Mn(η5 - C5Me5)2 is an example of a metallocene.

Explanation:

• In the metallocene Mn(η5 - C5Me5)2, the two (η5 - C5Me5) ligands will contribute 5 electrons each (considering the zero oxidation state method). While the number of valence electrons contributed by a metal atom for total electron count in a zero-oxidation state is equal to the group number.
• Considering the zero-oxidation state, Mn will contribute 7 electrons as it belongs to group 7. So the total electron count for the metallocene Mn(η5 - C5Me5)2 is

= (7+10)

= 17 electrons.

• The 17 electrons in Mn(η5 - C5Me5)2 are arranged in the molecular orbital. Out of which 12 electrons are placed in the Ligand-centred bonding molecular orbitals.
• The manganocene can exist in two distinct forms, one in a high spin form with 5 unpaired electrons as in (Cp)₂Mn and in the other form it can exist in a low spin with one unpaired electron as in Mn(η5 - C5Me5)2. This is because of the higher ligand field strength of (η5 - C5Me5) compare to the Cp ring.    
• The rest 5 electrons are placed in the frontier MO's of Mn(η5 - C5Me5)₂ as follows:

        ​

Conclusion:

• Hence, The correct electronic configuration of frontier MO's of Mn(η5 - C5Me5)2 is e22ga11ge21g

Concept:

• The trigonal bipyramidal complexes are sp³d hybridization.
• TBP complexes have five M-L bonds which are not equivalent. The bonds can be classified either as axial or equatorial bond.

Explanation:

• Finding the point group of TBP complex (ML₅):
• To find the point we will start by checking the rotational axes of symmetry. TBP complex (ML₅) has an rotational axis of symmetry of order 3 (C₃) .Also it has three axes of symmetry of order 2(C₂). Point group can either be D_3h or D_3d

Next, we check the plane of symmetry. The complex has 3 vertical planes of symmetry as well as a  horizontal plane passing through equatorial ligands. So, the point group is D_3h

• Energy order of d-orbitals in TBP crystal field:
• The ligands are Considered to be approaching from cartesian coordinates x, y and z. The axial ligands approaches d_z² orbital directly and  undergoes maximum repulsion. Therefore, it has maximum energy. Then comes orbitals lying on x-y plane . d_x-z and d_y-z orbitals experience least repulsion and has lowest energy. So, the relative energy order of metal d-orbitals in TBP crystal field is:

dz2 > dx2 - y2, dxy > dxz, dyz

Conclusion:

Therefore, trigonal pyramidal complex (ML₅)  belongs to D_3h point group symmetry and the relative energy of metal d-orbitals in TBP crystal field is dz2 > dx2 - y2, dxy > dxz, dyz

• The transfer of an electron from an orbital primarily ligand character to one with primarily metal character is known as ligand-to-metal charge transfer or LMCT.
• LMCT occurs if a ligand that can be easily oxidized is bound to a metal center in a high oxidation state, which is readily reduced.
• It is a correlation between the energies of charge transfer absorptions and the electrochemical properties of metals and ligands.
• Charge transfer transitions are not restricted by the selection rules that involve ‘d–d ’ transitions, The probability of these electronic transitions is very high, and the absorption bands are therefore intense.

Explanation:

• Ligand-to-metal charge transfer may give rise to absorptions in the UV or visible region of the electronic spectrum depending upon the energy gap between the HOMO and LUMO.
• Now, for a complex, the higher the formal oxidation state of the metal ion, the lower will be the energy of the orbital that involves the LMCT transition (primarily of metal character)
• This decreases the energy gap between the orbital with primarily ligand character and the orbital with primarily metal character. Thus higher will be the wavelength for LMCT transition.
• For the oxo‐anions, VO43-, CrO42-, and MnO4- the oxidation state of V, Cr, and Mn is +5, +6, and +7 respectively.
• As higher the formal oxidation state of the metal ion, the lower will be the energy gap between the orbitals that are involved in LMCT transitions. Thus, the energy order of  orbital with primarily metal character is,

​Mn⁺⁷+6+5.

• The energy gap between the orbital with primarily ligand character and the orbital with primarily metal character will follow the order,

​VO43- > CrO42- > MnO4.

• Thus the wavelength of the transitions are in the order,

​VO43- < CrO42- < MnO4-

• In the case of oxo-anions, WO42-, MoO42-, and CrO42-, the oxidation state for Cr, Mo, and W is +6.
• Ongoing from Cr⁺⁶ to W⁺⁶ the energy gap between the orbital with primarily ligand character and the orbital with primarily metal character increases thus the wavelength of transition decreases.
• Thus, the wavelength of the transitions are in the order,

WO42- < MoO42- < CrO42-

Conclusion:-

• Hence, the wavelength of the transitions is in the order

​VO43- < CrO42- < MnO4- and WO42- < MoO42- < CrO42-

Concept:

• The reaction of NiBr2 with two equivalents of PPh3 in CS2 is giving a red-coloured diamagnetic complex, [NiBr2 (PPh3)2 ].
• Nickel (Ni) in the complex [NiBr2 (PPh3)2 ]. Ni is in a +2 oxidation state. Hence the electronic configuration is [Ar]3d8.
• Out of eight 3d electrons, 6 are paired and two are unpaired.
• Octahedral and tetrahedral complexes of Ni(II) is thus paramagnetic, whereas the square planar ones are diamagnetic with no unpaired electrons.
• NiBr₂ is a yellow coloured complex. During its preparation, it is dissolved in ethanol and heated, the colour of the solution changes to green.
• Phosphine is dissolved in alcohol and then NiBr₂ is added to the solution, an immediate dark green precipitate is formed.

Explanation:

• Nickel (Ni) in the complex [NiBr2 (PPh3)2 ] is in a +2 oxidation state and is red coloured. It is also said that it is diamagnetic which indicates that the complex is square planar in nature.

​​

• The complex  [NiBr2 (PPh3)2 ], immediately transforms to a green-coloured complex and it is said that the complex now is paramagnetic.
• So, we know that Ni (II) complexes will be paramagnetic only when they are tetrahedral and octahedral, So as this was a four coordinated complex, the complex now formed is also four coordinated and tetrahedral.
• Electron pairs from Br and PPh3 are donated into one 4s and three 4p empty orbitals of Nickel giving us sp3 hybridization.

​

• The structure is tetrahedral, and the number of unpaired electrons is two.

Hence, the geometry and the number of unpaired electrons in the green-coloured complex, respectively, are tetrahedral and 2.

Concept:

• Orgel Diagrams are actually correlation diagram that represents relative energy levels of electronic terms for octahedral and tetrahedral complexes.
• Orgel Diagrams are used to represent a transition in metal complexes.
• It is only applicable for high-spin complexes
• Transitions represented by the orgel Diagram occur only from the ground state.

Explanation:

• [Ni(H2O)6]2+ is a weak field octahedral complex as H2O is weak filed ligand
• Now, the ground state of Ni2+(d8) is t2g6eg2
• Total Orbital Momentum, L = |-1-2| = 3 ,i.e., F
• Total Spin Momentum, S =  = 1 (number of unpaired electron is 2)
• Spin Multiplicity = 2S+1 = = 3
• The, ground state term is = 3F
• The observed electronic spectra of d2, d3, d7, and d8 octahedral and tetrahedral complexes ion, is as follows:

• According to the Orgel diagram, a transition occurs only from the ground state.
• The ground state for the d8 octahedral complex is A2g.
• Therefore, the possible transitions are,

³T2g(F)←³A2g(F),

³T1g(F)← ³A2g(F),

and ³T1g (P)← ³A2g(F)

• The energy gap is highest for the transition ³T1g (P)←3A2g(F), followed by 3T1g(F)← 3A2g(F), and is lowest for 3T2g(F)←3A2g(F).
• Now, the energy gap between the two energy levels is inversely proportional to wavelength.
• So the three bands A (~400 nm), B (~690 nm) and C (~1070 nm) will be

A (~400 nm) = 3T1g (P)←3A2g(F),

B (~690 nm) = 3T1g(F)← 3A2g(F)

C (~1070 nm) = 3T2g(F)←3A2g(F)

​Conclusion:

• Hence, the transitions assigned to A, B, and C, respectively, are

T1g(P) ← A2g, T1g ← A2g, and T2g ← A2g

Concept:

The magnetic moment of coordination compounds is a property that provides insight into the magnetic behavior of these complex molecules. Coordination compounds are composed of a central metal atom or ion bonded to surrounding ligands. The magnetic moment is a measure of the magnetic strength and orientation of the electron distribution within the compound.

Explanation:

\(u = {g\sqrt{J(J+1)}}\)

The above equation is used to calculate the magnetic moment of heavier atoms. Eg. F block elements.

Where, \(g = {1+{S(S+1) - L(L+1) +J(J+1)\over 2J(J+1)}}\)

S = 1/2 +1/2 +1/2+ 1/2 +1/2

S = 5/2

L = (3 x 1) +(2 x 1) + (1 x 1) + (0 x 1) + (-1 x 1)

L = 5

J = |L - S|.  As f orbital is less than half filled

J = |5 - 5/2|

J = 5/2

Putting the values of S, L and J in the above equation of g

Hence, g = 2/7

Now, putting the value of g to calculate magnetic moment of f⁵ ion.

\(u ={ {2 \over7}}{\sqrt{{5\over2}({5 \over2}+1)}}\)

\(u = {\sqrt {35}\over 7}\)

Conclusion:

The magnetic moment of f⁵ ion is \({\sqrt {35}\over 7}\).

Concept:

• Orgel diagrams are correlation diagrams that show the relative energies of electronic terms in transition metal complexes.
• Orgel diagrams will, however, show the number of spin-allowed transitions, along with their respective symmetry designations.

Explanation:

• The observed electronic spectra of d², d³, d⁷, and d⁸ octahedral and tetrahedral complexes ion, is as follows:

• In case of [Cr(en)3]3+ ion, the oxidation state of Cr is +3 and the electronic configuration is d3. En (Ethylenediamine) is a bidentate ligand thus the complex is octahedral.
• Hence, the complex [Cr(en)3]3+ will show 3 electronic transitions. These are ⁴A_2g →⁴T_2g, ⁴A_2g→⁴T_1g(F), and ⁴A_2g→4T1g(P).
• In case of trans-[Cr(en)2F2]+ ion, the oxidation state of Cr is also +3 and the electronic configuration is d3. But it changes it symmetry to D4h due to axial F.
• Splitting of energy level takes place due to change into symmetry.

  • Oh	D4h
    A1g	A1g
    A2g	B1g
    Eg	A1g + B1g
    T1g	A2g + Eg
    T2g	B2g + Eg

     

• 
• Thus, the complex ion trans-[Cr(en)2F2]+ will show 6 electronic transitions.

​Conclusion:

• ​Hence, the number of expected electronic transitions in [Cr(en)3]3+ and trans-[Cr(en)2F2]+ at 4 K is, respectively 3 and 6.