Coordination Compounds - Practice Questions with Answers
60 free MCQs on Coordination Compounds with worked answers and explanations. Study of compounds where a central metal atom is bonded to surrounding ligands. Covers nomenclature, types of isomerism, bonding theories (VBT, CFT), and applications in medicine, photography, and industry.
Below are 60 practice questions on Coordination Compounds, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Coordination Compounds notes.
The three common coordination geometries: octahedral (6 ligands), tetrahedral (4 ligands), and square planar (4 ligands in one plane).
Easy - 20 questions
Q1.
A coordination compound consists of a central metal atom surrounded by:
A Ligands (ions or molecules donating electron pairs)
B Mainly covalent bonds with little donor-acceptor character
C Mainly ionic bonds formed by complete electron transfer
D No surrounding atoms or groups whatsoever as generally observed
Show answer & explanation
Answer: A. Ligands (ions or molecules donating electron pairs)
Why: In coordination compounds, ligands donate lone pairs to the central metal atom, forming coordinate (dative) bonds.
Q2.
What is a ligand?
A An ion or molecule that donates a lone pair to the metal
B The central metal atom around which the structure is built
C The overall charged species formed once the structure is assembled
D A neutral complex with no net charge on the whole assembly
Show answer & explanation
Answer: A. An ion or molecule that donates a lone pair to the metal
Why: Ligands are Lewis bases that donate electron pairs to the central metal atom/ion (Lewis acid).
Q3.
The coordination number of the metal in [Co(NH<sub>3</sub>)<sub>6</sub>]3+ is:
A 6
B 3
C 4
D 2
Show answer & explanation
Answer: A. 6
Why: Coordination number = number of ligand donor atoms directly bonded to the metal. 6 NH<sub>3</sub> ligands give CN = 6.
Q4.
What is the name of [Cu(NH<sub>3</sub>)4]2+?
A Tetraamminecopper(II) ion
B Copper tetraamine as widely reported
C Diammine copper in standard practice
D Cupric amide under most conditions encountered
Show answer & explanation
Answer: A. Tetraamminecopper(II) ion
Why: Ligands are named first (tetraammine = 4 NH<sub>3</sub>), then the metal with oxidation state in Roman numerals.
Q5.
EDTA is an example of which type of ligand?
A Hexadentate (6 donor atoms)
B Monodentate, donating through only a single atom
C Bidentate, donating through exactly two atoms
D Tridentate, donating through exactly three atoms
Show answer & explanation
Answer: A. Hexadentate (6 donor atoms)
Why: EDTA (ethylenediaminetetraacetate) has 6 donor atoms (2 N + 4 O) and is a hexadentate ligand.
Q6.
A bidentate ligand donates how many electron pairs?
A Two (from two donor atoms)
B One, from a single donor atom on the ligand
C Three, from three separate donor atoms on the ligand
D Six, from six separate donor atoms on the ligand
Show answer & explanation
Answer: A. Two (from two donor atoms)
Why: Bidentate ligands (e.g., ethylenediamine, oxalate) have two donor atoms and occupy two coordination positions.
Q7.
What is the oxidation state of cobalt in [Co(NH<sub>3</sub>)5Cl]2+?
A +3
B +2
C +1
D 0
Show answer & explanation
Answer: A. +3
Why: NH<sub>3</sub> = 0, Cl = -1, overall charge = +2. Let Co = x: x + 5(0) + (-1) = +2; x = +3.
Q8.
What is the IUPAC name for [Fe(CN)<sub>6</sub>]4-?
A Hexacyanoferrate(II) ion
B Hexacyanoferrate(III) ion
C Iron hexacyanide
D Ferrocyanide
Show answer & explanation
Answer: A. Hexacyanoferrate(II) ion
Why: Fe is in +2 oxidation state (hexacyanoferrate(II)): 6 CN- = -6; charge = -4; x - 6 = -4, x = +2.
Q9.
The shape of [Ni(CN)4]2- is:
A Square planar
B Tetrahedral
C Octahedral
D Linear
Show answer & explanation
Answer: A. Square planar
Why: [Ni(CN)4]2- is square planar; CN- is a strong field ligand that forces dsp<sup>2</sup> hybridisation on Ni<sup>2+</sup> (d<sup>8</sup> configuration).
Q10.
The chelate effect refers to:
A Extra stability of complexes formed with polydentate ligands due to increased entropy
B The particular colour displayed by a coordination complex once it forms
C A general magnetic property uniformly shown by every chelated complex
D Simply the total count of separate ligand molecules attached to the central metal
Show answer & explanation
Answer: A. Extra stability of complexes formed with polydentate ligands due to increased entropy
Why: Chelation of polydentate ligands releases more solvent molecules per ligand than monodentate ligands, greatly increasing entropy and stability.
Q11.
What does the term 'Werner complex' refer to?
A Classic coordination compounds proposed by Alfred Werner (Nobel 1913)
B A purely covalent organic compound unrelated to coordination chemistry
C A simple ionic solid lacking any coordinate bonds
D An organic polymer formed by repeating monomer units
Show answer & explanation
Answer: A. Classic coordination compounds proposed by Alfred Werner (Nobel 1913)
Why: Alfred Werner proposed the theory of coordination compounds and octahedral geometry for complexes in the late 1800s.
Q12.
CN- is a strong field ligand because:
A It is a good sigma and pi acceptor (back-bonding with metal d orbitals causes large Delta)
B It generally carries a negative charge like many weak field ligands also do as frequently observed in practice
C It has an unusually small ionic size compared to other ligands in many documented cases
D It happens to contain a nitrogen atom in its structure according to conventional understanding
Show answer & explanation
Answer: A. It is a good sigma and pi acceptor (back-bonding with metal d orbitals causes large Delta)
Why: CN- is a strong pi-acceptor; it accepts electron density from filled metal d orbitals into its pi* orbital (back-bonding), causing large crystal field splitting.
Q13.
Which of the following is NOT a ligand?
A Na+
B CN-
C NH<sub>3</sub>
D Cl-
Show answer & explanation
Answer: A. Na+
Why: Na+ has no lone pairs to donate; it cannot act as a ligand. CN-, NH<sub>3</sub>, and Cl- all have lone pairs.
Q14.
The formula for potassium hexacyanoferrate(III) is:
A K3[Fe(CN)<sub>6</sub>]
B K<sub>4</sub>[Fe(CN)<sub>6</sub>]
C K[Fe(CN)<sub>6</sub>]
D K2[Fe(CN)<sub>6</sub>]
Show answer & explanation
Answer: A. K3[Fe(CN)<sub>6</sub>]
Why: Fe is +3, each CN- is -1, so 6 CN- = -6; total complex charge = +3 - 6 = -3; 3 K+ needed to balance.
Q15.
Ethylenediamine (en) is a bidentate ligand because it has:
A Two NH<sub>2</sub> donor atoms
B One NH<sub>2</sub> group
C A carboxyl group
D A carbon-carbon double bond
Show answer & explanation
Answer: A. Two NH<sub>2</sub> donor atoms
Why: Ethylenediamine (H2N-CH<sub>2</sub>-CH<sub>2</sub>-NH<sub>2</sub>) has two amine N donors that can both coordinate to the same metal.
Q16.
The primary valence in Werner's theory corresponds to:
A Oxidation state (ionisable valence)
B The coordination number, which Werner instead termed secondary valence
C A separate secondary valence distinct from the primary valence
D Simply the total number of ligands bound to the metal centre
A Cisplatin [cis-Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>]
B [Co(NH<sub>3</sub>)<sub>6</sub>]Cl3 in routine practice
C [Fe(CN)<sub>6</sub>]4- overall in most cases
D [Cu(NH<sub>3</sub>)4]2+ under typical conditions
Show answer & explanation
Answer: A. Cisplatin [cis-Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>]
Why: Cisplatin (cis-diamminedichloroplatinum(II)) crosslinks DNA strands in cancer cells, inhibiting replication.
Q18.
Haemoglobin is a coordination complex in which the metal is:
A Fe<sup>2+</sup> (iron II)
B Mg<sup>2+</sup> according to standard textbooks
C Cu<sup>2+</sup> in general practice
D Co3+ as frequently described
Show answer & explanation
Answer: A. Fe<sup>2+</sup> (iron II)
Why: Haemoglobin's haem group contains Fe<sup>2+</sup> coordinated in a porphyrin ring; it reversibly binds O<sub>2</sub>.
Q19.
An ambidentate ligand is one that can coordinate through:
A Two different donor atoms (e.g., SCN- via S or N)
B Two identical donor atoms positioned on opposite ends of the ligand
C Multiple chelate rings formed simultaneously with one metal centre
D Oxygen exclusively, regardless of any other potential donor atoms
Show answer & explanation
Answer: A. Two different donor atoms (e.g., SCN- via S or N)
Why: Ambidentate ligands (like SCN-, NO<sub>2</sub>-) have two potential donor atoms and can bond to the metal through either one.
Q20.
The coordination number of Pt in [PtCl<sub>4</sub>]2- is:
A 4
B 2
C 6
D 8
Show answer & explanation
Answer: A. 4
Why: Four chloride ligands surround Pt<sup>2+</sup>, giving a coordination number of 4 (square planar).
Medium - 20 questions
Q21.
Ionisation isomers differ in:
A Ions inside and outside the coordination sphere
B Arrangement of ligands around the metal in typical laboratory settings
C The metal's oxidation state under usual circumstances
D The number of ligands according to most researchers
Show answer & explanation
Answer: A. Ions inside and outside the coordination sphere
Why: Ionisation isomers have the same formula but differ in which ions are inside the square brackets (coordination sphere) vs outside.
Q22.
[Co(NH<sub>3</sub>)5Cl]SO4 and [Co(NH<sub>3</sub>)5SO4]Cl are examples of:
A Ionisation isomers
B Geometrical isomers
C Optical isomers
D Linkage isomers
Show answer & explanation
Answer: A. Ionisation isomers
Why: These two compounds have the same formula but the Cl and SO4 have exchanged positions inside/outside the coordination sphere.
Q23.
Geometrical isomerism (cis-trans) is possible in square planar complexes with:
A MA2B2 type (e.g., [Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>])
B MA4 type complexes, which show no geometric isomerism in square planar geometry
C MA6 type complexes, which are not square planar at all
D MA2 type complexes, which have too few ligands to show cis-trans isomerism
Show answer & explanation
Answer: A. MA2B2 type (e.g., [Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>])
Why: Square planar MA2B2 complexes can have A and B groups either adjacent (cis) or diagonal (trans), giving two isomers.
Q24.
In an octahedral complex MA3B3, the two geometrical isomers are:
A Facial (fac) and meridional (mer)
B Cis and trans
C Clockwise and anticlockwise
D Alpha and beta
Show answer & explanation
Answer: A. Facial (fac) and meridional (mer)
Why: In MA3B3 octahedral: fac isomer has A ligands on one face of the octahedron; mer has A ligands in a plane through the centre.
Q25.
Optical isomers in coordination chemistry (enantiomers) are non-superimposable mirror images. A classic example is:
A [Co(en)3]3+ (tris-chelate)
B [Co(NH<sub>3</sub>)<sub>6</sub>]3+ in the majority of cases studied
C [Ni(CN)4]2- as widely reported
D [CrCl6]3- in standard practice
Show answer & explanation
Answer: A. [Co(en)3]3+ (tris-chelate)
Why: [Co(en)3]3+ is chiral (D and L forms); it lacks a plane, centre, or axis of improper rotation, giving non-superimposable mirror images.
Q26.
What does VBT (Valence Bond Theory) explain about coordination compounds?
A Hybridisation of metal orbitals and geometry; doesn't explain colour or magnetism well
B Every property of the complex, including its colour and magnetism in full detail
C Mainly the magnetic behaviour of the complex and nothing else as frequently observed in practice
D Mainly the kinetics and reaction rates of the complex's ligand exchange in many documented cases
Show answer & explanation
Answer: A. Hybridisation of metal orbitals and geometry; doesn't explain colour or magnetism well
Why: VBT explains shape and hybridisation of complexes but fails to satisfactorily explain colour (d-d transitions) and paramagnetism in a quantitative way.
Q27.
In VBT, inner orbital complexes use d orbitals from:
A The inner (n-1)d subshell (e.g., 3d with 4s, 4p for period 4 metals)
B The outer nd subshell instead of the inner (n-1)d subshell under most conditions encountered
C 4f orbitals borrowed from the lanthanide series as frequently observed in practice
D 5d orbitals regardless of which period the metal belongs to in many documented cases
Show answer & explanation
Answer: A. The inner (n-1)d subshell (e.g., 3d with 4s, 4p for period 4 metals)
Why: Inner orbital (low spin) complexes: metal uses (n-1)d, ns, and np orbitals for hybridisation (e.g., d<sup>2</sup>sp<sup>3</sup> for octahedral).
Q28.
In CFT, the crystal field stabilisation energy (CFSE) for d<sup>6</sup> high spin in octahedral field is:
A -0.4 Delta_o (4 in t<sub>2g</sub> × -0.4 + 2 in eg × +0.6 = -1.6 + 1.2 = -0.4 Delta_o)
B Zero, since the high-spin d<sup>6</sup> arrangement gives no net stabilisation
C -1.2 Delta_o, the value corresponding instead to a d<sup>3</sup> configuration
D -2.4 Delta_o, the value corresponding instead to a low-spin d<sup>6</sup> configuration
Show answer & explanation
Answer: A. -0.4 Delta_o (4 in t<sub>2g</sub> × -0.4 + 2 in eg × +0.6 = -1.6 + 1.2 = -0.4 Delta_o)
A The donor atom through which an ambidentate ligand coordinates (e.g., NO<sub>2</sub>- via N or O)
B Their overall molecular formula, which differs between the two isomers according to conventional understanding
C The oxidation state assigned to the central metal atom in routine practice overall
D The total number of electrons present in the complex in most cases under typical conditions
Show answer & explanation
Answer: A. The donor atom through which an ambidentate ligand coordinates (e.g., NO<sub>2</sub>- via N or O)
Why: Linkage isomers: same formula but the ambidentate ligand bonds through different atoms. Example: [Co(NO<sub>2</sub>)(NH<sub>3</sub>)5]2+ vs [Co(ONO)(NH<sub>3</sub>)5]2+.
Q30.
Prussian blue is:
A Fe4[Fe(CN)<sub>6</sub>]3 (mixed valence Fe<sup>2+</sup>/Fe<sup>3+</sup> complex with CN- bridges)
B A simple aqueous CuSO<sub>4</sub> solution with little iron present according to conventional understanding
C Plain FeCl<sub>3</sub> dissolved in water with little cyanide bridges
D K3[Fe(CN)<sub>6</sub>], the separate potassium ferricyanide salt in routine practice
Show answer & explanation
Answer: A. Fe4[Fe(CN)<sub>6</sub>]3 (mixed valence Fe<sup>2+</sup>/Fe<sup>3+</sup> complex with CN- bridges)
Why: Prussian blue is iron(III) hexacyanoferrate(II), a coordination network used as a blue pigment and in detoxification of heavy metals.
Q31.
The spectrochemical series for common ligands in order of increasing field strength:
A I- < Br- < Cl- < F- < OH- < H<sub>2</sub>O < NH<sub>3</sub> < en < CN- < CO
B CO < CN- < NH<sub>3</sub> < F- < Cl-, listed in the reverse field-strength order
C All ligands shown to have identical field strength toward any metal
D The field strength ranking depending entirely on which metal is used
Show answer & explanation
Answer: A. I- < Br- < Cl- < F- < OH- < H<sub>2</sub>O < NH<sub>3</sub> < en < CN- < CO
Why: Weak field (halides, water) to strong field (NH<sub>3</sub>, en, CN-, CO). CO is the strongest due to extensive pi-backbonding.
Q32.
How does the EDTA complex [EDTA-Ca]2- help in water softening?
A It sequesters Ca<sup>2+</sup> and Mg<sup>2+</sup> by forming very stable (high K) chelates, preventing them from causing hardness
B It precipitates the calcium and magnesium ions out of solution as insoluble solid salts according to standard textbooks
C It promotes additional ionisation of the surrounding bulk water molecules in general practice as frequently described
D It generally raises the overall pH of the water without complexing any of the ions in most textbook accounts
Show answer & explanation
Answer: A. It sequesters Ca<sup>2+</sup> and Mg<sup>2+</sup> by forming very stable (high K) chelates, preventing them from causing hardness
Why: EDTA forms very stable 1:1 chelate complexes with Ca<sup>2+</sup> and Mg<sup>2+</sup>, effectively removing them from solution (sequestration).
Q33.
Coordination compounds with the same formula and donor atoms but different arrangements are called:
A Stereoisomers (including geometric and optical isomers)
B Constitutional (structural) isomers, which differ in connectivity instead
C Resonance structures of a single fixed arrangement of atoms
D Tautomers related by a simple proton-shift equilibrium
Show answer & explanation
Answer: A. Stereoisomers (including geometric and optical isomers)
Why: Stereoisomers have the same connectivity but different spatial arrangements; this includes cis/trans (geometric) and optical isomers.
Q34.
The effective atomic number (EAN) rule by Sidgwick states:
A Metal in a complex acquires electrons from ligands to achieve the electron count of the next noble gas
B The metal instead loses electrons to its surrounding ligands in the complex during normal conditions
C Ligands serve mainly to fix the oxidation state with little electron-count rule involved as generally observed
D The coordination number is usually numerically equal to the oxidation state in typical laboratory settings
Show answer & explanation
Answer: A. Metal in a complex acquires electrons from ligands to achieve the electron count of the next noble gas
Why: Sidgwick's EAN rule: the metal + ligand electron donation brings the metal to the same electron count as the nearest noble gas.
Q35.
Which of the following is an outer orbital complex (uses nd orbitals, sp<sup>3</sup>d<sup>2</sup>)?
A [CoF6]3- (F- is weak field, uses 4d orbitals, sp<sup>3</sup>d<sup>2</sup>, high spin)
B [Co(CN)6]3-, an inner orbital low-spin complex using 3d orbitals
C [Co(NH<sub>3</sub>)<sub>6</sub>]3-, an inner orbital low-spin complex using 3d orbitals
D [Fe(CO)<sub>5</sub>], a zero-valent carbonyl complex with trigonal bipyramidal geometry
Show answer & explanation
Answer: A. [CoF6]3- (F- is weak field, uses 4d orbitals, sp<sup>3</sup>d<sup>2</sup>, high spin)
Why: [CoF6]3-: F- is weak field; outer orbital hybridisation uses 4s, 4p, 4d (not 3d); it is high spin with 4 unpaired electrons.
Q36.
The stability constant (formation constant) Kf of a complex indicates:
A The extent to which the complex forms in solution (higher Kf = more stable)
B The specific colour the complex displays once largely formed under usual circumstances
C The melting point of the solid complex once isolated according to most researchers
D The overall molecular weight of the formed complex in the majority of cases studied
Show answer & explanation
Answer: A. The extent to which the complex forms in solution (higher Kf = more stable)
Why: Kf = [complex] / ([metal ion][ligand]<sup>n</sup>); a high Kf means the complex dissociates very little and is very stable.
Q37.
Which complex shows the trans effect most prominently?
A Square planar Pt(II) complexes
B Octahedral Co(III) as widely reported
C Tetrahedral Ni(II) in standard practice
D Linear Au(I) under most conditions encountered
Show answer & explanation
Answer: A. Square planar Pt(II) complexes
Why: The trans effect (kinetic lability of a ligand trans to another) is most pronounced and best studied in square planar Pt(II) complexes.
Q38.
Coordination isomers differ by:
A Exchange of ligands between two complex centres in a compound with both + and - complex ions
B The oxidation state formally assigned to each central metal ion present in the salt
C The total number of ligands individually attached to each of the two metal centres
D Geometry alone, while keeping an identical ligand distribution shared between centres
Show answer & explanation
Answer: A. Exchange of ligands between two complex centres in a compound with both + and - complex ions
Why: Coordination isomers: in a compound with two complex ions, ligands redistribute between the cationic and anionic complex. E.g., [Co(en)3][Cr(CN)6] vs [Cr(en)3][Co(CN)6].
Q39.
The role of EDTA in analytical chemistry (complexometric titration) is:
A It forms 1:1 stable complexes with metal ions at controlled pH, allowing accurate quantification
B It functions generally as a strong acid that lowers the solution's pH as frequently observed in practice
C It precipitates the target metal ions as an insoluble solid in many documented cases according to conventional understanding
D It acts mainly as a visual indicator that changes colour at the endpoint in routine practice
Show answer & explanation
Answer: A. It forms 1:1 stable complexes with metal ions at controlled pH, allowing accurate quantification
Why: EDTA complexometric titration: EDTA forms stable 1:1 complexes with most metal ions; endpoint is detected by metallochromic indicators (e.g., Eriochrome Black T).
Q40.
Which coordination compound was historically important in proving Werner's theory?
A [Co(NH<sub>3</sub>)<sub>6</sub>]Cl3 and its series of Co-NH<sub>3</sub>-Cl complexes
B NaCl, a simple ionic salt with no coordinate covalent bonding
C [Pt(CN)4]2-, a square planar complex unrelated to Werner's original series
D K<sub>4</sub>[Fe(CN)<sub>6</sub>], the separate potassium ferrocyanide salt
Show answer & explanation
Answer: A. [Co(NH<sub>3</sub>)<sub>6</sub>]Cl3 and its series of Co-NH<sub>3</sub>-Cl complexes
Why: Werner's studies on cobalt-ammonia complexes (e.g., [Co(NH<sub>3</sub>)<sub>6</sub>]Cl3, [Co(NH<sub>3</sub>)5Cl]Cl<sub>2</sub>, etc.) established the concept of primary and secondary valence.
Hard - 20 questions
Q41.
Using CFSE, predict whether [Co(CN)6]4- (Co2+, d7) is high or low spin:
A Low spin: CN- is strong field, large Delta causes maximum pairing (t<sub>2g</sub><sup>6</sup> eg1, 1 unpaired)
B High spin with three unpaired electrons distributed across t<sub>2g</sub> and eg in the majority of cases studied
C Largely diamagnetic with zero unpaired electrons in this d7 ion as widely reported
D High spin with five unpaired electrons, the maximum possible for d7 in standard practice
Show answer & explanation
Answer: A. Low spin: CN- is strong field, large Delta causes maximum pairing (t<sub>2g</sub><sup>6</sup> eg1, 1 unpaired)
Why: Co2+ (d7) with strong field CN-: low spin configuration is t<sub>2g</sub><sup>6</sup> eg1 with only 1 unpaired electron (mu ≈ 1.73 BM).
Q42.
The stability of chelate complexes is explained by:
A Chelate effect = combination of enhanced entropy (more particles released on complex formation) and entropic contribution to Gibbs energy
B Each individual metal-donor bond within the chelate ring being inherently far stronger than usual under most conditions encountered
C The chelate generally containing a much larger total number of covalent bonds overall as frequently observed in practice in many documented cases
D The chelate possessing a far greater degree of ionic character than ordinary monodentate complexes according to conventional understanding
Show answer & explanation
Answer: A. Chelate effect = combination of enhanced entropy (more particles released on complex formation) and entropic contribution to Gibbs energy
Why: When a chelate forms from a polydentate ligand, more solvent molecules are released than with an equivalent number of monodentate ligands, giving positive Delta_S and more negative Delta_G.
Q43.
The trans effect series in Pt(II) chemistry ranks ligands by their ability to labilise the trans position. The order includes:
Why: Strong pi-acceptors (CO, CN-, NO+) and strong sigma-donors (H-, alkyl) have the greatest trans-labilising effect in Pt(II) kinetics.
Q44.
Which of the following octahedral complexes shows optical isomerism?
A [Co(en)<sub>2</sub>Cl<sub>2</sub>]+ (cis form is chiral)
B [Co(NH<sub>3</sub>)<sub>6</sub>]3+, a perfectly symmetric homoleptic complex with a plane of symmetry
C [CoCl<sub>6</sub>]3-, a perfectly symmetric homoleptic complex with a plane of symmetry
D trans-[Co(en)<sub>2</sub>Cl<sub>2</sub>]+, which possesses an internal mirror plane and is achiral
Show answer & explanation
Answer: A. [Co(en)<sub>2</sub>Cl<sub>2</sub>]+ (cis form is chiral)
Why: The cis-[Co(en)<sub>2</sub>Cl<sub>2</sub>]+ has no plane of symmetry and shows Delta (right-handed) and Lambda (left-handed) enantiomers. The trans form has a plane of symmetry and is achiral.
Q45.
The pi-backbonding (back-donation) in metal carbonyl complexes: CO donates sigma electrons to the metal, and the metal:
A Donates electron density from filled d orbitals back into the CO pi* orbital (backbonding), strengthening M-C and weakening C-O
B Instead generally accepts additional pi electron density directly back from the filled CO pi orbital in routine practice overall
C Transfers most of its electrons to the CO ligand mainly through an ionic electron-transfer mechanism in most cases under typical conditions
D Forms highly no pi-type interaction with CO, relying mainly on simple sigma bonding according to standard textbooks in general practice
Show answer & explanation
Answer: A. Donates electron density from filled d orbitals back into the CO pi* orbital (backbonding), strengthening M-C and weakening C-O
Why: Metal-CO: CO sigma-donates to M, and M pi-backdonates from filled d orbitals into CO pi*; this strengthens M-C bond and weakens C-O (IR shows lower C-O stretch vs free CO).
Q46.
Why is [PtCl<sub>4</sub>]2- square planar while [NiCl<sub>4</sub>]2- is tetrahedral?
A Pt<sup>2+</sup> (5d<sup>8</sup>) has larger CFSE favouring square planar; Ni<sup>2+</sup> (3d<sup>8</sup>) has smaller Delta and the pairing energy penalty for square planar is not compensated
B Platinum's much greater atomic mass relative to nickel is what forces the square planar shape on its own as frequently described in most textbook accounts
C The chloride ligand bound to nickel is somehow physically larger than the same chloride ligand bound to platinum during normal conditions as generally observed
D Nickel is supplied with extra coordinating chloride ligands in solution that platinum is largely denied access to in typical laboratory settings
Show answer & explanation
Answer: A. Pt<sup>2+</sup> (5d<sup>8</sup>) has larger CFSE favouring square planar; Ni<sup>2+</sup> (3d<sup>8</sup>) has smaller Delta and the pairing energy penalty for square planar is not compensated
Why: Square planar (d<sup>8</sup>) requires a large crystal field splitting. 5d orbitals of Pt<sup>2+</sup> are larger and interact more strongly with ligands, giving large Delta favouring square planar. 3d Ni<sup>2+</sup> with Cl- (weak field) favours tetrahedral.
Q47.
The spectrochemical series order is explained by sigma and pi donation/acceptance. Why does H<sub>2</sub>O cause smaller splitting than NH<sub>3</sub>?
A H<sub>2</sub>O is a pi-donor (lone pairs on O donate into empty metal d orbitals), decreasing effective Delta; NH<sub>3</sub> has no pi-donor capacity
B Water is actually claimed to be a far stronger sigma donor than ammonia toward the metal centre under usual circumstances according to most researchers
C Ammonia is instead claimed to accept pi electron density back from the filled metal d orbitals in the majority of cases studied
D Water's oxygen donor atom is generally claimed to be much larger than ammonia's nitrogen donor atom as widely reported in standard practice
Show answer & explanation
Answer: A. H<sub>2</sub>O is a pi-donor (lone pairs on O donate into empty metal d orbitals), decreasing effective Delta; NH<sub>3</sub> has no pi-donor capacity
Why: H<sub>2</sub>O donates both sigma (into metal's empty orbital) and pi (from O lone pair back into metal d), partially cancelling sigma donor effect. NH<sub>3</sub> is a pure sigma donor giving higher net Delta.
Q48.
Prussian blue reaction: K<sub>4</sub>[Fe(CN)<sub>6</sub>] (potassium ferrocyanide) added to FeCl<sub>3</sub> gives:
A Prussian blue (KFe[Fe(CN)<sub>6</sub>]) intensely blue precipitate
B A pale yellow precipitate forming instead of any blue colour
C A deep red colouration with no precipitate forming at all
D No visible reaction occurring between the two reagents
Show answer & explanation
Answer: A. Prussian blue (KFe[Fe(CN)<sub>6</sub>]) intensely blue precipitate
Why: Fe<sup>3+</sup> (from FeCl<sub>3</sub>) + [Fe(CN)<sub>6</sub>]4- → KFe[Fe(CN)<sub>6</sub>] (Prussian blue), a mixed-valence Fe<sup>2+</sup>/Fe<sup>3+</sup> complex with intense blue charge-transfer band.
Q49.
The Irving-Williams series describes:
A The stability order of divalent transition metal complexes: Mn<sup>2+</sup> < Fe<sup>2+</sup> < Co2+ < Ni<sup>2+</sup> < Cu<sup>2+</sup> > Zn<sup>2+</sup>
B The spectrochemical series ranking ligands by field strength instead under most conditions encountered
C The trans effect series ranking ligands by their labilising power in Pt(II) complexes as frequently observed in practice
D The general stability ranking of different oxidation states for one metal in many documented cases
Show answer & explanation
Answer: A. The stability order of divalent transition metal complexes: Mn<sup>2+</sup> < Fe<sup>2+</sup> < Co2+ < Ni<sup>2+</sup> < Cu<sup>2+</sup> > Zn<sup>2+</sup>
Why: The Irving-Williams series reflects increasing Lewis acidity (smaller ionic radius, increasing Zeff) across the first row transition metals for +2 ions, culminating in Cu<sup>2+</sup> (Jahn-Teller distortion helps additional stability) then dropping at Zn<sup>2+</sup> (d<sup>10</sup>).
Q50.
Which of the following has the most CFSE stabilisation in an octahedral field?
A d<sup>3</sup> (t<sub>2g</sub><sup>3</sup>, CFSE = -1.2 Delta_o) and d<sup>6</sup> low spin (t<sub>2g</sub><sup>6</sup>, CFSE = -2.4 Delta_o)
B d<sup>0</sup>, which by definition has zero d electrons and therefore zero CFSE
C d<sup>10</sup>, which has all orbitals fully occupied giving zero net CFSE
D d<sup>5</sup> high spin, which has one electron in every orbital giving zero net CFSE
Show answer & explanation
Answer: A. d<sup>3</sup> (t<sub>2g</sub><sup>3</sup>, CFSE = -1.2 Delta_o) and d<sup>6</sup> low spin (t<sub>2g</sub><sup>6</sup>, CFSE = -2.4 Delta_o)
Why: d<sup>6</sup> low spin (t<sub>2g</sub><sup>6</sup> e<sub>g</sub><sup>0</sup>): CFSE = 6(-0.4) = -2.4 Delta_o, the maximum CFSE for any dn in octahedral field. d<sup>3</sup> gives -1.2 Delta_o (all in t<sub>2g</sub>).
Q51.
The reaction [PtCl<sub>4</sub>]2- + 2NH<sub>3</sub> → cis-[Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>] uses the trans effect of Cl- to control the product. The mechanism:
A First NH<sub>3</sub> goes to any position; second NH<sub>3</sub> goes cis to the first Cl- (Cl- labilises trans site, so second substitution is trans to Cl-)
B Both ammonia ligands are said to substitute in a largely random, loosely controlled order overall in most cases under typical conditions
C The second NH<sub>3</sub> ligand is said to substitute specifically trans to the first NH<sub>3</sub> ligand instead according to standard textbooks in general practice
D Chloride is said to act generally as a passive spectator ion with little influence on the order as frequently described in most textbook accounts
Show answer & explanation
Answer: A. First NH<sub>3</sub> goes to any position; second NH<sub>3</sub> goes cis to the first Cl- (Cl- labilises trans site, so second substitution is trans to Cl-)
Why: Step 1: [PtCl<sub>4</sub>]2- + NH<sub>3</sub> → [Pt(NH<sub>3</sub>)Cl<sub>3</sub>]-. Step 2: Cl- trans to the NH<sub>3</sub> is now trans-labilised; the second NH<sub>3</sub> replaces this Cl-, giving the cis product.
Q52.
Which of the following statements about EDTA (ethylenediaminetetraacetate) is correct?
A It is hexadentate (2 N + 4 O donor atoms) and forms thermodynamically very stable 1:1 chelates with most metal ions
B It is generally bidentate, donating through mainly two of its six potential donor atoms according to conventional understanding
C It is said to bind mainly to transition metal ions and not to any s-block cations in routine practice overall
D It is said to become notably unstable once the surrounding solution pH rises above 7 in most cases under typical conditions
Show answer & explanation
Answer: A. It is hexadentate (2 N + 4 O donor atoms) and forms thermodynamically very stable 1:1 chelates with most metal ions
Why: EDTA<sup>4-</sup> is hexadentate with 6 donor atoms; it forms kinetically stable, soluble chelate complexes (log Kf for Ca<sup>2+</sup>: 10.7; Fe<sup>3+</sup>: 25.1) over a wide pH range.
Q53.
In coordination chemistry, the 'symbiosis principle' states:
A Hard (sigma-donor) ligands stabilise hard metal centres; soft (pi-acceptor) ligands stabilise soft centres (HSAB principle applied to complexes)
B Every ligand type is said to behave in an largely equivalent way regardless of hard or soft donor character according to standard textbooks
C Mainly strong field ligands such as CN- are said to ever be capable of forming any reasonably stable complex in general practice as frequently described
D The coordination number on its own is said to be the single factor that largely determines complex stability in most textbook accounts during normal conditions
Show answer & explanation
Answer: A. Hard (sigma-donor) ligands stabilise hard metal centres; soft (pi-acceptor) ligands stabilise soft centres (HSAB principle applied to complexes)
Why: HSAB principle: hard metal centres (high charge, low polarisability) bind preferentially with hard ligands (F-, OH-, O donors); soft metals (Pt, Pd, Hg) bind soft ligands (I-, S donors, CO, CN-).
Q54.
The fluxional behaviour of [Fe(CO)<sub>5</sub>] (trigonal bipyramidal) refers to:
A Rapid exchange of axial and equatorial CO positions at room temperature via Berry pseudorotation
B A largely fixed, rigid structure with little exchange between CO positions as generally observed
C Permanent loss of one CO ligand followed by its slow recoordination in typical laboratory settings
D Spontaneous ring formation among the five coordinated CO ligands under usual circumstances
Show answer & explanation
Answer: A. Rapid exchange of axial and equatorial CO positions at room temperature via Berry pseudorotation
Why: Berry pseudorotation in Fe(CO)<sub>5</sub>: a concerted movement interconverts axial and equatorial CO positions; 13C NMR shows all 5 CO equivalent even at room temperature.
Q55.
For a d<sup>8</sup> metal ion in square planar field, the dx2-y2 orbital is highest in energy because:
A It points directly at the four ligands (sigma interaction); the other d orbitals are less directly aligned with ligands
B It generally happens to contain more electrons than the other four d orbitals combined according to most researchers
C The d<sup>8</sup> electron configuration is said to fill this particular orbital last by convention in the majority of cases studied
D The four surrounding ligands are positioned specifically so as to avoid this orbital largely as widely reported in standard practice
Show answer & explanation
Answer: A. It points directly at the four ligands (sigma interaction); the other d orbitals are less directly aligned with ligands
Why: In square planar geometry, dx2-y2 is destabilised most strongly as it lies in the xy plane pointing at all four ligands; dz<sup>2</sup> is intermediate; dxy is next; dxz and dyz are lowest.
Q56.
Cisplatin's anticancer activity is due to:
A Cross-linking DNA by displacing Cl- with purine N-donors (N7 of guanine), distorting DNA helix
B Generally releasing free toxic platinum ions that poison the entire cell under most conditions encountered
C Directly inhibiting a specific metabolic enzyme without touching DNA as frequently observed in practice
D Binding to and disrupting the lipid bilayer of the cell membrane in many documented cases
Show answer & explanation
Answer: A. Cross-linking DNA by displacing Cl- with purine N-donors (N7 of guanine), distorting DNA helix
Why: Cisplatin undergoes aquation (Cl- replaced by H<sub>2</sub>O), then bifunctionally binds N7 of adjacent guanines, forming 1,2-d(GpG) intrastrand cross-links that kink DNA and block replication.
Q57.
The total number of isomers (including geometric and optical) for [Co(en)(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>]+ is:
A Five (cis and trans geometric isomers, with the cis form giving three: two cis diastereomers of en orientation and optical isomers)
B Just two isomers in total, counting mainly a single cis form and a single trans form according to conventional understanding in routine practice
C Exactly three isomers in total, largely ignoring any optical activity within the cis form overall in most cases under typical conditions
D Mainly one single possible isomer, since the en ligand is said to fix a unique overall geometry according to standard textbooks
Show answer & explanation
Answer: A. Five (cis and trans geometric isomers, with the cis form giving three: two cis diastereomers of en orientation and optical isomers)
Why: This type of complex is complex enough for JEE: cis and trans geometric isomers exist; the cis isomer can have en coordinating in different ways and may show optical isomerism.
Q58.
In MO theory treatment of octahedral complexes, which MO is responsible for the colour (d-d transition)?
A t<sub>2g</sub> to eg* transition (technically t<sub>2g</sub> to eg MO; the antibonding eg* in MO theory corresponds to eg in CFT)
B A simple sigma bonding to sigma antibonding transition occurring within the ligand framework in general practice
C A pi bonding to pi antibonding transition localised largely on the surrounding ligand atoms as frequently described
D A deep core 1s to 2s electronic transition occurring within the central metal atom itself in most textbook accounts
Show answer & explanation
Answer: A. t<sub>2g</sub> to eg* transition (technically t<sub>2g</sub> to eg MO; the antibonding eg* in MO theory corresponds to eg in CFT)
Why: In MO theory for octahedral complexes, the t<sub>2g</sub> MOs (non-bonding) and eg* MOs (antibonding, metal dx2-y2 and dz<sup>2</sup> character) are involved in the d-d transition responsible for colour.
Q59.
The spectrochemical series and the nephelauxetic series differ in that:
A Spectrochemical series ranks field strength (Delta); nephelauxetic series ranks the ability to expand d orbital size (reduce electron repulsion, measured by beta)
B The two series are claimed to be in fact largely identical in highly everything that each one measures during normal conditions as generally observed in typical laboratory settings
C The spectrochemical series is claimed to exist mainly to classify complex colour and to serve no other purpose under usual circumstances according to most researchers
D The nephelauxetic series is claimed to instead classify coordination complexes mainly by their geometric isomer type in the majority of cases studied as widely reported
Show answer & explanation
Answer: A. Spectrochemical series ranks field strength (Delta); nephelauxetic series ranks the ability to expand d orbital size (reduce electron repulsion, measured by beta)
Why: Nephelauxetic (cloud-expanding) effect: covalent bonding with ligands delocalises metal d electrons, reducing electron-electron repulsion. beta = B(complex)/B(free ion); B is the Racah parameter.
Q60.
In the complex [Co(NH₃)₆]³⁺, cobalt(III) with the strong-field NH₃ ligand uses the hybridisation:
A sp³ tetrahedral
B dsp² square planar
C sp³d² outer orbital
D d²sp³ inner orbital
Show answer & explanation
Answer: D. d²sp³ inner orbital
Why: NH₃ is a strong-field ligand, so Co³⁺ (d⁶) pairs its electrons and uses inner (n−1)d orbitals: d²sp³, giving a low-spin octahedral complex.