NeetCrackers Chemistry Practice
Practice Set

Coordination Compounds — Practice Questions with Answers

30 fully solved questions covering Werner's theory, IUPAC nomenclature, isomerism, and bonding theories (VBT, CFT).

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Questions written and solutions verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Checked against NCERT and previous years' papers. Last reviewed September 2026.
10MCQs · 1 mark
10Short Answer · 3 marks
10Long Answer · 5 marks
90Total Marks

📑 On This Page

  1. Section A — 10 MCQs (1 mark each)
  2. Section B — 10 Short Answer Questions (3 marks each)
  3. Section C — 10 Long Answer Questions (5 marks each)
Section A

🎯 Multiple Choice Questions

10 questions · 1 mark each · Correct answer highlighted with explanation

Q1 The coordination number of a central metal ion refers to:
Correct Answer: (b). Coordination number is the number of ligand donor atoms directly bonded to the central metal ion/atom.
Q2 A ligand that can donate only one pair of electrons (bonds through only one donor atom) is called:
Correct Answer: (a) Unidentate (monodentate) — examples include $NH_3$, $H_2O$, $Cl^-$.
Q3 EDTA (ethylenediaminetetraacetate) is an example of a:
Correct Answer: (c). EDTA can bind through six donor atoms simultaneously, making it a hexadentate ligand, widely used to sequester metal ions.
Q4 According to crystal field theory, in an octahedral complex, the d-orbitals split into:
Correct Answer: (a). In an octahedral field, d-orbitals split into the lower-energy $t_{2g}$ set (three orbitals) and higher-energy $e_g$ set (two orbitals).
Q5 A ligand that can bond through either of two different donor atoms, but only one at a time, is called:
Correct Answer: (b) Ambidentate — an example is the thiocyanate ion ($SCN^-$), which can bond through either sulfur or nitrogen.
Q6 Werner's coordination theory proposed that metal ions exhibit two distinct types of valency:
Correct Answer: (a). Werner distinguished between primary valency (satisfied by anions, ionisable, equal to oxidation state) and secondary valency (satisfied by ligands, fixed, equal to coordination number).
Q7 Complexes exhibiting geometrical isomerism (cis-trans) most commonly occur in complexes with coordination number:
Correct Answer: (b). Square planar (coordination number 4) and octahedral (coordination number 6) complexes commonly exhibit cis-trans (geometrical) isomerism, given appropriate ligand arrangements.
Q8 A chelate is formed when:
Correct Answer: (a). Chelation involves a polydentate ligand wrapping around and binding to a single metal centre through multiple donor atoms, forming a stable ring structure.
Q9 Complexes with strong field ligands (causing large crystal field splitting) tend to be:
Correct Answer: (a). Strong field ligands cause large orbital splitting, favouring electron pairing in the lower-energy orbitals before populating the higher-energy set, resulting in a low-spin configuration.
Q10 In IUPAC nomenclature of coordination compounds, ligands are named in the complex ion in:
Correct Answer: (a). Ligands are cited in alphabetical order within the complex ion's name, regardless of whether they are anionic, neutral, or cationic ligands.
Section B

✍️ Short Answer Questions

10 questions · 3 marks each · Full model answers included

Q1. State Werner's postulates regarding coordination compounds.
3 Marks
Answer

Werner proposed that metal ions exhibit two types of valency: primary valency (ionisable, satisfied by negative ions, corresponding to the metal's oxidation state) and secondary valency (non-ionisable, fixed, satisfied by ligands directly bonded to the metal, corresponding to the coordination number).

He further proposed that secondary valencies are directed towards fixed positions in space around the central metal ion, giving the complex a definite geometrical shape (e.g., octahedral for coordination number 6).

Q2. Distinguish between a double salt and a coordination compound (complex), with an example of each.
3 Marks
Answer

A double salt loses its identity completely in aqueous solution, dissociating fully into its simple constituent ions — for example, Mohr's salt, $FeSO_4\cdot(NH_4)_2SO_4\cdot6H_2O$, dissociates completely into $Fe^{2+}$, $NH_4^+$, and $SO_4^{2-}$ ions in solution.

A coordination compound (complex) retains its identity in solution, since the complex ion does not dissociate into its individual constituent ligands — for example, $[Fe(CN)_6]^{4-}$ (ferrocyanide) does not release free $Fe^{2+}$ or $CN^-$ ions in solution, instead behaving as a single, distinct ionic species.

Q3. Define "chelate effect," and explain why chelate complexes are generally more stable than analogous complexes with unidentate ligands.
3 Marks
Answer

The chelate effect refers to the enhanced thermodynamic stability of metal complexes formed with polydentate (chelating) ligands, compared to similar complexes formed with an equivalent number of separate unidentate ligands.

This enhanced stability arises primarily from a favourable entropy effect — when a single polydentate ligand displaces several separate unidentate ligands, the overall number of independent particles in solution increases (since one chelating ligand replaces multiple individual ligand molecules), resulting in a favourable (positive) increase in entropy, which contributes to a more negative (more favourable) overall Gibbs free energy change, and hence greater stability of the chelate complex.

Q4. Explain the concept of "crystal field splitting" in an octahedral complex, and how the magnitude of splitting relates to ligand field strength.
3 Marks
Answer

In a free (isolated) metal ion, all five d-orbitals have equal energy. When surrounded by ligands in an octahedral arrangement, electrostatic repulsion from the approaching ligand electron pairs affects the d-orbitals unequally (depending on their specific spatial orientation relative to the ligands) — this causes the d-orbitals to split into two energy levels: a lower-energy $t_{2g}$ set and a higher-energy $e_g$ set.

The energy gap between these two sets (called the crystal field splitting energy, $\Delta_o$) depends on the specific ligand's "field strength" — strong field ligands (like $CN^-$, $CO$) cause a large splitting energy, while weak field ligands (like halide ions, $H_2O$) cause a comparatively small splitting energy, as summarised by the spectrochemical series.

Q5. Distinguish between high-spin and low-spin complexes, relating each to crystal field splitting energy and pairing energy.
3 Marks
Answer

A high-spin complex forms when the crystal field splitting energy ($\Delta_o$) is smaller than the electron pairing energy — electrons preferentially occupy all five d-orbitals singly (following Hund's rule) before any pairing occurs, resulting in the maximum possible number of unpaired electrons. This is typical with weak field ligands.

A low-spin complex forms when $\Delta_o$ is larger than the pairing energy — electrons preferentially pair up within the lower-energy $t_{2g}$ orbitals before occupying the higher-energy $e_g$ set, resulting in fewer unpaired electrons than the free ion. This is typical with strong field ligands.

Q6. Explain the difference between ionisation isomerism and hydrate (solvate) isomerism in coordination compounds.
3 Marks
Answer

Ionisation isomerism occurs when two complexes differ in which ion is present inside the coordination sphere (bonded to the metal) versus outside as a simple counter-ion — for example, $[Co(NH_3)_5Br]SO_4$ versus $[Co(NH_3)_5SO_4]Br$, which produce different ions in solution upon dissociation.

Hydrate (solvate) isomerism is a specific type of ionisation isomerism where water (or another solvent) molecules are distributed differently between the coordination sphere and outside it — for example, $[Cr(H_2O)_6]Cl_3$ versus $[Cr(H_2O)_5Cl]Cl_2\cdot H_2O$, which differ in how many water molecules are directly coordinated to the metal versus present as simple water of crystallisation.

Q7. Explain linkage isomerism, with an example involving the nitrite ion ($NO_2^-$).
3 Marks
Answer

Linkage isomerism occurs specifically with ambidentate ligands (which possess two different potential donor atoms), where the same ligand can bind to the metal through either of its two different donor atoms, producing distinct isomers.

Example: The nitrite ion ($NO_2^-$) can bind to a metal through its nitrogen atom (forming a "nitro" complex, e.g., $[Co(NH_3)_5(NO_2)]^{2+}$) or through one of its oxygen atoms (forming a "nitrito" complex) — these two distinct binding modes of the same ligand, attached to the identical metal centre, produce genuinely different linkage isomers with distinct chemical/physical properties.

Q8. State the basic postulates of valence bond theory (VBT) as applied to coordination compounds.
3 Marks
Answer

According to valence bond theory, the central metal ion makes available a specific number of empty (vacant) orbitals, equal to its coordination number, for accepting electron pairs donated by the surrounding ligands.

These empty metal orbitals undergo hybridisation (mixing) to form new, equivalent hybrid orbitals with specific geometric orientations (e.g., $sp^3d^2$ or $d^2sp^3$ hybridisation for octahedral geometry) — each ligand then donates an electron pair into one of these hybrid orbitals, forming a coordinate covalent bond, with the specific type of hybridisation directly determining the overall geometric shape of the resulting complex.

Q9. Explain optical isomerism in octahedral complexes, using $[Co(en)_3]^{3+}$ as an example.
3 Marks
Answer

Optical isomerism occurs when a complex and its mirror image are non-superimposable on each other (analogous to a pair of left and right hands) — these two non-superimposable mirror-image forms are called enantiomers, and they rotate plane-polarised light in opposite directions.

$[Co(en)_3]^{3+}$ (where "en" is the bidentate ethylenediamine ligand) exists as a pair of non-superimposable mirror-image (optical) isomers, since the three chelating "en" ligands wrap around the octahedral cobalt centre in a specific, chiral (handed) spatial arrangement that lacks an internal plane of symmetry, giving rise to this characteristic optical isomerism.

Q10. Explain the significance/application of coordination compounds in biological systems, citing one specific example.
3 Marks
Answer

Coordination compounds play crucial roles in numerous biological processes, since many essential biomolecules involve a metal ion coordinated within a specific organic ligand framework.

Example: Haemoglobin, essential for oxygen transport in blood, contains an iron(II) ion coordinated within a porphyrin ring structure (forming the "heme" group) — the iron centre can reversibly bind and release an oxygen molecule as an additional ligand, allowing haemoglobin to efficiently pick up oxygen in the lungs and release it to tissues throughout the body, a process fundamentally dependent on the coordination chemistry of the central iron ion.

Section C

📘 Long Answer / Numerical Questions

10 questions · 5 marks each · Detailed step-by-step solutions

Q1. Determine the oxidation state of the central metal ion in $[Co(NH_3)_5Cl]Cl_2$, and write the IUPAC name of this complex.
5 Marks
Answer

Let the oxidation state of Co be $x$. $NH_3$ is neutral (contributes 0); each $Cl$ contributes $-1$.

Overall, the compound must be neutral: $x + 5(0) + 1(-1)[\text{coordinated Cl}] + 2(-1)[\text{outer Cl}] = 0$

$x - 1 - 2 = 0 \Rightarrow x = +3$

IUPAC name: pentaamminechloridocobalt(III) chloride

Q2. Determine the oxidation state of the central metal ion in the complex ion $[Fe(CN)_6]^{4-}$, and write the number of unpaired electrons if this complex is known to be low-spin (given Fe²⁺ has a $d^6$ configuration).
5 Marks
Answer

Let oxidation state of Fe be $x$. Each $CN^-$ contributes $-1$ (6 total = $-6$). Overall charge is $-4$:

$x + 6(-1) = -4 \Rightarrow x - 6 = -4 \Rightarrow x = +2$

Oxidation state of Fe = +2 (i.e., $Fe^{2+}$, $d^6$ configuration)

Since $CN^-$ is a strong field ligand, this is a low-spin complex — all 6 electrons pair up within the lower-energy $t_{2g}$ orbitals (which can hold up to 6 electrons, 2 per orbital × 3 orbitals): 0 unpaired electrons (diamagnetic).

Q3. Write the formula for the coordination compound "hexaamminecobalt(III) chloride," and determine its coordination number and the total charge on the complex ion.
5 Marks
Answer

"Hexaammine" indicates 6 ammonia ($NH_3$) ligands; "cobalt(III)" indicates $Co^{3+}$; "chloride" indicates $Cl^-$ counter-ions balancing the charge.

Since $NH_3$ is neutral and Co is $+3$, the complex ion charge is $+3$: $[Co(NH_3)_6]^{3+}$ — requiring 3 chloride counter-ions to balance this charge.

Formula: $[Co(NH_3)_6]Cl_3$

Coordination number = 6 (six ammine ligands directly bonded to Co); charge on complex ion = +3

Q4. Write the IUPAC name for the coordination compound $K_4[Fe(CN)_6]$ (potassium ferrocyanide).
5 Marks
Answer

Let oxidation state of Fe be $x$; each $CN^-$ contributes $-1$ (6 total = $-6$); 4 potassium ions each contribute $+1$ outside the complex ion.

Overall neutral compound: $4(+1) + x + 6(-1) = 0 \Rightarrow 4 + x - 6 = 0 \Rightarrow x = +2$

Since the complex ion is anionic (overall negative), the metal name uses the Latin-derived suffix "-ate": IUPAC name: potassium hexacyanidoferrate(II)

Q5. Determine the number of unpaired electrons and the spin-only magnetic moment for the complex $[NiCl_4]^{2-}$ (tetrahedral, $Ni^{2+}$ has a $d^8$ configuration, and tetrahedral complexes are typically high-spin).
5 Marks
Answer

$Ni^{2+}$ has configuration $3d^8$. In a tetrahedral field (typically high-spin, with small crystal field splitting), electrons fill according to Hund's rule with minimal pairing.

Distributing 8 electrons among 5 d-orbitals (following maximum unpairing consistent with 8 electrons — each orbital gets at least 1, with 3 orbitals then getting a second electron): this leaves 2 unpaired electrons.

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx \mathbf{2.83\text{ BM}}$

Q6. Explain, with a structural description, the type of isomerism shown by $[Pt(NH_3)_2Cl_2]$ (square planar), and name the two isomers.
5 Marks
Answer

$[Pt(NH_3)_2Cl_2]$ is a square planar complex with two $NH_3$ and two $Cl$ ligands — this composition allows for geometrical (cis-trans) isomerism.

Cis isomer: The two chloride ligands (and correspondingly, the two ammine ligands) are positioned adjacent to each other, at 90° to one another.

Trans isomer: The two chloride ligands are positioned directly opposite each other, at 180° to one another (with the two ammine ligands also opposite each other).

These two distinct spatial arrangements of the same ligands around the same square planar metal centre give rise to two chemically and physically distinguishable geometrical isomers — notably, the cis isomer (cisplatin) has significant anticancer therapeutic activity, while the trans isomer does not.

Q7. Calculate the effective atomic number (EAN) of the central metal in $[Co(NH_3)_6]^{3+}$, given cobalt's atomic number is 27, and comment on its significance.
5 Marks
Answer

EAN $= Z - (\text{oxidation state}) + 2\times(\text{coordination number})$

$EAN = 27 - 3 + 2(6) = 24 + 12$

$EAN = \mathbf{36}$

This EAN of 36 exactly matches the atomic number of krypton (Kr), the nearest noble gas — according to the effective atomic number rule, complexes achieving an EAN equal to that of the nearest noble gas tend to show enhanced stability, consistent with $[Co(NH_3)_6]^{3+}$ being a notably stable, well-known complex.

Q8. Using valence bond theory, determine the hybridisation and geometry of $[Ni(CN)_4]^{2-}$, given that this complex is known to be diamagnetic (all electrons paired), and $Ni^{2+}$ has a $d^8$ configuration.
5 Marks
Answer

Since the complex is diamagnetic (no unpaired electrons), and $CN^-$ is a strong field ligand capable of pairing up the metal's own d-electrons, the 8 d-electrons of $Ni^{2+}$ must fully pair up within just 4 of the 5 d-orbitals, freeing up 1 empty d-orbital for use in hybridisation.

This freed inner d-orbital combines with one s and two p orbitals to form $dsp^2$ hybridisation (using an inner d-orbital), giving rise to a square planar geometry — consistent with the observed diamagnetism and the well-known square planar structure of $[Ni(CN)_4]^{2-}$.

Q9. Write the formula of the complex formed between $Cu^{2+}$ and EDTA (a hexadentate ligand), and explain why this complex is exceptionally stable.
5 Marks
Answer

EDTA (ethylenediaminetetraacetate, typically as the anion $EDTA^{4-}$) binds through all six of its donor atoms (2 nitrogen + 4 oxygen) to a single metal centre.

Formula: $[Cu(EDTA)]^{2-}$ (with $Cu^{2+}$ and $EDTA^{4-}$ combining to give an overall $2-$ charge)

Stability reasoning: Since EDTA forms multiple fused chelate rings simultaneously with the single copper centre (due to its hexadentate nature), this complex benefits substantially from the chelate effect (favourable entropy increase upon displacing multiple separate water/solvent ligands with the single, multi-donor EDTA molecule) — this is precisely why EDTA-metal complexes are exceptionally stable, a property widely exploited in analytical chemistry (e.g., EDTA titrations) and industrial/medical applications (e.g., treating heavy metal poisoning).

Q10. Given the crystal field stabilisation energy (CFSE) formula for an octahedral complex, $CFSE = [-0.4n(t_{2g})+0.6n(e_g)]\Delta_o$, calculate the CFSE for a $d^6$ low-spin octahedral complex (all 6 electrons in $t_{2g}$).
5 Marks
Answer

For a low-spin $d^6$ complex, all 6 electrons occupy the lower-energy $t_{2g}$ set (which can hold a maximum of 6 electrons): $n(t_{2g})=6$, $n(e_g)=0$

$CFSE = [-0.4(6) + 0.6(0)]\Delta_o$

$CFSE = [-2.4 + 0]\Delta_o$

$CFSE = \mathbf{-2.4\,\Delta_o}$

This large, negative CFSE value indicates a substantial stabilisation of the low-spin $d^6$ configuration relative to the hypothetical unsplit (spherical field) case — consistent with the notably high stability observed for low-spin $d^6$ complexes such as $[Co(NH_3)_6]^{3+}$ and $[Fe(CN)_6]^{4-}$.

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