Transition metals form coordination compounds far more readily than s-block or p-block elements. This tendency — complex formation — is the reason you see deeply coloured solutions, catalytic cycles, and biologically critical metalloenzymes throughout d-block chemistry.
Why d-block? Three properties converge: (1) small ionic radii with high charge density, (2) vacant or partially filled d-orbitals available to accept lone pairs from ligands, and (3) variable oxidation states that let the same metal accommodate different ligand sets. NCERT Class 12 Chemistry Chapter 4 (page 104) explicitly notes that the tendency to form complexes is a general characteristic of transition elements, arising from their small size, high ionic charge, and availability of d-orbitals for bond formation.
Charge density argument. A high charge-to-radius ratio polarises incoming ligands strongly, stabilising the metal–ligand bond. Compare: Na⁺ (large, +1) forms few stable complexes; Fe³⁺ (small, +3) forms hundreds.
d-orbital availability. Ligand lone pairs donate into empty or half-filled d-orbitals (σ-donation) and in some cases accept electron density back (π-back-bonding in carbonyls). Main-group ions lack accessible, energetically suitable empty orbitals of comparable energy, so their complex formation is limited.
Variable oxidation states matter. Because different oxidation states change the number of available d-orbitals and the effective nuclear charge, the same metal can stabilise both hard (F⁻, OH⁻) and soft (CN⁻, CO) ligands depending on its oxidation state.
Watch-out for NEET: Questions often ask why a specific d-block element forms more complexes than its s/p-block neighbour, or why higher oxidation states favour complex formation. The answer always routes back to charge density + orbital availability — not merely "has d-electrons."
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following is NOT a reason for the high tendency of transition metals to form coordination compounds?
Answer: D. In the ionic state, transition metals lose their outermost s-electrons first, so the s-orbital is empty — not filled. The three valid reasons are small size/high charge, vacant d-orbitals, and variable oxidation states (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: A describes the charge-density factor, which is a genuine reason for complex formation tendency — high charge/small radius polarises ligands effectively.
Why B is wrong: B describes d-orbital availability for lone pair acceptance, which is a core reason transition metals form complexes readily.
Why C is wrong: C describes how variable oxidation states allow different ligand environments, which is a genuine factor listed by NCERT for complex formation tendency.
Among the following ions, which would have the greatest tendency to form stable coordination complexes?
Answer: C. Fe³⁺ has a small ionic radius, high charge (+3), and vacant d-orbitals — all three factors maximise complex formation tendency. The s-block ions (Na⁺, K⁺, Ca²⁺) lack accessible d-orbitals and have lower charge density (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: Na⁺ has low charge (+1), large radius, and no available d-orbitals of suitable energy — very poor complex-forming ability.
Why B is wrong: Ca²⁺ has +2 charge and moderate radius but no vacant d-orbitals accessible at bonding energies — forms far fewer complexes than d-block ions.
Why D is wrong: K⁺ has the lowest charge density among these options (+1, very large radius) and no d-orbital availability — essentially no stable complex formation.
The complex formation tendency of transition metals is attributed to their high charge density. Which factor directly increases the charge density of a cation?
Answer: A. Charge density = charge/volume. A higher oxidation state increases the numerator (charge) while a smaller ionic radius decreases the denominator (volume), both raising charge density and hence complex-forming ability.
Why B is wrong: Paired d-electrons relate to magnetic properties, not directly to charge density. Charge density is determined by ionic charge and ionic radius, not electron pairing.
Why C is wrong: Increasing atomic number alone doesn't guarantee higher charge density — it depends on which electrons are removed and the resulting ionic radius.
Why D is wrong: Lower oxidation state means less charge, and larger radius means more volume — both reduce charge density, weakening complex formation tendency.
Why do main-group elements like Na and Mg form far fewer complexes than transition elements like Fe and Co?
Answer: B. The key difference is orbital availability: main-group cations (Na⁺, Mg²⁺) have no vacant d-orbitals at accessible energies to accept ligand lone pairs, whereas transition metal ions do (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: Main-group elements actually tend to have lower ionisation energies than many transition metals — and ionisation energy is not the direct reason for complex formation tendency.
Why C is wrong: Transition metals in their common ionic states have partially filled (not completely filled) d-orbitals — the vacant portions accept ligand electrons rather than repelling them.
Why D is wrong: While main-group elements show less oxidation state variability, some (like Sn, Pb) do show multiple states. The primary limitation is orbital availability, not merely oxidation state restriction.
Consider two ions: Cr³⁺ (ionic radius 62 pm) and Cr²⁺ (ionic radius 73 pm). Which forms more stable complexes with a given set of ligands, and why?
Answer: C. Cr³⁺ has charge +3 with radius 62 pm, giving a much higher charge density than Cr²⁺ (+2, 73 pm). Higher charge density polarises ligands more strongly, forming more stable metal–ligand bonds.
Why A is wrong: While Cr²⁺ does have one more d-electron, back-bonding is relevant only with specific π-acceptor ligands (CO, CN⁻). For a general ligand set, charge density is the dominant factor favouring Cr³⁺.
Why B is wrong: Although the number of d-orbitals is the same, the charge density differs dramatically — this is not an equal contest. Stability tracks charge/radius ratio, not orbital count alone.
Why D is wrong: Lower inter-electronic repulsion in Cr²⁺ doesn't compensate for its reduced charge density. Complex stability with general ligands correlates primarily with the metal's polarising power (charge density).
Sc³⁺ (d⁰ configuration, ionic radius 75 pm) has all five d-orbitals vacant — more empty d-orbitals than Fe³⁺ (d⁵ configuration, ionic radius 65 pm, effectively zero fully vacant d-orbitals in its high-spin complexes). Yet Fe³⁺ forms a far wider variety of stable complexes than Sc³⁺. Which statement best explains this?
Answer: A. Complex-forming ability is not set by vacant-orbital count alone — it is charge density (charge/size) that governs how strongly the metal ion polarises and holds ligands, and orbital availability merely supplies the acceptor site. Sc³⁺ (+3, 75 pm) and Fe³⁺ (+3, 65 pm) carry the same charge, but Sc³⁺'s larger radius gives it lower charge density, so it polarises ligands less strongly even though it has more empty d-orbitals to receive them. Both factors — charge density AND orbital availability — must be considered together; having only the orbital advantage does not outweigh a weaker electrostatic pull.
Why B is wrong: B is wrong because coordinate bonds are donor–acceptor bonds — the ligand supplies the lone pair and the metal ion only needs a vacant orbital to accept it into. Sc³⁺'s d⁰ configuration in fact gives it the maximum possible number of vacant d-orbitals, and it does form some complexes on that basis; the metal contributing its own electrons is not required.
Why C is wrong: C is wrong because natural abundance has no bearing on the electrostatic and orbital factors that govern complex stability; the actual reason is the charge-density difference between the two ions, not how common the element is.
Why D is wrong: D is wrong because orbital availability is one of the two documented factors behind complex-formation tendency, alongside charge density; charge alone is not sufficient either, since Sc³⁺ and Fe³⁺ carry the identical +3 charge yet differ sharply in complex-forming ability because of their different radii.
Zn²⁺ has a completely filled d-orbital (d¹⁰ configuration). Despite this, it forms complexes like [Zn(NH₃)₄]²⁺. Which factor primarily enables Zn²⁺ to form complexes?
Answer: A. In Zn²⁺ (d¹⁰), d-orbitals are fully occupied and cannot accept further electron pairs. However, the vacant 4s and 4p orbitals (and in tetrahedral geometry, sp³ hybridisation) provide the acceptor orbitals needed for ligand lone-pair donation.
Why B is wrong: Zn²⁺ has zero unpaired d-electrons (d¹⁰ fully paired). σ-bonding in its complexes uses vacant s and p orbitals, not d-orbitals.
Why C is wrong: While Zn²⁺ has reasonable charge density, charge density alone doesn't explain how bonding occurs when d-orbitals are full. The key is the availability of alternative acceptor orbitals (4s, 4p).
Why D is wrong: Zinc exhibits essentially only +2 oxidation state in its chemistry — it does not show +1 or +3 under normal conditions. Variable oxidation state is not a factor here.
Among Ti²⁺ (d²), V²⁺ (d³), Fe²⁺ (d⁶), and Cu²⁺ (d⁹), which ion has the maximum number of vacant d-orbitals available for accepting ligand electrons?
Answer: D. Ti²⁺ has d² configuration: 2 electrons occupy 2 of the 5 d-orbitals (one each, by Hund's rule), leaving 3 d-orbitals completely vacant. V²⁺ (d³) has 2 vacant, Fe²⁺ (d⁶) has 0 fully vacant in high-spin, and Cu²⁺ (d⁹) has 0 fully vacant d-orbitals.
Why A is wrong: Cu²⁺ (d⁹) has 9 electrons distributed across 5 d-orbitals — all 5 orbitals contain at least one electron, so zero are fully vacant for accepting ligand lone pairs.
Why B is wrong: V²⁺ (d³) has 3 electrons in 3 separate orbitals, leaving only 2 d-orbitals fully vacant — fewer than Ti²⁺'s 3 vacant orbitals.
Why C is wrong: Fe²⁺ (d⁶) in high-spin octahedral has electrons in all 5 d-orbitals (one orbital doubly occupied), leaving zero fully vacant d-orbitals.
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Given
Consider Co³⁺ and Co²⁺. Co³⁺ ionic radius = 55 pm, Co²⁺ ionic radius = 65 pm. Both have partially filled d-orbitals.
Required
Determine which ion has the greater tendency to form stable complexes, and by what factor the charge densities differ (approximating ions as spheres).
Concept
Complex formation tendency correlates with charge density (charge/volume). Higher charge density → stronger polarisation of ligand electron clouds → more stable metal–ligand bonds.
Formula
Charge density ∝ Z/r³ (for spherical ions, where Z = ionic charge, r = ionic radius).
Ratio = (Z₁/r₁³) ÷ (Z₂/r₂³)
Substitution
For Co³⁺: Z = 3, r = 55 pm
For Co²⁺: Z = 2, r = 65 pm
Ratio (Co³⁺ / Co²⁺) = (3/55³) ÷ (2/65³)
Calculation
55³ = 166,375
65³ = 274,625
Co³⁺ charge density ∝ 3/166,375 = 1.803 × 10⁻⁵
Co²⁺ charge density ∝ 2/274,625 = 7.284 × 10⁻⁶
Ratio = 1.803 × 10⁻⁵ / 7.284 × 10⁻⁶ = 2.47
Note: The ionic radii (55, 65 pm) and charges (3, 2) are exact given values in this problem — they do not limit significant figures. The ratio is reported to 3 significant figures.
Final answer
Co³⁺ has approximately 2.5 times the charge density of Co²⁺, and therefore a significantly greater tendency to form stable coordination complexes.
Common trap
Students sometimes compare only the charges (3 vs 2 → 1.5× difference) and forget that the radius difference contributes cubically. The actual advantage is ~2.5×, not 1.5×.
Similar NEET-style question
"Arrange Fe²⁺, Fe³⁺, and Na⁺ in decreasing order of tendency to form complexes. Justify using charge density and orbital availability."
Small size + high charge density + empty d-orbitals → strong coordination tendency. Form coordinate bonds with ligands via lone pairs. Ligand types: monodentate, bidentate, polydentate.
-- NCERT Class 12 Chemistry, Ch. 4, p. 104More in The d and f Block Elements: 3 exam traps and mistakes · 3 formulas · 3 question patterns from its other lessons.
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