Difficulty: Intermediate to Advanced | Prerequisites: Atomic structure (orbitals, electron configurations), basics of bonding and hybridisation.
This topic combines two CBSE chapters: the d- and f-block elements (Ch. 8) and coordination compounds (Ch. 9). They appear together because coordination chemistry is built on the electronic behaviour of transition metals. Together these chapters carry 8 to 10 marks on a typical paper. Expect IUPAC naming, isomerism identification, inner/outer orbital classification, and reasoning about why certain ions are strong oxidisers or reducers.
TL;DR
Transition metals form coloured, paramagnetic complexes because of partially filled d-orbitals. IUPAC naming follows a strict ligands-first, alphabetical-order system. Whether a complex is inner or outer orbital depends on which d-orbitals (3d or 4d) the metal uses for bonding. Lanthanide contraction causes the 4d and 5d elements to have nearly identical ionic radii.
Coordination entity
A central metal ion bonded to a fixed number of ligands in a definite spatial arrangement. Written in square brackets.
Ligand
A molecule or ion that donates a lone pair to the central metal. Monodentate ligands (Cl-, NH3, H2O) donate through one atom; bidentate ligands (ethylenediamine, "en") donate through two.
Linkage isomerism
Occurs when an ambidentate ligand (one that can bind through different atoms) coordinates to the metal in two ways. Classic example: NO2- can bind through N (nitro) or O (nitrito).
Inner orbital complex
A complex that uses the metal's inner d-orbitals (3d) for hybridisation, typically d2sp3. Produces low-spin, diamagnetic or weakly paramagnetic complexes.
Outer orbital complex
A complex that uses the metal's outer d-orbitals (4d) for hybridisation, typically sp3d2. Produces high-spin, paramagnetic complexes.
Diamagnetic
A species with all electrons paired. It is weakly repelled by a magnetic field.
Lanthanide contraction
The steady decrease in ionic radius across the lanthanide series (La3+ to Lu3+), caused by poor shielding of the nuclear charge by 4f electrons.
Electrode potential
A measure of a metal's tendency to lose or gain electrons. Metals with positive electrode potentials (like Cu) do not readily lose electrons to displace hydrogen from dilute acids.
Copper (Cu, Z=29): [Ar] 3d10 4s1, not [Ar] 3d9 4s2. The fully filled 3d subshell provides extra stability, so one 4s electron is promoted.
Chromium (Cr, Z=24): [Ar] 3d5 4s1, not [Ar] 3d4 4s2. The half-filled 3d subshell is similarly stabilised.
These exceptions appear as standalone MCQs: "The element having [Ar] 3d10 4s1 is ____."
Cr2+ has configuration [Ar] 3d4. It readily loses one electron to reach [Ar] 3d3 (Cr3+), which is half-filled and stable. This makes Cr2+ a strong reducing agent.
Mn3+ also has [Ar] 3d4. It readily gains one electron to reach [Ar] 3d5 (Mn2+), which is half-filled and stable. This makes Mn3+ a strong oxidising agent.
Both are driven by the stability of the d5 configuration.
Cu has a positive electrode potential (+0.34 V). Metals with positive electrode potentials cannot displace hydrogen from dilute mineral acids because they are less willing to oxidise than hydrogen.
The assertion "Cu cannot liberate H2" is true. The reason "Cu has positive electrode potential" is also true. However, in the CBSE answer key for this paper, the stated answer is (4), treating the assertion as false. This is a debated question; know both arguments.
Decreasing ionic radii across La3+ to Lu3+ due to increasing effective nuclear charge (poor 4f shielding).
Similarity between 4d and 5d elements (e.g. Zr and Hf have nearly identical radii), making their chemical separation difficult.
Zinc (Zn) has the lowest enthalpy of atomisation, owing to its completely filled 3d10 4s2 configuration and weak metallic bonding (no unpaired d-electrons to contribute to bonding).
Note: the source paper names copper, but the standard NCERT answer is zinc. Be prepared for either in exam discussions.
FeCr2O4 + Na2CO3 (fused in air) --> Na2CrO4 (yellow, compound A).
Na2CrO4 + acid --> Na2Cr2O7 (orange, compound B). Chromates are yellow in alkaline solution; dichromates are orange in acidic solution.
Na2Cr2O7 + KCl --> K2Cr2O7 (compound C, orange).
Acidified K2Cr2O7 oxidises Na2SO3 to Na2SO4 (compound D).
Name ligands first in alphabetical order. Use prefixes (di-, tri-, bis-, tris-) to indicate multiples.
The metal follows, with its oxidation state in Roman numerals in parentheses.
For anionic complexes, the metal name takes the suffix "-ate" (nickel becomes nickelate, iron becomes ferrate).
Examples from the paper:
[Fe(en)2Cl2]+: bis(ethylenediamine)dichloridoiron(III) ion
[Co(NH3)4(H2O)Br]SO4: tetraaqua bromidocobalt(III) sulfate (more precisely, tetraamminaquabromidocobalt(III) sulfate)
[Ni(CN)4]2-: tetracyanidonickelate(II) ion
[Co(NH3)5(NO2)]2+ and [Co(NH3)5(ONO)]2+ are linkage isomers. NO2- binds through N in the first (nitro) and through O in the second (nitrito).
[Co(NH3)6]3+: Co3+ is d6. With strong-field NH3, electrons pair in 3d, freeing two 3d orbitals for d2sp3 hybridisation. Inner orbital complex, low-spin, diamagnetic.
[Ni(NH3)6]2+: Ni2+ is d8. Even with NH3, eight d-electrons cannot free two 3d orbitals. The complex uses 4d orbitals for sp3d2 hybridisation. Outer orbital complex, high-spin, paramagnetic.
[Ni(CN)4]2- is diamagnetic: Ni2+ (d8) with strong-field CN- forces all electrons to pair, giving a square planar dsp2 geometry with zero unpaired electrons.
[NiCl4]2- is paramagnetic (weak-field Cl-, tetrahedral, 2 unpaired electrons).
[Fe(CN)6]3- is paramagnetic (Fe3+ is d5; even with CN-, one unpaired electron remains).
[CoF6]3- is paramagnetic (weak-field F-, high-spin d6, 4 unpaired electrons).
Students often assume that Cu should react with dilute HCl because it is a "metal." Electrode potential determines this, not whether something is a metal. Cu has a positive E0, so it does not displace H2.
IUPAC naming errors: students list ligands in the order they appear in the formula instead of alphabetical order. Always alphabetise.
Students confuse linkage isomerism with ionisation isomerism. Linkage isomerism is about the same ligand binding through different atoms; ionisation isomerism is about swapping a ligand and a counter-ion.
⚠️ IUPAC naming is a guaranteed 3-mark question. Practise naming at least 10 complexes before the exam.
⚠️ "Which is diamagnetic?" MCQs test whether you can count unpaired electrons for each complex. Know the d-electron count for common metals in their common oxidation states.
⚠️ Inner vs outer orbital complex explanation is a standard 2-mark question. State the hybridisation and which d-orbitals are used.
⚠️ Chromate/dichromate identification chains appear in Section B (2 marks). Memorise: yellow = chromate (alkaline), orange = dichromate (acidic).
The electronic configuration [Ar] 3d10 4s1 corresponds to ______.
True or false: [Ni(CN)4]2- is paramagnetic. ______
Linkage isomers differ in how a ______ ligand binds to the metal.
The charge required to reduce 1 mol of MnO4- to MnO2 is ______ F.
True or false: Lanthanide contraction makes Zr and Hf chemically similar. ______
Q: Write the IUPAC name of [Fe(en)2Cl2]+.
A: Bis(ethylenediamine)dichloridoiron(III) ion.
Q: Explain why [Co(NH3)6]3+ is an inner orbital complex while [Ni(NH3)6]2+ is an outer orbital complex.
A: Co3+ (d6) pairs its electrons in the presence of strong-field NH3, freeing two 3d orbitals for d2sp3 hybridisation (inner orbital). Ni2+ (d8) has too many d-electrons to vacate 3d orbitals even with NH3, so it uses 4d orbitals for sp3d2 hybridisation (outer orbital).
Q: The charge required for the reduction of 1 mol MnO4- to MnO2 is?
A: Mn goes from +7 to +4, a change of 3 electrons per atom. Wait, that is 3F. But the CBSE answer key states 5F. Recheck: MnO4- has Mn in +7; MnO2 has Mn in +4. Change = 7 - 4 = 3 electrons. The paper's answer of 5F appears to be an error in the source; the correct answer by standard chemistry is 3F. Be aware this question has been debated.
Q: Why is Cr2+ a strong reducing agent?
A: Cr2+ has configuration [Ar] 3d4. It readily oxidises to Cr3+ ([Ar] 3d3), which is a stable half-filled t2g configuration. The drive toward this stable state makes Cr2+ a strong reducing agent.
Coordination compounds connect to Electrochemistry (Ch. 4) through electrode potentials and the Nernst equation, both of which use oxidation states determined here. The concept of crystal field splitting (not directly in this paper but in NCERT) also underpins the colour and magnetic behaviour tested in these questions.
Tags: coordination compounds, d-block elements, f-block, transition metals, IUPAC naming, linkage isomerism, inner orbital complex, outer orbital complex, diamagnetic, paramagnetic, lanthanide contraction, chromate, dichromate, electrode potential, copper, Cr2+, Mn3+, CBSE Class 12 Chemistry, Chapter 8, Chapter 9