Coordination Compounds Formulas For JEE 2026
Coordination Compounds is an important chapter in JEE Chemistry, especially in the inorganic chemistry section. It explains how metal ions combine with ligands to form coordination complexes. In this topic, students study important concepts such as Werner’s theory, different types of ligands, and IUPAC nomenclature, which help them understand how coordination compounds are formed and how they are named correctly.
The chapter also includes topics like isomerism, bonding theories such as Valence Bond Theory (VBT) and Crystal Field Theory (CFT), crystal field splitting, and the spectrochemical series. These concepts help students understand the bonding, structure, and properties of coordination compounds, which are frequently asked in exams. For quick revision during preparation, students can also refer to a well-structured JEE Mains Chemistry Formula PDF to review key formulas and important concepts easily.
Introduction to Coordination Compounds
A coordination compound (or complex compound) consists of a central metal ion surrounded by molecules or ions that donate their electron pairs to it. Examples include the deep blue solution when excess ammonia is added to CuSO$$_4$$, or the blood-red color when KSCN is added to FeCl$$_3$$. Coordination compounds are found everywhere — hemoglobin (Fe complex), chlorophyll (Mg complex), vitamin B$$_{12}$$ (Co complex), and many industrial catalysts.
Werner's Theory
Werner's Postulates
- Metals show two types of valency:
- Primary valency (ionizable) — satisfied by negative ions, corresponds to oxidation state
- Secondary valency (non-ionizable) — satisfied by ligands, corresponds to coordination number
- The secondary valency is fixed for a given metal ion and is directional (determines the geometry)
- Ligands satisfying secondary valency are arranged in a definite geometry around the metal
Worked Example
Explain CoCl$$_3 \cdot $$6NH$$_3$$ using Werner's theory.
Formula: $$[\text{Co(NH}_3\text{)}_6]\text{Cl}_3$$
- Co$$^{3+}$$ is the central metal ion (oxidation state = +3)
- Primary valency = 3, satisfied by 3 Cl$$^-$$ ions (ionizable, outside the bracket)
- Secondary valency = 6, satisfied by 6 NH$$_3$$ molecules (non-ionizable, inside the bracket)
- Geometry: octahedral
This compound gives 3 Cl$$^-$$ ions in solution (confirmed by AgNO$$_3$$ test giving 3 moles of AgCl).
Key Terminology
Central Metal Atom/Ion
The metal atom or ion at the center of the coordination entity that accepts electron pairs from ligands. Example: Fe$$^{3+}$$ in $$[\text{Fe(CN)}_6]^{3-}$$.
Ligand
A molecule or ion that donates at least one electron pair to the central metal atom/ion. Ligands are Lewis bases. Example: NH$$_3$$, Cl$$^-$$, H$$_2$$O, CN$$^-$$.
Coordination Number (CN)
The total number of donor atoms directly bonded to the central metal ion. For $$[\text{Co(NH}_3\text{)}_6]^{3+}$$, CN = 6.
Coordination Sphere
The central metal ion together with its ligands, written inside square brackets. Example: $$[\text{Cu(NH}_3\text{)}_4]^{2+}$$.
Oxidation State
The charge on the central metal if all ligands are removed along with their electron pairs.
Types of Ligands
Ligands are classified based on the number of donor atoms they use to bind to the metal.
Classification of Ligands by Denticity
| Type | Donor Atoms | Examples | Name |
|---|---|---|---|
| Monodentate | 1 | NH$$_3$$, Cl$$^-$$, H$$_2$$O, CN$$^-$$, CO | One "tooth" |
| Bidentate | 2 | en (ethylenediamine), ox$$^{2-}$$ (oxalate C$$_2$$O$$_4^{2-}$$) | Two "teeth" |
| Polydentate | 3 or more | EDTA$$^{4-}$$ (hexadentate, 6) | Many "teeth" |
Ambidentate Ligand
A ligand that can coordinate through two different atoms (but only one at a time). Examples: CN$$^-$$ (through C or N), NO$$_2^-$$ (through N or O), SCN$$^-$$ (through S or N).
Chelate
A complex in which a bidentate or polydentate ligand forms a ring with the metal ion. Chelates are extra stable. Example: $$[\text{Cu(en)}_2]^{2+}$$ (two 5-membered rings).
Tip: EDTA$$^{4-}$$ is the most common hexadentate ligand in JEE. It has 6 donor atoms (4 oxygen + 2 nitrogen) and wraps around the metal like a cage. One EDTA can replace 6 monodentate ligands.
Worked Example
Find the oxidation state and coordination number of Co in $$[\text{CoCl}_2\text{(en)}_2]^+$$.
Let oxidation state of Co $$= x$$.
Charge: $$x + 2(-1) + 2(0) = +1 \Rightarrow x = +3$$
Coordination number: Cl$$^-$$ is monodentate (1 donor atom each $$\times$$ 2 = 2), en is bidentate (2 donor atoms each $$\times$$ 2 = 4).
CN = 2 + 4 = 6. Oxidation state = +3.
IUPAC Nomenclature
IUPAC Naming Rules
- Cation first, then anion (like ionic compounds)
- Within the coordination sphere:
- Name ligands in alphabetical order (ignoring prefixes like di-, tri-)
- Anionic ligands end in -o (chlorido, cyanido, oxalato)
- Neutral ligands keep their names, except: H$$_2$$O = aqua, NH$$_3$$ = ammine, CO = carbonyl, NO = nitrosyl
- Use di-, tri-, tetra- for simple ligands; bis-, tris-, tetrakis- for complex ligand names
- Metal naming:
- If complex is cation/neutral → use normal metal name (cobalt, iron, copper)
- If complex is anion → metal name ends in -ate (cobaltate, ferrate, cuprate)
- Oxidation state of metal in Roman numerals in parentheses after metal name
Common Ligand Names for IUPAC
| Ligand | IUPAC Name | Type |
|---|---|---|
| Cl$$^-$$ | chlorido | Anionic |
| Br$$^-$$ | bromido | Anionic |
| OH$$^-$$ | hydroxido | Anionic |
| CN$$^-$$ | cyanido | Anionic |
| NO$$_2^-$$ | nitrito-N / nitrito-O | Anionic (ambidentate) |
| C$$_2$$O$$_4^{2-}$$ | oxalato | Anionic |
| NH$$_3$$ | ammine | Neutral |
| H$$_2$$O | aqua | Neutral |
| CO | carbonyl | Neutral |
| en | ethylenediamine | Neutral |
Worked Example
Name the following:
(a) $$[\text{Co(NH}_3\text{)}_5\text{Cl}]\text{Cl}_2$$
Cation: $$[\text{Co(NH}_3\text{)}_5\text{Cl}]^{2+}$$. Ligands in order: ammine (5) and chlorido (1). Metal: cobalt. Oxidation state: $$x + 5(0) + (-1) = +2 \Rightarrow x = +3$$.
Name: Pentaamminechloridocobalt(III) chloride
(b) $$\text{K}_3[\text{Fe(CN)}_6]$$
Anion: $$[\text{Fe(CN)}_6]^{3-}$$. Ligands: cyanido (6). Metal: ferrate (anionic complex). Oxidation state: $$x + 6(-1) = -3 \Rightarrow x = +3$$.
Name: Potassium hexacyanidoferrate(III)
Tip: When alphabetizing ligands, ignore the multiplying prefix. "Ammine" comes before "chlorido" alphabetically. Also, remember: there are no spaces between ligand names, but there is a space before the metal name only if ligands and metal name are separate words.
Isomerism in Coordination Compounds
Structural Isomerism
Types of Structural Isomerism
1. Ionization Isomerism: Isomers that produce different ions in solution.
Example: $$[\text{Co(NH}_3\text{)}_5\text{Br}]\text{SO}_4$$ vs. $$[\text{Co(NH}_3\text{)}_5\text{SO}_4]\text{Br}$$
First gives SO$$_4^{2-}$$ in solution; second gives Br$$^-$$ in solution.
2. Hydrate (Solvate) Isomerism: Difference in whether water is inside or outside the coordination sphere.
Example: $$[\text{Cr(H}_2\text{O)}_6]\text{Cl}_3$$ (violet) vs. $$[\text{Cr(H}_2\text{O)}_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}$$ (blue-green) vs. $$[\text{Cr(H}_2\text{O)}_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}$$ (green)
All have the formula CrCl$$_3 \cdot $$6H$$_2$$O but differ in the number of Cl$$^-$$ ions released.
3. Linkage Isomerism: Occurs with ambidentate ligands that can bind through different atoms.
Example: $$[\text{Co(NH}_3\text{)}_5\text{ONO}]^{2+}$$ (O-bonded, nitrito) vs. $$[\text{Co(NH}_3\text{)}_5\text{NO}_2]^{2+}$$ (N-bonded, nitro)
Stereoisomerism
Geometrical (cis-trans) Isomerism
Occurs when ligands can be arranged differently in space around the metal.
In square planar complexes (CN = 4): $$[\text{MA}_2\text{B}_2]$$ type
- cis: same ligands on the same side (adjacent, $$90°$$)
- trans: same ligands on opposite sides ($$180°$$)
Example: $$[\text{Pt(NH}_3\text{)}_2\text{Cl}_2]$$ — cis-platin (anti-cancer drug) and trans-platin
In octahedral complexes (CN = 6): $$[\text{MA}_4\text{B}_2]$$ type
- cis: two B ligands on the same side ($$90°$$ apart)
- trans: two B ligands on opposite sides ($$180°$$ apart)
$$[\text{MA}_3\text{B}_3]$$ type shows fac (facial — 3 same ligands on one triangular face) and mer (meridional — 3 same ligands in one plane).
Optical Isomerism
Optical isomers are non-superimposable mirror images of each other (like left and right hands). They are called enantiomers.
- One rotates plane-polarized light to the right (dextro, d or +)
- The other rotates it to the left (laevo, l or $$-$$)
- Requirement: the complex must lack a plane of symmetry
- Common in octahedral complexes with bidentate ligands: $$[\text{M(en)}_3]^{n+}$$, cis-$$[\text{M(en)}_2\text{Cl}_2]^{n+}$$
- The trans isomer of $$[\text{M(en)}_2\text{Cl}_2]^{n+}$$ does not show optical isomerism (has a plane of symmetry)
Tip: For JEE: cis-$$[\text{Pt(NH}_3\text{)}_2\text{Cl}_2]$$ (cisplatin) is a famous anti-cancer drug — this is frequently asked. Also, $$[\text{Co(en)}_3]^{3+}$$ is a classic example of an optically active complex.
Valence Bond Theory (VBT)
VBT explains bonding in complexes using the concept of hybridization. The central metal ion provides empty orbitals, and ligands donate electron pairs into these orbitals. The type of hybridization determines the geometry.
VBT — Hybridization and Geometry
| CN | Hybridization | Geometry | Example |
|---|---|---|---|
| 4 | $$sp^3$$ | Tetrahedral | $$[\text{NiCl}_4]^{2-}$$ |
| 4 | $$dsp^2$$ | Square planar | $$[\text{Ni(CN)}_4]^{2-}$$ |
| 6 | $$d^2sp^3$$ | Octahedral (inner) | $$[\text{Co(NH}_3\text{)}_6]^{3+}$$ |
| 6 | $$sp^3d^2$$ | Octahedral (outer) | $$[\text{CoF}_6]^{3-}$$ |
Inner Orbital Complex
Uses $$(n-1)d$$ orbitals for hybridization (e.g., $$d^2sp^3$$). Formed with strong field ligands. Usually low spin (fewer unpaired electrons) and diamagnetic or weakly paramagnetic.
Outer Orbital Complex
Uses $$nd$$ orbitals for hybridization (e.g., $$sp^3d^2$$). Formed with weak field ligands. Usually high spin (more unpaired electrons) and paramagnetic.
Worked Example
Predict the hybridization and magnetic nature of $$[\text{Fe(CN)}_6]^{4-}$$.
Fe$$^{2+}$$: $$[\text{Ar}] \; 3d^6$$ (6 d-electrons)
CN$$^-$$ is a strong field ligand $$\Rightarrow$$ it forces pairing of d-electrons.
After pairing: 3d has all 6 electrons in 3 orbitals (paired), leaving 2 empty 3d orbitals.
Hybridization: $$d^2sp^3$$ (inner orbital) $$\Rightarrow$$ octahedral
Unpaired electrons: 0 $$\Rightarrow$$ diamagnetic
Worked Example
Predict the hybridization and magnetic nature of $$[\text{FeF}_6]^{3-}$$.
Fe$$^{3+}$$: $$[\text{Ar}] \; 3d^5$$ (5 d-electrons, all unpaired)
F$$^-$$ is a weak field ligand $$\Rightarrow$$ no pairing of d-electrons.
All five 3d orbitals are singly occupied — none available for bonding.
Hybridization: $$sp^3d^2$$ (outer orbital, using 4d orbitals) $$\Rightarrow$$ octahedral
Unpaired electrons: 5 $$\Rightarrow$$ paramagnetic ($$\mu = \sqrt{35} = 5.92$$ BM)
Crystal Field Theory (CFT)
CFT treats the metal–ligand interaction as purely electrostatic: the negatively charged ligands create an electric field that affects the energies of the d-orbitals. In a free metal ion, all five d-orbitals have the same energy (degenerate). When ligands approach, they repel the d-electrons unequally, causing the d-orbitals to split into groups of different energy.
Crystal Field Splitting in Octahedral Complexes
In an octahedral complex (6 ligands along x, y, z axes):
- d-orbitals split into two groups:
- $$t_{2g}$$ set (lower energy): $$d_{xy}$$, $$d_{xz}$$, $$d_{yz}$$ — point between the axes
- $$e_g$$ set (higher energy): $$d_{x^2-y^2}$$, $$d_{z^2}$$ — point along the axes (toward ligands)
- The energy gap between $$t_{2g}$$ and $$e_g$$ is called $$\Delta_o$$ (octahedral splitting energy)
- $$t_{2g}$$ is lowered by $$-0.4\Delta_o$$ and $$e_g$$ is raised by $$+0.6\Delta_o$$ (relative to average)
Crystal Field Splitting in Tetrahedral Complexes
In a tetrahedral complex (4 ligands between axes):
- The splitting is inverted compared to octahedral:
- $$e$$ set (lower energy): $$d_{x^2-y^2}$$, $$d_{z^2}$$
- $$t_2$$ set (higher energy): $$d_{xy}$$, $$d_{xz}$$, $$d_{yz}$$
- The splitting energy $$\Delta_t$$ is much smaller: $$\Delta_t = \frac{4}{9}\Delta_o$$
- Because $$\Delta_t$$ is small, tetrahedral complexes are almost always high spin
Crystal Field Stabilization Energy (CFSE)
CFSE Calculation (Octahedral)
$$$\text{CFSE} = (-0.4 \times n_{t_{2g}} + 0.6 \times n_{e_g}) \Delta_o + P$$$
where $$n_{t_{2g}}$$ = number of electrons in $$t_{2g}$$, $$n_{e_g}$$ = number in $$e_g$$, and $$P$$ = pairing energy (added only if extra pairing occurs due to strong field).
Worked Example
Calculate the CFSE for $$d^6$$ in a strong octahedral field.
Strong field $$\Rightarrow$$ electrons fill $$t_{2g}$$ first (with pairing) before entering $$e_g$$.
$$d^6$$ strong field: $$t_{2g}^6 \; e_g^0$$ (all 6 in $$t_{2g}$$, fully paired)
CFSE $$= (-0.4 \times 6 + 0.6 \times 0)\Delta_o = -2.4\Delta_o$$ (+ pairing energy for 3 extra pairs)
This configuration has 0 unpaired electrons $$\Rightarrow$$ diamagnetic (e.g., $$[\text{Fe(CN)}_6]^{4-}$$).
Compare with weak field: $$t_{2g}^4 \; e_g^2$$ (high spin)
CFSE $$= (-0.4 \times 4 + 0.6 \times 2)\Delta_o = -0.4\Delta_o$$
This has 4 unpaired electrons $$\Rightarrow$$ paramagnetic (e.g., $$[\text{Fe(H}_2\text{O)}_6]^{2+}$$).
Spectrochemical Series
Spectrochemical Series
Weak field (small $$\Delta$$) $$\longrightarrow$$ Strong field (large $$\Delta$$):
$$\text{I}^- < \text{Br}^- < \text{S}^{2-} < \text{Cl}^- < \text{N}_3^- < \text{F}^- < \text{OH}^- < \text{ox}^{2-} < \text{H}_2\text{O} < \text{NCS}^- < \text{edta}^{4-} < \text{NH}_3 < \text{en} < \text{NO}_2^- < \text{CN}^- < \text{CO}$$
Key points:
- Halides are generally weak field (I$$^-$$ weakest, F$$^-$$ strongest among them)
- H$$_2$$O is moderate; NH$$_3$$ is moderately strong
- CN$$^-$$ and CO are very strong field ligands
Important
Strong field ligands cause large $$\Delta$$ $$\Rightarrow$$ electrons pair up rather than enter higher orbitals $$\Rightarrow$$ low spin complexes. Weak field ligands cause small $$\Delta$$ $$\Rightarrow$$ electrons prefer to remain unpaired $$\Rightarrow$$ high spin complexes.
Color of Coordination Compounds
The colors of transition metal complexes arise from d–d transitions. When white light falls on a complex, electrons absorb light of a specific wavelength (energy = $$\Delta$$) and jump from the lower d-orbital set to the higher one. The transmitted/reflected color is the complementary color of the absorbed light.
Color and d–d Transitions
- $$[\text{Ti(H}_2\text{O)}_6]^{3+}$$ ($$d^1$$): absorbs green/yellow light $$\Rightarrow$$ appears purple
- $$[\text{Cu(H}_2\text{O)}_6]^{2+}$$ ($$d^9$$): absorbs red light $$\Rightarrow$$ appears blue
- $$[\text{Ni(H}_2\text{O)}_6]^{2+}$$ ($$d^8$$): absorbs red light $$\Rightarrow$$ appears green
- Complexes with $$d^0$$ or $$d^{10}$$ are colorless (no d–d transition possible)
Magnetic Properties
Predicting Magnetic Behavior
- Determine the $$d^n$$ configuration of the metal ion
- Identify the ligand as strong field or weak field (spectrochemical series)
- Fill the d-orbitals accordingly:
- Strong field: fill $$t_{2g}$$ completely (with pairing) before $$e_g$$
- Weak field: fill all orbitals singly first, then pair (Hund's rule)
- Count unpaired electrons and use $$\mu = \sqrt{n(n+2)}$$ BM
Tip: The high-spin vs low-spin difference only matters for $$d^4$$, $$d^5$$, $$d^6$$, and $$d^7$$ octahedral complexes. For $$d^1$$–$$d^3$$ and $$d^8$$–$$d^{10}$$, the filling is the same regardless of field strength.
Applications of Coordination Compounds
Important Applications
- Extraction of metals: Gold is extracted using CN$$^-$$ (forms $$[\text{Au(CN)}_2]^-$$) — cyanide leaching
- Medicine: cis-platin ($$[\text{Pt(NH}_3\text{)}_2\text{Cl}_2]$$) is an anti-cancer drug; EDTA is used to treat heavy metal poisoning (chelation therapy)
- Analytical chemistry: Ni$$^{2+}$$ detected by DMG reagent; Fe$$^{3+}$$ detected by KSCN (blood-red $$[\text{Fe(SCN)}]^{2+}$$)
- Photography: AgBr dissolves in Na$$_2$$S$$_2$$O$$_3$$ (hypo) forming $$[\text{Ag(S}_2\text{O}_3\text{)}_2]^{3-}$$
- Electroplating: Gold and silver plating use cyanide complexes
- Biology: Hemoglobin (Fe), chlorophyll (Mg), vitamin B$$_{12}$$ (Co)
Tip: For JEE Mains, the most frequently tested topics are: (1) IUPAC naming, (2) finding oxidation state and CN, (3) identifying isomer types, (4) predicting hybridization (VBT) and magnetic nature, (5) spectrochemical series and high-spin vs low-spin, (6) CFSE calculation for octahedral complexes.
Coordination Compounds Formulas For JEE 2026: Conclusion
Coordination Compounds play a crucial role in understanding many important concepts in inorganic chemistry. Topics such as Werner’s theory, ligand classification, coordination number, and IUPAC nomenclature help students understand how metal ions interact with ligands to form stable complexes. In addition, concepts like isomerism, bonding theories, and hybridization provide deeper insight into the structure and geometry of these compounds.
For competitive exams like JEE, mastering ideas such as crystal field splitting, spectrochemical series, and magnetic properties is extremely important. Regular practice and quick revision of key formulas and concepts can help students solve questions more efficiently and improve their overall performance in the chemistry section.
