NEET Chemistry: Coordination Compounds – Complete Revision Guide, Formulas & Solved PYQs

Introduction to Coordination Compounds for NEET UG

Coordination Compounds represent one of the most vital and scoring chapters in the Class 12 inorganic chemistry syllabus for the NEET UG examination. These chemical species consist of a central metal ion or atom bonded to a surrounding group of bound molecules or ions called ligands via coordinate covalent (dative) bonds. Beyond theoretical chemistry, coordination complexes form the structural backbone of essential bio-inorganic molecules necessary for living organisms. For example, hemoglobin (an iron-containing complex) transports oxygen in human blood, chlorophyll (a magnesium complex) drives photosynthesis in plants, and Vitamin B12 (cyanocobalamin, a cobalt complex) is critical for red blood cell synthesis and neurological functioning.

For NEET aspirants, mastering Coordination Compounds is essential not only for inorganic chemistry mastery but also because it bridges concepts from chemical bonding, d-block elements, and biological systems. A clear understanding of ligand field strength, stereochemistry, hybridization, and magnetic behavior will guarantee high accuracy in NEET questions.

NEET Weightage & Expected Questions

In the NEET UG examination, Coordination Compounds carries an exceptionally high weightage. Historical trends from NTA NEET papers indicate that this single chapter consistently yields 3 to 4 questions annually, accounting for 12 to 16 marks in the Chemistry section. Nearly all questions are directly conceptual or numerical applications based strictly on the NCERT textbook syllabus.

• Expected Question Distribution:
  • 1 Question on IUPAC Nomenclature or Identification of Isomerism (Structural / Optical).
  • 1 Question on Bonding Theories (Valence Bond Theory hybridization, geometry, and magnetic moment calculation).
  • 1 Question on Crystal Field Theory (CFT splitting energy $\Delta_o$, $t_{2g}$ and $e_g$ electron configuration, spectrochemical series).
  • 1 Question on Werner's Theory, Synergic Bonding, or Organometallics.

Because these questions are standard and repeatable, achieving 100% accuracy in this chapter provides a massive boost to your All India Rank (AIR).

Core Concepts & Key Mechanisms Explained

1. Werner's Theory of Coordination Compounds

Alfred Werner proposed that transition metals exhibit two types of valencies:

• Primary Valence: Corresponds to the oxidation state of the central metal ion. It is ionizable, satisfied strictly by negative ions, and non-directional.
• Secondary Valence: Corresponds to the coordination number (number of donor atoms attached to the metal). It is non-ionizable, satisfied by neutral molecules or negative ions, and directional in space (dictating the geometry of the complex).

Example: In $\text{[Co(NH}_3\text{)}_6\text{]Cl}_3$, the primary valence is $+3$ (satisfied by $3\text{ Cl}^-$ ions outside the coordination sphere), while the secondary valence is $6$ (satisfied by $6\text{ NH}_3$ ligands inside the coordination sphere).

2. Ligands and Coordination Sphere

Ligands act as Lewis bases by donating electron pairs to the central metal (Lewis acid). Ligands are classified by denticity:

• Unidentate: Donates one electron pair (e.g., $\text{H}_2\text{O}, \text{NH}_3, \text{Cl}^-, \text{CN}^-$).
• Didentate: Donates two electron pairs from two donor atoms (e.g., Ethane-1,2-diamine / $en$, Oxalate ion / $\text{ox}^{2-}$).
• Polydentate: Donates multiple pairs (e.g., $\text{EDTA}^{4-}$, a hexadentate ligand).
• Ambidentate Ligands: Unidentate ligands possessing two different donor atoms that can bond through either atom (e.g., $\text{NO}_2^-$ bonding via $\text{N}$ or $\text{O}$; $\text{SCN}^-$ bonding via $\text{S}$ or $\text{N}$). This gives rise to linkage isomerism.
• Chelating Ligands: Di- or polydentate ligands that form ring structures with the metal ion, enhancing complex stability (Chelate Effect).

3. Isomerism in Coordination Compounds

Isomerism is divided into Structural Isomerism and Stereoisomerism:

A. Structural Isomerism

• Ionization Isomerism: Exchange of ions between coordination sphere and ionization sphere. Example: $\text{[Co(NH}_3\text{)}_5\text{SO}_4\text{]Br}$ (red) and $\text{[Co(NH}_3\text{)}_5\text{Br]SO}_4$ (violet).
• Solvate / Hydrate Isomerism: Difference in the number of solvent molecules inside vs. outside the coordination sphere. Example: $\text{[Cr(H}_2\text{O)}_6\text{]Cl}_3$ (violet) vs. $\text{[Cr(H}_2\text{O)}_5\text{Cl]Cl}_2\cdot\text{H}_2\text{O}$ (green).
• Linkage Isomerism: Occurs with ambidentate ligands (e.g., $\text{[Co(NH}_3\text{)}_5(\text{NO}_2)]\text{Cl}_2$ [nitrito-N] vs. $\text{[Co(NH}_3\text{)}_5(\text{ONO})]\text{Cl}_2$ [nitrito-O]).
• Coordination Isomerism: Interchange of ligands between cationic and anionic coordination entities (e.g., $\text{[Co(NH}_3\text{)}_6][\text{Cr(CN)}_6]$ and $\text{[Cr(NH}_3\text{)}_6][\text{Co(CN)}_6]$).

B. Stereoisomerism

• Geometrical Isomerism: Occurs in square planar $[MA_2B_2]$, $[MA_2BC]$ complexes (Cis/Trans) and octahedral $[MA_4B_2]$ or $[MA_3B_3]$ complexes (Facial/Meridional or fac/mer). Note: Square planar complexes of type $[MABCD]$ display 3 geometrical isomers. Tetrahedral complexes DO NOT show geometrical isomerism due to symmetrical spatial arrangement.
• Optical Isomerism: Occurs in chiral molecules that lack a plane of symmetry. Commonly seen in octahedral complexes with chelating ligands like $\text{[Co(en)}_3]^{3+}$ or $cis\text{-[Co(en)}_2\text{Cl}_2]^+$. Note that $trans\text{-[Co(en)}_2\text{Cl}_2]^+$ is optically inactive due to a plane of symmetry.

4. Valence Bond Theory (VBT)

VBT explains bonding via hybridization of atomic orbitals ($s, p, d$). Key features:

• $sp^3$ Hybridization: Tetrahedral geometry (e.g., $\text{[Ni(CO)}_4]$, $\text{[NiCl}_4]^{2-}$).
• $dsp^2$ Hybridization: Square Planar geometry (e.g., $\text{[Ni(CN)}_4]^{2-}, \text{[Pt(NH}_3\text{)}_4]^{2+}$).
• $d^2sp^3$ Hybridization: Octahedral Inner Orbital Complex (uses inner $(n-1)d$ orbitals, typically low spin, strong field ligands like $\text{CN}^-, \text{CO}$).
• $sp^3d^2$ Hybridization: Octahedral Outer Orbital Complex (uses outer $nd$ orbitals, high spin, weak field ligands like $\text{F}^-, \text{H}_2\text{O}$).

5. Crystal Field Theory (CFT)

CFT treats metal-ligand interactions as purely electrostatic. In an octahedral ligand field, the five degenerate d-orbitals split into two sets due to ligand repulsion:

• $t_{2g}$ set: Lower energy triply degenerate set ($d_{xy}, d_{yz}, d_{zx}$).
• $e_g$ set: Higher energy doubly degenerate set ($d_{z^2}, d_{x^2-y^2}$).
• The energy separation between $t_{2g}$ and $e_g$ is denoted as $\Delta_o$ (Octahedral crystal field splitting energy).

Spectrochemical Series: Orders ligands by field strength based on experimental absorption spectrum data:

$\text{I}^- < \text{Br}^- < \text{SCN}^- < \text{Cl}^- < \text{S}^{2-} < \text{F}^- < \text{OH}^- < \text{C}_2\text{O}_4^{2-} < \text{H}_2\text{O} < \text{NCS}^- < \text{EDTA}^{4-} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO}$

• Strong Field Ligands (SFL): $\Delta_o > P$ (where $P$ is electron pairing energy). Cause electron pairing, resulting in Low Spin Complexes.
• Weak Field Ligands (WFL): $\Delta_o < P$. Do not cause electron pairing; electrons enter $e_g$ before pairing in $t_{2g}$, resulting in High Spin Complexes.
• For tetrahedral complexes, splitting is reversed and magnitude is smaller: $\Delta_t = \frac{4}{9}\Delta_o$. Pairing never occurs in tetrahedral complexes because $\Delta_t$ is always less than $P$.

6. Synergic Bonding in Metal Carbonyls

In metal carbonyls such as $\text{Ni(CO)}_4$ or $\text{Fe(CO)}_5$, bonding consists of two components:

1. A $\sigma$-bond formed by donation of a lone pair from carbon of $\text{CO}$ into an empty $d$-orbital of the metal.
2. A $\pi$-back bond formed by donation of filled $d$-electrons from the metal into the empty antibonding $\pi^*$ orbital of $\text{CO}$.

This cooperative interaction is termed synergic bonding, which significantly strengthens the $\text{M-C}$ bond while weakening and lengthening the $\text{C-O}$ bond.

Important Formulas & Key Terms Table

Solved Step-by-Step Previous Years Questions (PYQs)

PYQ 1: NEET 2022

Question: What is the correct order of increasing field strength for the ligands according to the spectrochemical series?

(A) $\text{SCN}^- < \text{F}^- < \text{C}_2\text{O}_4^{2-} < \text{CN}^-$
(B) $\text{CN}^- < \text{C}_2\text{O}_4^{2-} < \text{F}^- < \text{SCN}^-$
(C) $\text{F}^- < \text{SCN}^- < \text{C}_2\text{O}_4^{2-} < \text{CN}^-$
(D) $\text{C}_2\text{O}_4^{2-} < \text{F}^- < \text{SCN}^- < \text{CN}^-$

Solution & Explanation:
According to NCERT spectrochemical series, halide ions and sulfur donors lie on the weak field side, oxalate lies in the middle, and cyanide ($ \text{CN}^-$) lies on the strong field side.
Correct sequence: $\text{SCN}^- < \text{F}^- < \text{C}_2\text{O}_4^{2-} < \text{CN}^-$.
Correct Option: (A)

PYQ 2: NEET 2020

Question: The calculated spin-only magnetic moment of $\text{[Cr(H}_2\text{O)}_6]^{2+}$ is:

(A) $3.87\text{ BM}$
(B) $4.90\text{ BM}$
(C) $5.92\text{ BM}$
(D) $2.84\text{ BM}$

Solution & Explanation:
Step 1: Find oxidation state of Chromium ($ \text{Cr}$) in $\text{[Cr(H}_2\text{O)}_6]^{2+}$.
$\text{H}_2\text{O}$ is neutral ligand, so $\text{Cr}$ is in $+2$ oxidation state.
Step 2: Electronic configuration of $\text{Cr}$ ($Z=24$) is $[\text{Ar}] 3d^5 4s^1$.
For $\text{Cr}^{2+}$, electron configuration is $[\text{Ar}] 3d^4 4s^0$.
Step 3: $\text{H}_2\text{O}$ is a weak field ligand ($\Delta_o < P$), so pairing does not occur. Number of unpaired electrons $n = 4$.
Step 4: Calculate spin-only magnetic moment $\mu$:
$$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\text{ BM}$$
Correct Option: (B)

PYQ 3: NEET 2021

Question: The geometry and magnetic behavior of $\text{[Ni(CO)}_4]$ are respectively:

(A) Square planar and diamagnetic
(B) Tetrahedral and diamagnetic
(C) Square planar and paramagnetic
(D) Tetrahedral and paramagnetic

Solution & Explanation:
Step 1: Oxidation state of $\text{Ni}$ ($Z=28$) in $\text{[Ni(CO)}_4]$ is $0$.
Ground state of $\text{Ni}$: $[\text{Ar}] 3d^8 4s^2$.
Step 2: $\text{CO}$ is a strong field ligand. It causes pairing of $4s$ electrons into the $3d$ orbitals.
Rearranged electronic configuration: $3d^{10} 4s^0$.
Step 3: The empty $4s$ and three $4p$ orbitals hybridize to form four equivalent $sp^3$ hybrid orbitals.
Geometry: Tetrahedral.
Since all $3d$ electrons are paired ($3d^{10}$), it is diamagnetic.
Correct Option: (B)

Common NEET Traps & Mistakes to Avoid

• Trap 1: Assuming $\text{NH}_3$ is ALWAYS a strong field ligand.
  Reality: $\text{NH}_3$ acts as a strong field ligand for $\text{Co}^{3+}, \text{Cu}^{2+}, \text{Ni}^{2+}$ but acts as a weak field ligand for $\text{Fe}^{2+}, \text{Fe}^{3+}, \text{Mn}^{2+}$. Always check the metal ion and charge!
• Trap 2: Expecting $trans\text{-[M(en)}_2\text{X}_2]$ complexes to show optical activity.
  Reality: The trans isomer contains a plane of symmetry and an inversion center, making it achiral and optically inactive. Only the cis isomer is optically active!
• Trap 3: Confusing Geometry of $\text{[NiCl}_4]^{2-}$ vs $\text{[Ni(CN)}_4]^{2-}$.
  Reality: $\text{[NiCl}_4]^{2-}$ involves $\text{Cl}^-$ (WFL) $\rightarrow sp^3$ (Tetrahedral, paramagnetic, $n=2$). $\text{[Ni(CN)}_4]^{2-}$ involves $\text{CN}^-$ (SFL) $\rightarrow dsp^2$ (Square planar, diamagnetic, $n=0$).
• Trap 4: Incorrect determination of ambidentate linkage site in IUPAC names.
  Reality: $-\text{NO}_2^-$ bound via Nitrogen is named nitrito-N; when bound via Oxygen ($-\text{ONO}^-$) it is named nitrito-O.
• Trap 5: C-O Bond Strength vs M-C Bond Strength in Metal Carbonyls.
  Reality: Increased synergic back-bonding strengthens the $\text{M-C}$ bond but weakens and lengthens the $\text{C-O}$ bond! Questions frequently test this inverse relationship.

High-Yield NEET Practice MCQs with Answer Keys

Question 1

Which among the following complexes is an outer orbital complex and exhibits paramagnetic behavior?

(A) $\text{[Fe(CN)}_6]^{3-}$
(B) $\text{[Ni(NH}_3\text{)}_6]^{2+}$
(C) $\text{[Co(NH}_3\text{)}_6]^{3+}$
(D) $\text{[Cr(NH}_3\text{)}_6]^{3+}$

Answer: (B)
Solution: In $\text{[Ni(NH}_3\text{)}_6]^{2+}$, $\text{Ni}^{2+}$ is $3d^8$. Outer orbital hybridization $sp^3d^2$ is utilized because inner $3d$ orbitals cannot be vacated without pairing down beyond 8 electrons. It has 2 unpaired electrons ($n=2$), making it paramagnetic and an outer orbital complex.

Question 2

The total number of geometrical isomers possible for a square planar complex $\text{[Pt(NH}_3)(\text{Br})(\text{Cl})(\text{Py})]$ is:

(A) 2
(B) 3
(C) 4
(D) 6

Answer: (B)
Solution: A square planar complex of the general formula $[MABCD]$ with four distinct unidentate ligands exhibits exactly 3 geometrical isomers (fixed one ligand and vary the ligand trans to it).

Question 3

Which complex ion has the highest Crystal Field Splitting Energy ($\Delta_o$)?

(A) $\text{[Co(H}_2\text{O)}_6]^{3+}$
(B) $\text{[Co(NH}_3\text{)}_6]^{3+}$
(C) $\text{[Co(CN)}_6]^{3-}$
(D) $\text{[Co(C}_2\text{O}_4)_3]^{3-}$

Answer: (C)
Solution: All complexes contain $\text{Co}^{3+}$. The splitting energy $\Delta_o$ depends on ligand field strength in the spectrochemical series. Since $\text{CN}^-$ is the strongest field ligand listed, $\text{[Co(CN)}_6]^{3-}$ possesses the highest $\Delta_o$.

Question 4

What is the magnetic moment of $\text{[FeF}_6]^{3-}$? (Atomic number of $\text{Fe} = 26$)

(A) $1.73\text{ BM}$
(B) $4.90\text{ BM}$
(C) $5.92\text{ BM}$
(D) $2.84\text{ BM}$

Answer: (C)
Solution: $\text{Fe}^{3+}$ has electronic configuration $3d^5$. Fluoride ion ($\text{F}^-$) is a weak field ligand, so no electron pairing occurs ($t_{2g}^3 e_g^2$). Number of unpaired electrons $n=5$.
$$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\text{ BM}$$

Question 5

Homoleptic complexes among the following are:
(i) $\text{[Co(NH}_3\text{)}_6]^{3+}$
(ii) $\text{[Co(NH}_3\text{)}_4\text{Cl}_2]^+$
(iii) $\text{[Ni(CN)}_4]^{2-}$
(iv) $\text{[Fe(H}_2\text{O)}_5\text{NO}]^{2+}$

(A) (i) and (ii)
(B) (i) and (iii)
(C) (ii) and (iv)
(D) (iii) and (iv)

Answer: (B)
Solution: Homoleptic complexes contain only one type of ligand donor group. $\text{[Co(NH}_3\text{)}_6]^{3+}$ (only $\text{NH}_3$) and $\text{[Ni(CN)}_4]^{2-}$ (only $\text{CN}^-$) are homoleptic. Complexes (ii) and (iv) contain more than one type of ligand and are heteroleptic.

Summary & Final NEET Revision Tips

• Master Spectrochemical Series: Memorize the order: $\text{Halides} < \text{Oxygen Donors} < \text{Nitrogen Donors} < \text{Carbon Donors}$. Strong field ligands cause electron pairing; weak field ligands do not.
• Quick Hybridization Trick for Coordination Number 4: Nickel complexes: $\text{Ni}^{2+} + \text{SFL} \rightarrow dsp^2$ (Square Planar, Diamagnetic); $\text{Ni}^{2+} + \text{WFL} \rightarrow sp^3$ (Tetrahedral, Paramagnetic).
• Identify Optical Activity Fast: Cis-isomers of bis- and tris-chelates (e.g., $cis\text{-[Co(en)}_2\text{Cl}_2]^+$) are non-superimposable on mirror images and optically active. Trans-isomers are optically inactive!
• Formula Recall: Always double-check oxidation state before computing magnetic moment using $\mu = \sqrt{n(n+2)}\text{ BM}$.
• NCERT Direct Examples: Re-read NCERT examples on synergic bonding in metal carbonyls and biological importance (Hemoglobin, Chlorophyll, Cyanocobalamin) 24 hours prior to the exam.