Coordination Compounds Notes, Formulas, Traps & PYQs | ExamiFly

Q: Why does $[Ni(CO)_4]$ exhibit $sp^3$ hybridisation and diamagnetic behavior despite Nickel having zero oxidation state?

In $[Ni(CO)_4]$, Nickel is in zero oxidation state with valence shell $3d^8 4s^2$. Carbon monoxide ($\text{CO}$) is an exceptionally strong $\pi$-acceptor ligand that forces the two $4s$ electrons to pair completely into the $3d$ subshell, resulting in a fully filled $3d^{10}$ configuration ($n = 0$, diamagnetic). The vacant $4s$ and three $4p$ orbitals then hybridise to form four equivalent $sp^3$ orbitals, yielding tetrahedral geometry.

Q: Why is $[Ti(H_2O)_6]^{3+}$ violet in color, whereas $[Sc(H_2O)_6]^{3+}$ is completely colorless?

Color in transition complexes stems from $d$-$d$ electronic transitions. In $[Ti(H_2O)_6]^{3+}$, $Ti^{3+}$ has a $3d^1$ configuration ($t_{2g}^1 e_g^0$). Absorption of yellow-green light excites this single electron from $t_{2g}$ to $e_g$, transmitting violet light. In $[Sc(H_2O)_6]^{3+}$, $Sc^{3+}$ has a $3d^0$ configuration; with zero $d$-electrons present, $d$-$d$ transitions are impossible, rendering the ion completely colorless.

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Coordination Compounds Notes, Formulas, Traps & PYQs | ExamiFly | Examifly

Solved High-Yield Sample Questions

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Question #1NEET 2024

Which of the following complex species is paramagnetic and forms an inner orbital octahedral complex according to Valence Bond Theory?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Determine metal oxidation states and valence configurations. In $[Fe(CN)_6]^{3-}$ , $Fe$ is in $+3$ state ( $3d^5$ ). In $[Co(NH_3)_6]^{3+}$ , $Co$ is $+3$ ( $3d^6$ ). In $[Fe(H_2O)_6]^{3+}$ , $Fe$ is $+3$ ( $3d^5$ ). In $[Ni(CN)_4]^{2-}$ , $Ni$ is $+2$ ( $3d^8$ square planar).

Step 2

Analyze ligand field strength and hybridisation. $\text{CN}^-$ is a strong field ligand, forcing pairing of $3d^5$ electrons in $[Fe(CN)_6]^{3-}$ to give $t_{2g}^5$ with $n = 1$ unpaired electron (paramagnetic). It uses inner $3d$ orbitals for $d^2sp^3$ hybridisation (inner orbital complex).

Step 3

$[Co(NH_3)_6]^{3+}$ has $3d^6$ completely paired ( $t_{2g}^6$ ), making it diamagnetic ( $n = 0$ ). $[Fe(H_2O)_6]^{3+}$ contains weak ligand $\text{H}_2\text{O}$ , forming an outer orbital $sp^3d^2$ complex with $n = 5$ . Thus, $[Fe(CN)_6]^{3-}$ is both paramagnetic and an inner orbital complex.

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Question #2JEE Main 2023

The spin-only magnetic moment value of $[Cr(H_2O)_6]^{2+}$ is $X \, \text{B.M.}$ and that of $[Mn(CN)_6]^{4-}$ is $Y \, \text{B.M.}$ . The ratio of $X$ to $Y$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

For $[Cr(H_2O)_6]^{2+}$ , central metal is $Cr^{2+}$ with $3d^4$ configuration. Water is a weak field ligand ( $\Delta_o < P$ ), giving high-spin arrangement $t_{2g}^3 e_g^1$ with $n = 4$ unpaired electrons. $X = \sqrt{4(4+2)} = \sqrt{24} \, \text{B.M.}$

Step 2

For $[Mn(CN)_6]^{4-}$ , central metal is $Mn^{2+}$ with $3d^5$ configuration. Cyanide is a strong field ligand ( $\Delta_o > P$ ), forcing pairing to give low-spin arrangement $t_{2g}^5 e_g^0$ with $n = 1$ unpaired electron. $Y = \sqrt{1(1+2)} = \sqrt{3} \, \text{B.M.}$

Step 3

Compute the required ratio $\frac{X}{Y} = \frac{\sqrt{24}}{\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$ . The exact option representation given is $\frac{\sqrt{24}}{\sqrt{3}}$ .

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Frequently Asked Questions

Why does

[Ni(CO)_4]

exhibit

sp^3

hybridisation and diamagnetic behavior despite Nickel having zero oxidation state?

In $[Ni(CO)_4]$ , Nickel is in zero oxidation state with valence shell $3d^8 4s^2$ . Carbon monoxide ( $\text{CO}$ ) is an exceptionally strong $\pi$ -acceptor ligand that forces the two $4s$ electrons to pair completely into the $3d$ subshell, resulting in a fully filled $3d^{10}$ configuration ( $n = 0$ , diamagnetic). The vacant $4s$ and three $4p$ orbitals then hybridise to form four equivalent $sp^3$ orbitals, yielding tetrahedral geometry.

How do I rapidly decide whether a complex is high-spin or low-spin under exam conditions?

First inspect the transition series: for $4d$ and $5d$ metals (e.g., $\text{Pt}^{2+}, \text{Pd}^{2+}, \text{Rh}^{3+}$ ), ALL ligands act as strong-field ligands due to massive $Z_{\text{eff}}$ , forming low-spin complexes. For $3d$ metals, examine metal oxidation state ( $+4$ or higher increases $\Delta_o$ significantly) and position of ligand in spectrochemical series: $\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO}$ . If $\Delta_o > P$ , low-spin; if $\Delta_o < P$ , high-spin.

Why is

[Ti(H_2O)_6]^{3+}

violet in color, whereas

[Sc(H_2O)_6]^{3+}

is completely colorless?

Color in transition complexes stems from $d$ - $d$ electronic transitions. In $[Ti(H_2O)_6]^{3+}$ , $Ti^{3+}$ has a $3d^1$ configuration ( $t_{2g}^1 e_g^0$ ). Absorption of yellow-green light excites this single electron from $t_{2g}$ to $e_g$ , transmitting violet light. In $[Sc(H_2O)_6]^{3+}$ , $Sc^{3+}$ has a $3d^0$ configuration; with zero $d$ -electrons present, $d$ - $d$ transitions are impossible, rendering the ion completely colorless.

Mastering Coordination Compounds: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Effective Atomic Number (EAN)

Spin-Only Magnetic Moment (\mu_s)

Crystal Field Splitting Energy (CFSE) for Octahedral Complexes

Tetrahedral vs Octahedral Splitting Energy Relation

Overall Formation Constant (\beta_n)

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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