$[Cr(NH_3)_6]^{3+}$ is paramagnetic while $[Ni(CN)_4]^{2-}$ is diamagnetic. Explain why?

> [!TLDR] > The magnetic behaviour depends on the **crystal field splitting** and the **electronic configuration** of the central metal ion. $[Cr(NH_3)_6]^{3+}$ is paramagnetic because it has **three unpaired electrons** in a weak-field octahedral environment, while $[Ni(CN)_4]^{2-}$ is diamagnetic because it has **zero unpaired electrons** in a strong-field square planar geometry. ## Why This Approach Works The key to predicting magnetic behaviour lies in understanding how ligands influence the d-orbital splitting of the central metal ion. Paramagnetism arises from **unpaired electrons** — the more unpaired electrons, the stronger the paramagnetic effect. Diamagnetism, on the other hand, occurs when **all electrons are paired**. Two factors determine whether electrons remain unpaired or get forced into pairs: 1. **Crystal field splitting energy** ($\Delta$) — how much the d-orbitals split in energy 2. **Pairing energy** ($P$) — the energy cost to put two electrons in the same orbital When $\Delta P$ (strong field), electrons pair up in lower-energy orbitals before occupying higher ones — giving fewer or zero unpaired electrons. Let's apply this to each complex. --- ## Step-by-Step Reasoning ### 1. Determine the oxidation state and d-electron count for each complex **For $[Cr(NH_3)_6]^{3+}$:** - $NH_3$ is neutral, so the charge comes entirely from Cr. - Cr is in +3 oxidation state. - Cr atomic number = 24. Electronic configuration: $[Ar]\,3d^5\,4s^1$. - Cr$^{3+}$ loses three electrons: the 4s electron and two 3d electrons. - So Cr$^{3+}$ has $d^3$ configuration. **For $[Ni(CN)_4]^{2-}$:** - $CN^-$ is a −1 ligand. Four CN⁻ give −4 charge. - Overall complex charge is −2, so Ni must be in +2 oxidation state. - Ni atomic number = 28. Configuration: $[Ar]\,3d^8\,4s^2$. - Ni$^{2+}$ loses the two 4s electrons → $d^8$ configurati

[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]²⁻ is diamagnetic. Explain why?

Exercises · 5.19

[Cr(NH_3)_6]^{3+}

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Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

Imagine you're holding a ball on a hill. If you place it exactly at the top, it's balanced — but the slightest push sends it rolling down. That's an unstable position. If you place it in a small dip on the hillside, it stays put even if nudged — that's stable. Oxidation states work the same way: some are like the hilltop (easily changed), others like the dip (hard to change).

The Core Intuition

An oxidation state is just a number we assign to an atom to track how many electrons it has gained or lost compared to its neutral state. But atoms don't "want" to stay in arbitrary oxidation states — they want to reach a configuration that minimises their energy.

Stability here means: how reluctant is that oxidation state to change under normal conditions? A stable oxidation state resists being oxidised further or reduced further. An unstable one readily changes into something else.

The Precise Statement

Stability of an oxidation state refers to the tendency of an element to maintain that particular oxidation state under given conditions (temperature, pH, presence of other reagents). A stable oxidation state is one that does not easily undergo redox reactions — it is neither easily oxidised nor easily reduced.

This depends on three key factors:

Electronic configuration – Half-filled and fully-filled d or f subshells confer extra stability (e.g., $\text{Fe}^{3+}$ with $d^5$ is more stable than $\text{Fe}^{2+}$ with $d^6$ in some contexts).
Inert pair effect – Heavier p-block elements (like Tl, Pb, Bi) show lower oxidation states (e.g., +1 for Tl) as more stable than higher ones (+3 for Tl), because the s-electrons become reluctant to participate.
Disproportionation tendency – Some oxidation states are unstable because they spontaneously convert into two other states (e.g., $\text{Cu}^+$ in aqueous solution gives $\text{Cu}^{2+}$ and $\text{Cu}$ ).

Important

Stability is relative — it depends on the environment. $\text{Mn}^{2+}$ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.

Examples That Make It Concrete

Transition metals – $\text{Cr}^{3+}$ ( $d^3$ ) and $\text{Mn}^{2+}$ ( $d^5$ ) are exceptionally stable because half-filled/half-filled-like configurations have low energy. $\text{Cr}^{2+}$ ( $d^4$ ) is easily oxidised to $\text{Cr}^{3+}$ — it's unstable.

p-block elements – $\text{Pb}^{2+}$ is stable, $\text{Pb}^{4+}$ is a strong oxidising agent (unstable). $\text{Sn}^{2+}$ is a reducing agent (easily oxidised to $\text{Sn}^{4+}$ ), so $\text{Sn}^{4+}$ is more stable for tin.

Common pattern – For most elements, the most common oxidation state is the most stable one under standard conditions. But "most common" isn't always "most stable" — e.g., $\text{Fe}^{3+}$ is common but $\text{Fe}^{2+}$ is more stable in acidic solution.

Watch out

Do not confuse "stability" with "occurrence". $\text{Mn}^{7+}$ (as $\text{MnO}_4^-$ ) is common in the lab but is a powerful oxidising agent — it is not stable in the sense of resisting change. It readily accepts electrons.

How to Think About It in Exams …

Why this formula?

Stability of Oxidation States: Why It Works

This concept explains why certain oxidation states of an element are more stable than others — and why some states are never observed at all.

The Core Idea: Energy Minimisation

An oxidation state is stable when the total energy of the system is at a minimum. This depends on three competing factors:

Ionisation energy (energy needed to remove electrons)
Lattice energy (for ionic compounds) or bond energy (for covalent compounds)
Electronic configuration (half-filled / fully-filled subshells)

There is no single formula for stability — instead, we use trends and principles that act as "formulae" for prediction.

Key Principle 1: Inert Pair Effect (for p-block elements)

Why it holds:

For heavier elements (e.g., Tl, Pb, Bi), the 6s² electrons are held very tightly due to poor shielding and relativistic effects. They resist removal.

Result: Lower oxidation state (e.g., +1 for Tl, +2 for Pb) becomes more stable than the higher state (+3, +4).
Example: $\ce{Tl^3+}$ is a strong oxidising agent — it readily gains two electrons to become $\ce{Tl^+}$ .

Derivation logic:

The energy cost to remove the 6s² electrons is greater than the energy gained by forming additional bonds or lattice. So the system stays in the lower state.

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Why it holds:

A half-filled ( $d^5$ , $f^7$ ) or fully-filled ( $d^{10}$ , $f^{14}$ ) subshell has extra exchange energy and symmetry — making it unusually stable.

Example: $\ce{Mn^{2+}}$ ( $d^5$ ) is more stable than $\ce{Mn^{3+}}$ ( $d^4$ ). $\ce{Fe^{3+}}$ ( $d^5$ ) is more stable than $\ce{Fe^{2+}}$ ( $d^6$ ).

Derivation logic:

The exchange energy (Hund's rule) is maximum for half-filled configurations. Removing an electron from a half-filled shell costs extra energy — so the half-filled state is favoured.

Key Principle 3: Lattice Energy / Hydration Energy Compensation

For transition metals, stability of a particular oxidation state in aqueous solution depends on:

\Delta G^\circ = \Delta H^\circ_{\text{hydration}} - \Delta H^\circ_{\text{ionisation}}

Why it holds:

Higher oxidation states have higher ionisation energy (harder to remove electrons).
But they also have higher hydration energy (smaller, more charged ions attract water more strongly).
The balance determines which state is stable.

Example:

$\ce{Cu^+}$ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
$\ce{Cu^{2+}}$ is stable in water.

Key Principle 4: Disproportionation

Some oxidation states are unstable and spontaneously convert to two other states:

\ce{2Cu^+ -> Cu + Cu^{2+}}

Why it holds: …

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The magnetic behaviour depends on the crystal field splitting and the electronic configuration of the central metal ion. $[Cr(NH_3)_6]^{3+}$ is paramagnetic because it has three unpaired electrons in a weak-field octahedral environment, while $[Ni(CN)_4]^{2-}$ is diamagnetic because it has zero unpaired electrons in a strong-field square planar geometry.

Why This Approach Works

The key to predicting magnetic behaviour lies in understanding how ligands influence the d-orbital splitting of the central metal ion. Paramagnetism arises from unpaired electrons — the more unpaired electrons, the stronger the paramagnetic effect. Diamagnetism, on the other hand, occurs when all electrons are paired.

Two factors determine whether electrons remain unpaired or get forced into pairs:

Crystal field splitting energy ( $\Delta$ ) — how much the d-orbitals split in energy
Pairing energy ( $P$ ) — the energy cost to put two electrons in the same orbital

When $\Delta < P$ (weak field), electrons follow Hund's rule and occupy orbitals singly first — giving maximum unpaired electrons. When $\Delta > P$ (strong field), electrons pair up in lower-energy orbitals before occupying higher ones — giving fewer or zero unpaired electrons.

Let's apply this to each complex.

Step-by-Step Reasoning

1. Determine the oxidation state and d-electron count for each complex

For $[Cr(NH_3)_6]^{3+}$ :

$NH_3$ is neutral, so the charge comes entirely from Cr.
Cr is in +3 oxidation state.
Cr atomic number = 24. Electronic configuration: $[Ar]\,3d^5\,4s^1$ .
Cr $^{3+}$ loses three electrons: the 4s electron and two 3d electrons.
So Cr $^{3+}$ has $d^3$ configuration.

For $[Ni(CN)_4]^{2-}$ :

$CN^-$ is a −1 ligand. Four CN⁻ give −4 charge.
Overall complex charge is −2, so Ni must be in +2 oxidation state.
Ni atomic number = 28. Configuration: $[Ar]\,3d^8\,4s^2$ .
Ni $^{2+}$ loses the two 4s electrons → $d^8$ configuration.

Note

Always remember: in transition metal ions, the 4s electrons are lost before the 3d electrons when forming cations.

2. Identify the geometry and crystal field splitting pattern

$[Cr(NH_3)_6]^{3+}$ — six ligands → octahedral geometry.

In octahedral field, d-orbitals split into:
- Lower energy: $t_{2g}$ ( $d_{xy}, d_{xz}, d_{yz}$ ) — three orbitals
- Higher energy: $e_g$ ( $d_{x^2-y^2}, d_{z^2}$ ) — two orbitals
Splitting energy = $\Delta_o$

$[Ni(CN)_4]^{2-}$ — four ligands, and CN⁻ is a very strong field ligand → square planar geometry.

Square planar is derived from octahedral by removing two ligands along the z-axis.
The d-orbital splitting in square planar (energy increasing) is:
- Lowest: $d_{xz}, d_{yz}$ (degenerate)
- Then: $d_{z^2}$
- Then: $d_{xy}$
- Highest: $d_{x^2-y^2}$
The splitting between the lowest and highest is very large — much larger than $\Delta_o$ for the same ligand.

t_{2g} < e_g

3. Classify the ligand as weak or strong field

$NH_3$ is a moderate field ligand — it lies in the middle of the spectrochemical series. For Cr $^{3+}$ ( $d^3$ ), the pairing energy is relatively high because all three electrons are in different orbitals anyway. So even with NH₃, the field is effectively weak for this configuration — no pairing occurs.

$CN^-$ is a very strong field ligand — near the top of the spectrochemical series. It causes large splitting, so $\Delta$ is much larger than $P$ . This forces maximum pairing.

Tip

A quick memory aid: The spectrochemical series from weak to strong — I < Br < Cl < F < OH < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO. Anything to the right of NH₃ tends to be strong field for most ions.

4. Fill the d-orbitals and count unpaired electrons …

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Method: Crystal Field Theory (CFT) + Electronic Configuration Analysis

This method explains magnetic behaviour by examining:

The oxidation state of the central metal ion
Its d-electron count
The crystal field splitting caused by the ligand
Whether unpaired electrons remain (paramagnetic) or all electrons are paired (diamagnetic)

Step 1: Determine oxidation state and d-electron count for each complex

For $[Cr(NH_3)_6]^{3+}$ :

Ligand: $NH_3$ is neutral → charge on complex = +3 comes from Cr
So, Cr is in +3 oxidation state
Cr atomic number = 24 → Cr $^{3+}$ = $[Ar]\,3d^3$
d-electron count = 3

For $[Ni(CN)_4]^{2-}$ :

Ligand: $CN^-$ has charge –1; four ligands give –4
Complex charge = –2 → Ni must be in +2 oxidation state
Ni atomic number = 28 → Ni $^{2+}$ = $[Ar]\,3d^8$
d-electron count = 8

Step 2: Identify ligand field strength and geometry

Complex	Ligand	Field strength	Geometry
$[Cr(NH_3)_6]^{3+}$	$NH_3$	Moderate field (irrelevant for $d^3$ — no pairing choice exists)	Octahedral
$[Ni(CN)_4]^{2-}$	$CN^-$	Strong field (low spin)	Square planar

Key fact: $CN^-$ is a strong field ligand that causes large splitting; $NH_3$ is intermediate — and for Cr $^{3+}$ ( $d^3$ ) the distinction does not matter, since three electrons occupy the three $t_{2g}$ orbitals singly under any field strength.

Step 3: Fill d-orbitals according to Hund’s rule and splitting

For $[Cr(NH_3)_6]^{3+}$ (octahedral, d $^3$ ):

Octahedral splitting: $t_{2g}$ (lower energy) and $e_g$ (higher energy)
With 3 electrons: all go into $t_{2g}$ unpaired (Hund’s rule)
Unpaired electrons = 3 → Paramagnetic

For $[Ni(CN)_4]^{2-}$ (square planar, d $^8$ ): …

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Here’s a breakdown of the common mistakes students make on this exact question, and how to avoid each one.

Mistake 1: Forgetting to find the oxidation state of the metal first

Students often jump straight to the electronic configuration of the neutral atom (Cr or Ni) without adjusting for the charge on the complex.

Why it’s wrong: The number of electrons on the metal changes when it forms a complex. You must know the exact $d^n$ configuration of the metal ion inside the complex.
How to avoid: Always calculate the oxidation state of the metal first.
- For $[Cr(NH_3)_6]^{3+}$ : $NH_3$ is neutral. Let Cr be $x$ . $x + 6(0) = +3 \implies x = +3$ . So, Cr is in +3 state.
- For $[Ni(CN)_4]^{2-}$ : $CN^-$ has a -1 charge. Let Ni be $x$ . $x + 4(-1) = -2 \implies x = +2$ . So, Ni is in +2 state.

Mistake 2: Using the wrong electronic configuration for the ion

Once you have the oxidation state, students often write the configuration of the neutral atom and then remove electrons from the wrong orbitals (e.g., removing 4s electrons before 3d).

Why it’s wrong: In transition metal ions, the 4s orbital is higher in energy than the 3d orbital. Electrons are always removed from the 4s orbital first.
How to avoid: Write the configuration of the neutral atom, then remove the required number of electrons from the 4s orbital first, then the 3d orbital.
- Cr (Z=24): $[Ar] 3d^5 4s^1$ (remember the exception for half-filled stability).
- Cr $^{3+}$ : Remove 3 electrons. Remove the 1 from 4s, then 2 from 3d. Result: $[Ar] 3d^3$ .
- Ni (Z=28): $[Ar] 3d^8 4s^2$ .
- Ni $^{2+}$ : Remove 2 electrons. Remove both from 4s. Result: $[Ar] 3d^8$ .

Mistake 3: Ignoring the ligand field strength (the “why” of pairing)

This is the most critical mistake. Students correctly identify the $d^8$ configuration for Ni $^{2+}$ but then say it must be paramagnetic because $d^8$ has two unpaired electrons in the free ion.

Why it’s wrong: The magnetic property depends on the geometry and ligand strength. $CN^-$ is a strong field ligand that causes pairing.
How to avoid: Always check the ligand and geometry.
- For $[Ni(CN)_4]^{2-}$ : $CN^-$ is strong field. The geometry is square planar (common for $d^8$ with strong ligands). In square planar, the energy gap is so large that all 8 electrons pair up. Result: 0 unpaired electrons → Diamagnetic.
- For $[Cr(NH_3)_6]^{3+}$ : $NH_3$ is a moderate field ligand. The geometry is octahedral. For a $d^3$ configuration, even in a strong field, you cannot pair electrons because you need at least 4 electrons to start pairing (Hund’s rule). Result: 3 unpaired electrons → Paramagnetic.

Mistake 4: Confusing “paramagnetic” with “diamagnetic” based on the metal alone …

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Q.[Cr(NH3)6]3+[Cr(NH_3)_6]^{3+}[Cr(NH3​)6​]3+ is paramagnetic while [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2− is diamagnetic. Explain why?

Why This Approach Works

Step-by-Step Reasoning

1. Determine the oxidation state and d-electron count for each complex

2. Identify the geometry and crystal field splitting pattern

3. Classify the ligand as weak or strong field

4. Fill the d-orbitals and count unpaired electrons …

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Q. $[Cr(NH_3)_6]^{3+}$ is paramagnetic while $[Ni(CN)_4]^{2-}$ is diamagnetic. Explain why?

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