Q.The spin only magnetic moment of $[MnBr_4]^{2-}$ is 5.9 BM. Predict the geometry of the complex ion?

Concept: Magnetic Moment Calculation – The spin-only formula $\mu = \sqrt{n(n+2)}$ BM links the observed magnetic moment to the number of unpaired electrons $n$, which then determines the geometry via the metal’s $d$-orbital splitting.

1. Given $\mu = 5.9$ BM, solve for $n$:

   $5.9 = \sqrt{n(n+2)} \implies n(n+2) \approx 34.8 \implies n \approx 5$.

2. Mn in $[MnBr_4]^{2-}$: Mn is in +2 oxidation state (Br⁻ is $-1$, overall $-2$).

   Mn²⁺ has $d^5$ configuration. With 5 unpaired electrons, the complex is high spin. …

The key to this problem is connecting the observed magnetic moment to the number of unpaired electrons, and then using that number to decide the geometry.

Why magnetic moment tells us geometry

The spin-only formula $\mu = \sqrt{n(n+2)}$ BM gives the number of unpaired electrons $n$ in a complex. Different geometries split the $d$-orbitals differently, which determines whether electrons pair up or stay unpaired. For a $d^5$ ion like $Mn^{2+}$, the geometry decides whether we get 5 unpaired electrons (high-spin) or just 1 (low-spin).

Step 1: Find the oxidation state and $d$-electron count

Bromide ($Br^-$) is a monovalent anion. The complex is $[MnBr_4]^{2-}$, so:

• Let oxidation state of Mn be $x$.
• $x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2$.

So Mn is in the $+2$ state. The electronic configuration of Mn ($Z=25$) is $[Ar]\,3d^5 4s^2$. Removing two electrons (the $4s^2$ first, as is standard for transition metals) gives $Mn^{2+}$: $[Ar]\,3d^5$.

Thus, we have a $d^5$ system.

Step 2: Determine the number of unpaired electrons from the magnetic moment

Given $\mu = 5.9$ BM. Using the spin-only formula:

Square both sides:

Solve the quadratic $n^2 + 2n - 34.81 = 0$:

Taking the positive root: $n \approx \frac{9.97}{2} \approx 4.99 \approx 5$.

So there are 5 unpaired electrons.

Step 3: Interpret 5 unpaired electrons for a $d^5$ ion

Five unpaired electrons means all five $d$-electrons are in different orbitals, all with parallel spins (Hund's rule). This is only possible if the crystal field splitting is small — so small that it does not force pairing. This is the high-spin configuration.

For $d^5$, the high-spin configuration is $t_{2g}^3 e_g^2$ (in octahedral field) or $e^2 t_2^3$ (in tetrahedral field). Both give 5 unpaired electrons. So how do we choose between geometries?

Step 4: Consider the possible geometries for a 4-coordinate complex

A $[MX_4]^{2-}$ ion can be:

• Tetrahedral — all four ligands equivalent, bond angles ~109.5°.
• Square planar — ligands in a plane at 90° angles.

For a $d^5$ ion: …

Method: Spin-Only Magnetic Moment → Geometry Prediction

This uses the Spin-Only Magnetic Moment Formula combined with Crystal Field Theory.

Step 1: Recall the formula

The spin-only magnetic moment ($\mu$) is given by:

where $n$ = number of unpaired electrons.

Step 2: Find the number of unpaired electrons

Given $\mu = 5.9 \ \text{BM}$:

Square both sides:

Solve the quadratic:

Taking the positive root:

So, the complex has 5 unpaired electrons.

Step 3: Determine the metal ion's electron configuration

• Mn atomic number = 25

• Oxidation state in $[MnBr_4]^{2-}$:

  Let Mn oxidation state = $x$

  $x + 4(-1) = -2 \implies x = +2$

• Mn$^{2+}$ configuration:

  Mn: $[Ar]\,3d^5\,4s^2$

  Mn$^{2+}$: $[Ar]\,3d^5$

Step 4: Match unpaired electrons with geometry

For $d^5$ configuration:

🧠 The Core Concept First

The spin-only magnetic moment formula is:

Where $n$ = number of unpaired electrons.

Given $\mu = 5.9 \ \text{BM}$, we solve:

Squaring both sides:

Solving the quadratic $n^2 + 2n - 34.81 = 0$ gives $n \approx 5$ (since $5 \times 7 = 35$).

So the complex has 5 unpaired electrons.

✗ Common Mistake #1: Forgetting to Find $n$ First

The error: Students jump straight to geometry without calculating the number of unpaired electrons.

Why it's wrong: Geometry depends on the d-electron configuration (which depends on $n$), not on the magnetic moment value directly.

How to avoid: Always compute $n$ from $\mu$ first. Memorize the common values:

Here $5.9 \approx 5.92$, so $n = 5$.

✗ Common Mistake #2: Wrong Oxidation State of Mn

The error: Assuming Mn is in +2 state without checking.

Why it's wrong: The complex is $[MnBr_4]^{2-}$. Each $Br^-$ has charge $-1$, so:

So Mn is indeed in +2 state. But many students forget to verify.

How to avoid: Always calculate oxidation state systematically: sum of charges = overall charge.

✗ Common Mistake #3: Wrong d-Electron Count

The error: For Mn$^{2+}$, students write $d^5$ but then forget that $n=5$ means high spin in a weak field ligand environment.

Why it's wrong: $Br^-$ is a weak field ligand (low in spectrochemical series). So for $d^5$, the configuration is $t_{2g}^3 e_g^2$ (all unpaired) — giving $n=5$.

How to avoid: Remember:

• Weak field → high spin → maximum unpaired electrons
• Strong field → low spin → minimum unpaired electrons

For $d^5$, weak field gives $n=5$, strong field gives $n=1$.

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