Magnetic Moments (Numerical)

CHEColour & Magnetic Behaviour

Paramagnetism of transition-metal ions comes from unpaired $d$ electrons. ISC numericals use the spin-only formula $\mu = \sqrt{n(n+2)}$ BM, so the workflow is: write the ion configuration, count unpaired electrons $n$ , then substitute.

Spin-only magnetic moment

\mu = \sqrt{n(n+2)}\ \text{BM}

n

is the number of unpaired electrons; BM = Bohr magneton; orbital contribution is neglected at ISC level.

Worked values for common ions

n=1:1.73 \quad n=2:2.83 \quad n=3:3.87 \quad n=4:4.90 \quad n=5:5.92\ \text{BM}

obtained by substituting each

n

into

\mu=\sqrt{n(n+2)}

Number of unpaired electrons from a measured moment

n = -1 + \sqrt{1 + \mu^2} \quad\text{i.e. solve}\quad n^2 + 2n - \mu^2 = 0

rearranging

\mu^2 = n(n+2)

; e.g.

\mu = 3.87 \Rightarrow n = 3

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons (orbital contribution is ignored at ISC level).
$Fe^{2+}$ is $3d^6$ with $n=4$ , so $\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90$ BM.
$Mn^{2+}$ is $3d^5$ with $n=5$ , so $\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92$ BM (the maximum for a $3d$ ion).
$Cr^{3+}$ is $3d^3$ ( $n=3$ ): $\mu = \sqrt{15} \approx 3.87$ BM; $Ti^{3+}$ is $3d^1$ ( $n=1$ ): $\mu = \sqrt{3} \approx 1.73$ BM.
A moment of $3.87$ BM means $n=3$ unpaired electrons; matching $3d$ ions include $Cr^{3+}$ ( $3d^3$ ) and $V^{2+}$ ( $3d^3$ ), whereas $Ni^{2+}$ ( $3d^8$ , $n=2$ ) gives only $2.83$ BM, so it is not this one.
Diamagnetic ions have $n=0$ (no unpaired electrons): $Sc^{3+}$ ( $3d^0$ ), $Ti^{4+}$ ( $3d^0$ ), $Zn^{2+}$ ( $3d^{10}$ ) and $Cu^+$ ( $3d^{10}$ ) are diamagnetic; $Fe^{3+}$ ( $3d^5$ ) is strongly paramagnetic.
Workflow for any numerical: (1) write the ion's $3d$ configuration, (2) count unpaired electrons by Hund's rule, (3) substitute $n$ into $\mu = \sqrt{n(n+2)}$ .
The greater the number of unpaired electrons, the larger $\mu$ and the more strongly paramagnetic the ion; magnetic moment thus rises to a maximum at $d^5$ then falls.
Paramagnetic substances are attracted into a magnetic field (unpaired electrons); diamagnetic substances are weakly repelled (all electrons paired).
At ISC level we always use weak-field/free-ion electron counts for $3d$ ions, so $Fe^{3+}$ has $n=5$ and $Co^{2+}$ ( $3d^7$ ) has $n=3$ giving $\mu \approx 3.87$ BM.
Observed moments for many first-series ions match the spin-only value closely, which is why the orbital term is safely ignored for $3d$ transition ions.

Where the marks go

Using the number of $d$ electrons as $n$ instead of the number of unpaired electrons: $Fe^{2+}$ is $3d^6$ but $n=4$ , not 6.
Writing $\mu = \sqrt{n}\,(n+2)$ or $\sqrt{n(n-2)}$ — the formula is $\sqrt{n(n+2)}$ with a plus sign inside the root.
Forgetting to remove $4s$ before $3d$ when building the ion, leading to the wrong $n$ (e.g. miscounting $Mn^{2+}$ ).
Reporting the moment in wrong units or as a plain number; the answer carries Bohr magneton (BM) units.
Assuming an ion is diamagnetic just because the metal is in a high oxidation state — check the actual $d$ -electron count ( $Fe^{3+}$ , $d^5$ , is strongly paramagnetic).

How the board asks it

Numericalthe spin-only formula $\mu = \sqrt{n(n+2)}$
Calculate the spin-only magnetic moment (in BM) of the $Fe^{2+}$ ion. (Atomic number of $Fe = 26$ .)
Predict the productunpaired-electron count and $d$ -configuration
Predict which of $Mn^{2+}$ ( $3d^5$ ) and $Ni^{2+}$ ( $3d^8$ ) will have the larger magnetic moment, and calculate the value (in BM) for each.
Identify / classifyreverse use of the spin-only formula
A divalent ion of a first-transition-series metal has a spin-only magnetic moment of $3.87$ BM. Find the number of unpaired electrons and identify the ion.
Distinguishparamagnetic vs diamagnetic behaviour
How will you distinguish between $Zn^{2+}$ and $Cu^{2+}$ ions on the basis of their behaviour in a magnetic field? Justify using their $3d$ configurations.
Assertion–Reasonmagnetic moment maximum at $d^5$
Assertion: $Mn^{2+}$ has the highest spin-only magnetic moment among the divalent $3d$ ions. Reason: $Mn^{2+}$ has the maximum number of five unpaired electrons. State whether each is true and whether the reason correctly explains the assertion.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.