Question

What would be the calculated magnetic moment of [Ni(H 2 O) 6 ] 2+ ?

Answer 1

Answer on Question #54386 – Chemistry – Inorganic Chemistry

Question:

What would be the calculated magnetic moment of $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right) 6\right]^{2+}$?

Answer:

The magnetic moment (spin only) for $\mathrm{Ni}^{2+}$ ion can be found according to the equation:

$$\mu = \left[ n (n + 2) \right] ^ {1 / 2}, \text{ where } n - \text{ the number of unpaired electrons}.$$

Being in octahedral coordination environment, Ni(II) has the following electronic configuration of d-electrons:

$$3 d ^ {8}: e _ {g} ^ {2} t _ {2 g} ^ {6}$$

Thus, electrons sitting on the $e_g$ are unpaired, and those, which are on the $t_{2g}$, form 3 pairs. Therefore, the number of unpaired electrons is only 2.

Finally, the magnetic moment equals:

$$\mu = \left[ 2 (2 + 2) \right] ^ {1 / 2} = 2.83 \mu_B \text{ (the Bohr magneton)}$$

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