In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?
(i) [FeF$_6$]$^{3-}$ (ii) [Co(NH$_3$)$_6$]$^{3+}$ (iii) [NiCl$_4$]$^{2-}$ (iv) [Cu(NH$_3$)$_4$]$^{2+}$
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To determine the number of unpaired electrons in a coordination complex, follow these steps: 1. Find the oxidation state of the metal. 2. Write the d-electron configuration. 3. Determine the geometry (octahedral, tetrahedral, etc.). 4. Consider the ligand strength (strong-field for low spin, weak-field for high spin). 5. Fill the d-orbitals according to the splitting diagram.
The spin-only magnetic moment ($\mu$) is calculated using the formula $\mu = \sqrt{n(n+2)}$ Bohr Magnetons (BM), where 'n' is the number of unpaired electrons. A larger 'n' leads to a larger magnetic moment. So, we need to find the number of unpaired electrons in each complex.
(i) [FeF$_6$]$^{3-}$: Iron is in the +3 oxidation state (Fe$^{3+}$), with an electron configuration of [Ar]3d$^5$. F$^-$ is a weak-field ligand, so this is a high-spin octahedral complex. The five d-electrons will occupy the orbitals singly. Thus, n = 5.
(ii) [Co(NH$_3$)$_6$]$^{3+}$: Cobalt is in the +3 oxidation state (Co$^{3+}$), with an electron configuration of [Ar]3d$^6$. NH$_3$ is a strong-field ligand, so this is a low-spin octahedral complex. The six d-electrons will pair up in the lower energy t$_{2g}$ orbitals. Thus, there are no unpaired electrons, n = 0.
(iii) [NiCl$_4$]$^{2-}$: Nickel is in the +2 oxidation state (Ni$^{2+}$), with an electron configuration of [Ar]3d$^8$. This is a tetrahedral complex (as Cl$^-$ is a weak-field ligand). In a tetrahedral field, the d-orbitals split into e (lower) and t$_2$ (higher). Filling the d$^8$ configuration gives (e)$^4$(t$_2$)$^4$. The last two electrons in the t$_2$ orbitals are unpaired. Thus, n = 2.
(iv) [Cu(NH$_3$)$_4$]$^{2+}$: Copper is in the +2 oxidation state (Cu$^{2+}$), with an electron configuration of [Ar]3d$^9$. This is a square planar complex. The d$^9$ configuration will always have one unpaired electron, regardless of the geometry or ligand strength. Thus, n = 1.
The number of unpaired electrons are:
(i) n = 5
(ii) n = 0
(iii) n = 2
(iv) n = 1
The decreasing order of the number of unpaired electrons (and thus the magnetic moment) is:
n=5>n=2>n=1>n=0
Which corresponds to the order of complexes: (i)>(iii)>(iv)>(ii).
