Question

How many of the following complexes are paramagnetic?

• A. K$_3$[Fe(CN)$_6$]
• B. [Co(ox)$_3$]$^{3-}$
• C. [Cu(H$_2$O)$_4$]SO$_4$
• D. [NiCl$_4$]$^{2-}$
• E. [Ni(CO)$_4$]
• F. FeSO$_4$.7H$_2$O
• G. [Ni(H$_2$O)$_6$]$^{2-}$
• H. [PtCl$_4$]$^{2-}$
• I. [Co(H$_2$O)$_6$]$^{2+}$

Answer 1

5

Answer 2

To determine if a complex is paramagnetic, we need to find the number of unpaired electrons in the central metal ion.

1. K$_3$[Fe(CN)$_6$]: Fe$^{3+}$ ($d^5$), strong field CN$^-$ $\rightarrow$ low spin $t_{2g}^5$ $\rightarrow$ 1 unpaired electron. Paramagnetic.
2. [Co(ox)$_3$]$^{3-}$: Co$^{3+}$ ($d^6$), strong field ox$^{2-}$ $\rightarrow$ low spin $t_{2g}^6$ $\rightarrow$ 0 unpaired electrons. Diamagnetic.
3. [Cu(H$_2$O)$_4$]SO$_4$: Cu$^{2+}$ ($d^9$) $\rightarrow$ 1 unpaired electron. Paramagnetic.
4. [NiCl$_4$]$^{2-}$: Ni$^{2+}$ ($d^8$), weak field Cl$^-$, tetrahedral $\rightarrow$ 2 unpaired electrons. Paramagnetic.
5. [Ni(CO)$_4$]: Ni$^0$ ($3d^8 4s^2$), strong field CO, tetrahedral $\rightarrow$ 0 unpaired electrons. Diamagnetic.
6. FeSO$_4$.7H$_2$O: Fe$^{2+}$ ($d^6$), weak field H$_2$O, octahedral $\rightarrow$ high spin $t_{2g}^4 e_g^2$ $\rightarrow$ 4 unpaired electrons. Paramagnetic.
7. [Ni(H$_2$O)$_6$]$^{2-}$: Ni$^{2-}$ ($3d^{10} 4s^2$) $\rightarrow$ $3d^{10}$ is filled $\rightarrow$ 0 unpaired electrons. Diamagnetic.
8. [PtCl$_4$]$^{2-}$: Pt$^{2+}$ ($d^8$), square planar $\rightarrow$ 0 unpaired electrons. Diamagnetic.
9. [Co(H$_2$O)$_6$]$^{2+}$: Co$^{2+}$ ($d^7$), weak field H$_2$O, octahedral $\rightarrow$ high spin $t_{2g}^5 e_g^2$ $\rightarrow$ 3 unpaired electrons. Paramagnetic.

The paramagnetic complexes are (1), (3), (4), (6), and (9). There are 5 paramagnetic complexes.