Which of the following species would you expect tobe diamagnetic and which paramagnetic? (a) \(\mathrm{K}^{+}=\) (b) \(\mathrm{Cr}^{3+} ;\) (c) \(\mathrm{Zn}^{2+} ;\) (d) \(\mathrm{Cd} ;\) (e) \(\mathrm{Co}^{3+} ;\) (f) \(\mathrm{Sn}^{2+} ;\) (g) Br.

Write down the electronic configuration for each species. \n- The electronic configuration of \(K\) is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1\).\nSo, \(K^+\) has lost one electron and will have the configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6\).\n- \(Cr^{3+}\) is formed by losing three electrons from \(Cr\), whose configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1\).\n- \(Zn^{2+}\) and \(Cd\) have the same configuration as \(Cd\), which is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2\).\n- The electronic configuration of \(Co\) is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2\), so \(Co^{3+}\) is formed by losing three electrons.\n- The electronic configuration for \(Sn^{2+}\) is gotten by first writing that of \(Sn\) and removing two electrons.\n- \(Br\)'s electronic configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^5\).

Check each species' electron configuration to see if there are unpaired electrons.\n- \(K^+\), \(Zn^{2+}\), \(Cd\), and \(Br\) all have paired electrons, so they're diamagnetic.\n- \(Cr^{3+}\), \(Co^{3+}\), and \(Sn^{2+}\) have unpaired electrons, so they're paramagnetic.