Question

which of the following metal ions will repel an external magnetic field? ni2+ mn2+ fe2+ zn2+ cu2+

Explanation:

Step1: Determine the electron configuration of each ion

• For \(Ni^{2+}\):
• The atomic number of \(Ni\) is \(28\). The electron configuration of \(Ni\) is \([Ar]3d^{8}4s^{2}\).
• For \(Ni^{2+}\), we remove 2 electrons (from the \(4s\) orbital first). So the electron configuration is \([Ar]3d^{8}\). There are unpaired electrons (\(3d^{8}\) has 2 unpaired electrons), so it is paramagnetic (attracted to a magnetic field).
• For \(Mn^{2+}\):
• The atomic number of \(Mn\) is \(25\). The electron configuration of \(Mn\) is \([Ar]3d^{5}4s^{2}\).
• For \(Mn^{2+}\), we remove 2 electrons (from the \(4s\) orbital). So the electron configuration is \([Ar]3d^{5}\). There are unpaired electrons (\(3d^{5}\) has 5 unpaired electrons), so it is paramagnetic (attracted to a magnetic field).
• For \(Fe^{2+}\):
• The atomic number of \(Fe\) is \(26\). The electron configuration of \(Fe\) is \([Ar]3d^{6}4s^{2}\).
• For \(Fe^{2+}\), we remove 2 electrons (from the \(4s\) orbital). So the electron configuration is \([Ar]3d^{6}\). There are unpaired electrons (\(3d^{6}\) has 4 unpaired electrons), so it is paramagnetic (attracted to a magnetic field).
• For \(Zn^{2+}\):
• The atomic number of \(Zn\) is \(30\). The electron configuration of \(Zn\) is \([Ar]3d^{10}4s^{2}\).
• For \(Zn^{2+}\), we remove 2 electrons (from the \(4s\) orbital). So the electron configuration is \([Ar]3d^{10}\). There are no unpaired electrons (\(3d^{10}\) has all electrons paired), so it is diamagnetic (repelled by a magnetic field).
• For \(Cu^{2+}\):
• The atomic number of \(Cu\) is \(29\). The electron configuration of \(Cu\) is \([Ar]3d^{10}4s^{1}\).
• For \(Cu^{2+}\), we remove 2 electrons (1 from the \(4s\) and 1 from the \(3d\) orbital). So the electron configuration is \([Ar]3d^{9}\). There are unpaired electrons (\(3d^{9}\) has 1 unpaired electron), so it is paramagnetic (attracted to a magnetic field).

Answer:

\(Zn^{2+}\)