Question

which of the following is paramagnetic? fe³+ zn ag+ cl¯

Explanation:

Paramagnetic substances have unpaired electrons. Let's analyze each option:

• \(Fe^{3 +}\):
• The electron configuration of \(Fe\) (atomic number \(26\)) is \([Ar]3d^{6}4s^{2}\).
• For \(Fe^{3+}\), we remove 3 electrons (\(2\) from \(4s\) and \(1\) from \(3d\)), so the electron configuration is \([Ar]3d^{5}\).
• In the \(3d\) sub - shell (\(n = 3,l = 2\)), according to Hund's rule, each of the five \(d\) orbitals has one unpaired electron.
• \(Zn\):
• The electron configuration of \(Zn\) (atomic number \(30\)) is \([Ar]3d^{10}4s^{2}\).
• All the electrons in \(3d\) and \(4s\) orbitals are paired.
• \(Ag^{+}\):
• The electron configuration of \(Ag\) (atomic number \(47\)) is \([Kr]4d^{10}5s^{1}\).
• For \(Ag^{+}\), we remove 1 electron from \(5s\), so the electron configuration is \([Kr]4d^{10}\).
• All the electrons in \(4d\) orbitals are paired.
• \(Cl^{-}\):
• The electron configuration of \(Cl\) (atomic number \(17\)) is \([Ne]3s^{2}3p^{5}\).
• For \(Cl^{-}\), we add 1 electron to \(3p\) sub - shell, so the electron configuration is \([Ne]3s^{2}3p^{6}\).
• All the electrons in \(3s\) and \(3p\) orbitals are paired.

Answer:

\(Fe^{3 +}\)