Question

The metal complex that is diamagnetic is (Atomic number : \(Fe, 26; Cu, 29\))

• A. \(K_3[Cu(CN)_4]\)
• B. \(K_2[Cu(CN)_4]\)
• C. \(K_3[Fe(CN)_4]\)
• D. \(K_4[FeCl_6]\)

Answer 1

The Correct Option is A

To determine which metal complex is diamagnetic, we need to evaluate the electronic configuration of each metal complex. Diamagnetism occurs when all electrons in the compound are paired.

1. Let's examine each complex:
   • \(K_3[Cu(CN)_4]\):
     • Copper (Cu) in this complex is in the +1 oxidation state, as indicated by the charge balance in the compound: \(K_3[Cu(CN)_4] \rightarrow 3K^+ + [Cu(CN)_4]^{3-}\).
     • The electronic configuration of Cu is: \([Ar] 3d^{10} 4s^1\).
     • In the +1 oxidation state, Cu becomes: \([Ar] 3d^{10}\), where all electrons are paired.
     • All electrons are paired, making this complex diamagnetic.
   • \(K_2[Cu(CN)_4]\):
     • With similar logic, we assume it would be Cu(II), which results in an electronic configuration of \([Ar] 3d^9\), leaving an unpaired electron and making it paramagnetic.
   • \(K_3[Fe(CN)_4]\):
     • Iron (Fe) is in the +3 oxidation state here. The electronic configuration of Fe is: \([Ar] 3d^6 4s^2\).
     • For Fe(III), this turns into \([Ar] 3d^5\), having unpaired electrons, indicating paramagnetic behavior.
   • \(K_4[FeCl_6]\):
     • Assuming Fe(II), the electronic configuration becomes \([Ar] 3d^6\). This also results in unpaired electrons, thus it is paramagnetic.
2. Based on the electronic configurations and logic, the only complex that exhibits diamagnetism due to all paired electrons is \(K_3[Cu(CN)_4]\).

Therefore, the metal complex that is diamagnetic is \(K_3[Cu(CN)_4]\).