The calculated spin-only magnetic moment values in \(\mathrm{BM}\) for \(\left[\mathrm{FeCl}_4\right]^{-}\)and \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\) are

• Correct answer
  A

  \(5.9 \mathrm{~BM},~ 1.732 \mathrm{~BM}~\)
• B

  \(4.89 \mathrm{~BM},~ 1.732 \mathrm{~BM}~\)
• C

  \(3.87 \mathrm{~BM},~ 1.732 \mathrm{~BM}~\)
• D

  \(1.732 \mathrm{~BM},~ 2.82 \mathrm{~BM}~\)

Solution

Hint : \(\mathrm{Fe}^{+3}=[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0\) \(\left[\mathrm{FeCl}_4\right]^{-}\)
No pairing as \(\mathrm{Cl}^{-}\)is weak field ligand, hence have five unpaired electron \((n=5)\). \(\begin{aligned} & \therefore \quad \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} \text { B.M }=\sqrt{5(5+2)} \text { B.M }=5.9 \text { B.M } \\ & {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-3}}\end{aligned}\)
Pairing takes place as \(\mathrm{CN}^{-}\)is strong field ligand but has one unpaired electron \((n=1)\) \(\therefore \mu=\sqrt{n(n+2)} \text { B.M }=\sqrt{1(1+2)}=1.732 \text { B.M }\)

Hence, the correct answer is option: 'A'.