Question

(i)${{K}_{4}}[Fe{{(CN)}_{6}}]$(ii) ${{K}_{3}}[Cr{{(CN)}_{6}}]$(iii)${{K}_{3}}[Fe{{(CN)}_{6}}]$(iv)${{K}_{2}}[Ni{{(CN)}_{4}}]$choose the complex which is paramagnetic [Kerala CET $2005$]
A. (i),(ii) and (iii)
B. (i), (iii) and (iv)
C. (ii) and (iii)
D. (i), (ii) and (iv)
E. (ii) and (iv)

Answer 1

Hint: The magnetic properties of a complex can be determined by the size of the atoms and their electronic configuration. The number of unpaired electrons in a particular complex indicates the paramagnetism of any complex. Here we have to find an unpaired electron from the electronic configuration of the central metal in the given complex.

Complete answer:Transition metals ($Sc,Ti,V,Cr,Mn,Ni$etc.) have a special ability to form magnets. Metal complexes that have paired electrons in their valence shell are called diamagnetic complexes and metal complexes that have unpaired electrons are called paramagnetic complexes. The last electron in transition metal complexes resides on the d-orbital and when there are unpaired electrons residing on the d-orbital is called the paramagnetic complex. With increasing unpaired electrons the paramagnetic effect increases.
Here we have four transition metal complexes. Cyanide $(C{{N}^{-}})$is a strong ligand and it causes the pairing of unpaired $3d-$electrons in an excited state. Let’s check their unpaired electrons one by one.
(i) ${{K}_{4}}[Fe{{(CN)}_{6}}]$
The atomic number of iron,$Fe=26$ and the electronic configuration of $Fe=[Ar]3{{d}^{6}}4{{s}^{2}}$
Here $F{{e}^{2+}}=[Ar]3{{d}^{6}}4{{s}^{0}}$
$Fe=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow & \uparrow & \uparrow & \uparrow \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   \uparrow \downarrow \\
\end{matrix}}}\,$ in ground state
$Fe=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & {} & {} \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   \uparrow \downarrow \\
\end{matrix}}}\,$ in excited state
$F{{e}^{2+}}=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & {} & {} \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   {} \\
\end{matrix}}}\,$
$F{{e}^{2+}}$has zero unpaired electron, hence diamagnetic.
 (ii)${{K}_{3}}[Cr{{(CN)}_{6}}]$
The atomic number of chromium, $Cr=24$and electronic configuration of $Cr=[Ar]3{{d}^{4}}4{{s}^{2}}$
Here $C{{r}^{3+}}=[Ar]3{{d}^{3}}4{{s}^{0}}$
$Cr=\underset{3d}{\mathop{\begin{matrix}
   \uparrow & \uparrow & \uparrow & \uparrow & {} \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   \uparrow \downarrow \\
\end{matrix}}}\,$ in ground state
$Cr=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow & \uparrow & {} & {} \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   \uparrow \downarrow \\
\end{matrix}}}\,$in excited state
$C{{r}^{3+}}=\underset{3d}{\mathop{\begin{matrix}
   \uparrow & \uparrow & \uparrow & {} & {} \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   {} \\
\end{matrix}}}\,$
$C{{r}^{3+}}$has three unpaired electrons in $3d-$orbital and hence paramagnetic.
(iii)${{K}_{3}}[Fe{{(CN)}_{6}}]$
The atomic number of iron,$Fe=26$ and the electronic configuration of $Fe=[Ar]3{{d}^{6}}4{{s}^{2}}$
Here $F{{e}^{3+}}=[Ar]3{{d}^{5}}4{{s}^{0}}$
$Fe=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow & \uparrow & \uparrow & \uparrow \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   \uparrow \downarrow \\
\end{matrix}}}\,$in ground state
$Fe=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & {} & {} \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   \uparrow \downarrow \\
\end{matrix}}}\,$in excited state
$F{{e}^{3+}}=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow \downarrow & \uparrow & {} & {} \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   {} \\
\end{matrix}}}\,$
$F{{e}^{3+}}$has one unpaired electron in $3d-$orbital and hence paramagnetic.
 (iv)${{K}_{2}}[Ni{{(CN)}_{4}}]$
Here $N{{i}^{2+}}=[Ar]3{{d}^{8}}4{{s}^{0}}$
$Ni=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \uparrow & \uparrow \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   \uparrow \downarrow \\
\end{matrix}}}\,$in ground state
$Ni=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & {} \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   \uparrow \downarrow \\
\end{matrix}}}\,$in excited state
$N{{i}^{2+}}=\underset{3d}{\mathop{\begin{matrix}
   \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & {} \\
\end{matrix}}}\,\underset{4s}{\mathop{\begin{matrix}
   {} \\
\end{matrix}}}\,$
$N{{i}^{2+}}$has no unpaired electron, hence it is diamagnetic.
Hence ${{K}_{3}}[Cr{{(CN)}_{6}}]$and ${{K}_{3}}[Fe{{(CN)}_{6}}]$are paramagnetic.

Hence, option (C) is correct.

Note: A permanent magnetic moment exists in paramagnetic substances and these metal complexes are strongly attracted by the magnetic field because of unpaired electrons present in them. Therefore to calculate the magnetic moment one must remember the atomic number of transition metals.