Which of the following complexes exhibits the highest paramagnetic behaviour? Where gly = glycine, en =…
Which of the following complexes exhibits the highest paramagnetic behaviour?
Where gly = glycine, en = ethylenediamine and bpy = bipyridyl moities). (At. number $\mathrm{Ti}$ $=22, \mathrm{~V}=23, \mathrm{Fe}=26, \mathrm{Co}=27$ )
The electronic configuration of
\(V(23)=[A r] 4 s^2, 3 d^3\)
Let in \(\left[V(\mathrm{gly})_2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+}\)oxidation state of \(V\) is \(x\).
\(\begin{aligned}
& x+(-1) \times 2+(-1) 2+(0 \times 2)=+1 \\
& x=+5 \\
& V^{5+}=[A r] 4 s^0, 3 d^0 \text { (no unpaired electrons) }
\end{aligned}\)
The electronic configuration of
\(F e(26)=[A r] 4 s^2, 3 d^6\)
Let the oxidation state of \(F e\) in \(\left[\mathrm{Fe}(e n)(p p y)\left(\mathrm{NH}_3\right)_2\right]^{2+}\) is \(x\)
\(\begin{aligned}
& {[x+(0)+(0)+(0) \times 2]=+2} \\
& x=+2 \\
& F e^{2+}=[\mathrm{Ar}] 3 d^6(\because 4 \text { unpaired electron })
\end{aligned}\)
but, bpy, en and \(\mathrm{NH}_3\) all are strong field ligands, so pairing occurs and thus, \(F e^{2+}\) contains no unpaired electron.
The electronic configuration of
\(C o(27)=[A r] 4 s^2, 3 d^7\)
Oxidation state of Co in \(\left[\mathrm{Co}(\mathrm{Ox})_2(\mathrm{OH})_2\right]^{-}\)
\(\begin{aligned}
& x+(-2) \times 2+(-1) \times 2=-1 \\
& x=+5
\end{aligned}\)
\(C o^{5+}=[A r], 3 d^4\) [4 unpaired electrons]
ox and \(O H\) are weak field ligands.
The electronic configuration of
\(T i(22)=[A r] 4 s^2, 3 d^2\)
Oxidation state of \(T i\) in \(\left(T i\left(\mathrm{NH}_3\right)_6\right]^{3+}\) is 3.
\(T i^{3+}=[A r] 3 d^1 \text { (one unpaired electron) }\)
Hence, complex \(\left[\mathrm{Co}(\mathrm{Ox})_2(\mathrm{OH})_2\right]^{-}\) has maximum number of unpaired electrons, thus show maximum paramagnetism.