Q62 / 200 · NEET 2008 (Screening) · Chemistry
Which of the following complexes exhibits the highest paramagnetic behaviour?
where gly = glycine, en = ethylenediamine and bpy = bipyridyl moities
(At no: $\mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Fe}=26, \mathrm{Co}=27$ )
- A$\left[\mathrm{V}(\mathrm{gly})_2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+}$
- B$\left[\mathrm{Fe}(\mathrm{en})\left(\text { bpy) }\left(\mathrm{NH}_3\right)_2\right]^{2+}\right.$
- Correct answerC$\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2\right]^{-}$
- D$\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
Solution
Key Idea : Greater is the number of unpaired electrons, larger is the paramagnetism.
$\begin{aligned}
& {\left[\mathrm{V}(\mathrm{gly})_2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+}} \\
& \mathrm{V}_{23}=[\mathrm{Ar}] 4 s^2, 3 d^3
\end{aligned}$
Oxidation state of $\mathrm{V}$ in $\left[\mathrm{V}(\mathrm{gly})_2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+}$ is
$\begin{gathered}
x+(-1) \times 2+(-1) \times 2+(0) \times 2=+1 \\
x=+5 \\
\mathrm{~V}^{5+}=[\mathrm{Ar}] 3 d^0
\end{gathered}$
(No unpaired electron in option (A))
So, option (B),
$\begin{aligned}
& {\left[\mathrm{Fe}\left(\text { en) }(\text { bpy })\left(\mathrm{NH}_3\right)_2\right]^{2+}\right.} \\
& \mathrm{Fe}_{26}=[\mathrm{Ar}] 4 s^2, 3 d^6
\end{aligned}$
Oxidation state of $\mathrm{Fe}$ in $\left[\mathrm{Fe}(\mathrm{en}) \text { (bpy) }\left(\mathrm{NH}_3\right)_2\right]^{2+}$ is
$\begin{gathered}
x+(0)+(0)+(0) \times 2=+2 \\
x=+2 \\
\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6
\end{gathered}$
But en, bpy and $\mathrm{NH}_3$ all are strong field ligands so pairing occurs, thus no unpaired electrons.
So, option (C),
$\begin{gathered}
{\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH}]_2\right]^{-}} \\
\mathrm{Co}_{27}=[\mathrm{Ar}] 4 s^2, 3 d^7
\end{gathered}$
Oxidation state of $\mathrm{Co}$ in $\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2\right]^{-}$is
$x+(-2) \times 2+(-1) \times 2=-1$
$\begin{aligned}
& x-6=-1 \\
& x=+5 \\
& \mathrm{Co}^{5+}=[\mathrm{Ar}] 3 d^4 \\
& \text{(4 paired electrons)} \\
& \text{Now, option (D),} \\
& {\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}} \\
& \mathrm{Ti}_{22}=[\mathrm{Ar}] 4 s^2, 3 d^2
\end{aligned}$
$\mathrm{Ti}_{22}=[\mathrm{Ar}] 4 s^2, 3 d^2$
Oxidation state of Ti in $\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is +3, thus it contains 1 unpaired electron.
Hence, $\quad\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2\right]^{-}$has highest paramagnetic behaviour.
Hence, the correct answer is option: 'C'.