Which of the following complexes exhibits the highest paramagnetic behaviour? [AIPMT (S) 2008] |
where gly = glycine, en = ethylenediamine and bpy = bipyridyl moities |
(At no : Ti = 22, V = 23, Fe = 26, Co = 27) |
A) \[{{[V{{(gly)}_{2}}{{(OH)}_{2}}{{(N{{H}_{3}})}_{2}}]}^{+}}\]
B) \[{{[Fe(en)(bpy){{(N{{H}_{3}})}_{2}}]}^{2+}}\]
C) \[{{[Co{{(OX)}_{2}}{{(OH)}_{2}}]}^{-}}\]
D) \[{{[Ti{{(N{{H}_{3}})}_{6}}]}^{3+}}\]
Correct Answer: C
Solution :
[c] Key Idea: Greater is the number of unpaired electrons, larger is the paramagnetism. |
\[{{[V{{(gly)}_{2}}{{(OH)}_{2}}{{(N{{H}_{3}})}_{2}}]}^{+}}\] |
\[{{V}_{23}}=[Ar]4{{s}^{2}},3{{d}^{3}}\] |
Oxidation state of V in |
\[{{[V{{(gly)}_{2}}{{(OH)}_{2}}{{(N{{H}_{3}})}_{2}}]}^{+}}\] is |
\[x+(-1)\times 2+(-1)\times 2+(0)\times 2=+1\] |
\[x=+5\] |
\[{{V}^{5+}}=[Ar]3{{d}^{0}}\](No unpaired electron) |
\[{{[Fe(en)(bpy){{(N{{H}_{3}})}_{2}}]}^{2+}}\] |
\[F{{e}_{26}}=[Ar]4{{s}^{2}},3{{d}^{6}}\] |
Oxidation state of Fe in \[{{[Fe(en)(bpy){{(N{{H}_{3}})}_{2}}]}^{2+}}\]is |
\[x+(0)+(0)+(0)\times 2=+2\]\[x=+2\] |
\[x=+2\] |
\[F{{e}^{2+}}=[Ar]3{{d}^{6}}\] |
But en, \[bpy\] and \[N{{H}_{3}}\] all are strong field ligands, so pairing occurs, thus no unpaired electrons. |
\[{{[Co{{(OX)}_{2}}(OH)]}_{2}}{{]}^{-}}\] |
\[C{{o}_{27}}=[Ar]4{{s}^{2}},3{{d}^{7}}\] |
Oxidation state of Co in \[{{[Co{{(OX)}_{2}}{{(OH)}_{2}}]}^{-}}\] is |
\[x+(-2)\times 2+(-1)\times 2=-1\] |
\[x-6=-1\] |
\[x=+5\] |
\[C{{o}^{5+}}=[Ar]3{{d}^{4}}\](4 unpaired electrons) |
\[{{[Ti{{(N{{H}_{3}})}_{6}}]}^{3+}}\]\[T{{i}_{22}}=[Ar]4{{s}^{2}},3{{d}^{2}}\] |
Oxidation state of Ti in \[{{[Ti{{(N{{H}_{3}})}_{6}}]}^{3+}}\] is +3 thus it contains 1 unpaired electron. |
Hence, \[{{[Co{{(OX)}_{2}}{{(OH)}_{2}}]}^{-}}\]has highest paramagnetic behaviour. |
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