| 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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