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 :
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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