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Which of the following complexes exhibits the highest paramagnetic behaviour ?
Where gly = glycine, en = ethylenediamine and bpy = bipyridyl moieties
(Atomic number Ti = 22, V = 23, Fe = 26, Co = 27)
Where gly = glycine, en = ethylenediamine and bpy = bipyridyl moieties
(Atomic number Ti = 22, V = 23, Fe = 26, Co = 27)
A
$[Co(ox)_2(OH)_2]^-$
B
$[Ti(NH_3)_6]^{3+}$
C
$[V(gly)_2(OH)_2(NH_3)_2]^+$
D
$[Fe(en)(bpy)(NH_3)_2]^{2+}$
Detailed Solution
Paramagnetic character depends on the number of unpaired electrons: $\mu = \sqrt{n(n + 2)}$ BM. More the number of unpaired electrons, more will be the paramagnetic character.
$[Co(ox)_2(OH)_2]^-$: oxalate is $-2$ and hydroxide is $-1$, so $x + 2(-2) + 2(-1) = -1 \Rightarrow x = +5$. $Co^{5+}$ is $3d^4$, which has 4 unpaired electrons (the source remarks that this unusually high oxidation state is not really feasible).
$[Ti(NH_3)_6]^{3+}$: the oxidation state of Ti is +3. $Ti^{3+}$ is $3d^1$ and has just one unpaired electron.
$[V(gly)_2(OH)_2(NH_3)_2]^+$: glycinate is $-1$ and hydroxide is $-1$, so $x + 2(-1) + 2(-1) = +1 \Rightarrow x = +3$. $V^{3+}$ is $3d^2$, so the number of unpaired electrons is 2.
$[Fe(en)(bpy)(NH_3)_2]^{2+}$: all ligands are neutral, so the oxidation state of Fe is +2 ($3d^6$). With the strong field ligands en, bpy and $NH_3$ the electrons pair up ($t_{2g}^6$), leaving no unpaired electron in the low spin state.
Number of unpaired electrons: Co complex = 4, Ti complex = 1, V complex = 2.
So the complex taken to show the highest paramagnetic behaviour is $[Co(ox)_2(OH)_2]^-$.
$[Co(ox)_2(OH)_2]^-$: oxalate is $-2$ and hydroxide is $-1$, so $x + 2(-2) + 2(-1) = -1 \Rightarrow x = +5$. $Co^{5+}$ is $3d^4$, which has 4 unpaired electrons (the source remarks that this unusually high oxidation state is not really feasible).
$[Ti(NH_3)_6]^{3+}$: the oxidation state of Ti is +3. $Ti^{3+}$ is $3d^1$ and has just one unpaired electron.
$[V(gly)_2(OH)_2(NH_3)_2]^+$: glycinate is $-1$ and hydroxide is $-1$, so $x + 2(-1) + 2(-1) = +1 \Rightarrow x = +3$. $V^{3+}$ is $3d^2$, so the number of unpaired electrons is 2.
$[Fe(en)(bpy)(NH_3)_2]^{2+}$: all ligands are neutral, so the oxidation state of Fe is +2 ($3d^6$). With the strong field ligands en, bpy and $NH_3$ the electrons pair up ($t_{2g}^6$), leaving no unpaired electron in the low spin state.
Number of unpaired electrons: Co complex = 4, Ti complex = 1, V complex = 2.
So the complex taken to show the highest paramagnetic behaviour is $[Co(ox)_2(OH)_2]^-$.