Coordination Compounds - Result Question 56
####63. Which of the following complexes exhibits the highest paramagnetic behaviour?
[2008]
(a) $[V(gly)_2(OH)_2(NH_3)_2]^{+}$
(b) $[Fe(en)(bpy)(NH_3)_2]^{2+}$
(c) $[Co(Ox)_2(OH)_2]^{-}$
(d) $[Ti(NH_3)_6]^{3+}$
where gly = glycine, en = ethylenediamine and bpy $=$ bipyridyl moities)
(At.nosTi=22, V=23, $Fe=26, Co=27$ )
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Solution:
- (c) More the number of unpaired electrons present in a complex more is its paramagnetic behaviour.
$x+2(-1)+2(-1)+2(0)=+1$
$\Rightarrow x=+5$
$V^{5+}=[Ar] ;$ no unpaired electron
$x=+2$
$Fe^{2+}=[Ar] 3 d^{6}$
But due to presence of strong field ligands, en, bpy and $NH_3$; the electrons will try to pair up. Thus, the complex will contain no unpaired electron.
(c) $.\stackrel{e}{[Co}(\stackrel{-2}{O x})_2(\stackrel{-1}{OH})_2]^{-}$
$x+2(-2)+2(-1)=-1$
$\Rightarrow x=+5$
$Co^{5+}=[Ar] 3 d^{4}$
Thus, 4 unpaired electrons.
$x=+3$
$Ti^{3+}=[Ar] 3 d^{1}$
Thus, 1 unpaired electron.
Hence, option (c) is the correct answer.
To find unpaired electrons let us calculate the oxidation states of elements in each complex and then write the electronic configuration for that oxidation state to find the number of unpaired electrons in it.
