Question

Which of the following complexes exhibits the highest paramagnetic behaviour?
where gly=glycine, en=ethylenediamine and bipy =bipyridyl
(At. no. `Ti=22, V=23, Fe=26, Co=27`)
A. `[Ti(NH_(3))_(6)]^(3+)`
B. `[V(gly)_(2)(OH)_(2)(NH_(3))_(2)]^(+)`
C. `[Fe(en)(bpy)(NH_(3))_(2)]^(2+)`
D. `[Co(o x)_(2)(OH)_(2)]^(-)`

Answer 1

In `[Ti(NH_(3))_(6)]^(3+)` , the oxidation state of Ti is `+3` . `Ti(3d^(2)4s^(2))implies Ti^(3+) (3d^(1))implies` one unpaired electron In `[V(gly)_(2)(OH)_(2)]^(+)` ,Let the oxidation no. of V be x, then `(x) + 2(-1) + 2 2(0) = + 1`
`x = +5`
`V(3d^(3)4s^(2))implies V^(5+) (3d^(0)4s^(0))implies` zero unpaired electrons In `[Fe(en)(bpy)(NH_(3))_(2)]^(2+)` , let the oxidation statre of Fe be x, then `(x) + (0) + (0) 2(0) = +2`
`x = +2`
`Fe(d^(6)4s^(2))implies Fe^(2+) (3d^(6)`. Since en, bpy are strong field ligands, all the d electron are paired. In `[Co(o x)_(2)(OH)_(2)]^(-)` , let the oxidation number of Co be x, then `(x) + 2(-2) + 2(-1) = -1`
`x = +5`
`Co(3d^(7)4s^(2)) implies Co^(5) (3d^(4))` . As ox and `OH^(-)` are weak field ligands, the four d electrons remain unpaired, leading to highest paramagnetic behaviour.