A) I and IV only
B) II, III and IV only
C) I, III and IV only
D) All of these
Correct Answer: D
Solution :
(I) High CFSE in low spin \[{{d}^{5}}\] configuration. It is reported square planar and contains one unpaired electron. (II) \[{{\Delta }_{0}}\]increases with increase in charge on central metal ion. (III) \[4{{d}^{8}}\] configuration \[\to \] high CFSE compels for pairing of electrons and thus squares planar (i.e., \[ds{{p}^{2}}\] and diamagnetic) red - violet colour may be accounted for charge transfer (of triphenyl phosphine group). (IV) \[SC{{N}^{-}}\] is weak field ligand and thus, no empty d-orbital is available for \[ds{{p}^{2}}\] hybridization. As a result of this Co (II) complex is \[s{{p}^{2}}\] hybridized.You need to login to perform this action.
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