Question

Identify the most likely transition metal M:

a)   K3[M(CN)6], in which M is a first-series transition metal and the complex has 3 unpaired electrons.

b)   [M(H2O)6]3+, in which M is a second-series transition metal and LFSE = –2.4 Do.

c)   Tetrahedral [MCl4]–, which has 5 unpaired electrons and M is a first-series transition metal.

d)   Square planar [MCl2(NH3)2], in which M is a d8 third-series transition metal.

Answer 1

(a) In K3[M(CN)6] , M is in +3 oxidation state and according to the question it has 3 unpaired electrons. We also know that CN^- is a strong field ligand which can pair up unpaired electrons during formation of either inner or outer sphere complex. Possible candidates are Cr(4s²3d⁶) and Ni(4s²3d⁸). We also know that Ni doesnot show +3 oxidation state. Thus the only other option is Cromium(Cr)

(b) Metal has oxidation state of +3 since water is a neutral ligand. Given that LFSE = -2.4 and belongs to second-series transition metal. This is possible when t_2g set has 6 electrons in a low spin complex i.e                     (-0.4 x 6= -2.4). This means the metal must have electronic configuration of 5s²4d⁷ i.e Rhodium (Rh)

(c) In [MCl₄]^- , M has oxidation state of +3 i.e it has lost 3 electrons. According to question, it has 5 unpaired electron. So the metal M belongs to d⁶ first tansition series. Since Cl is also a weak field ligand , So the M = Iron (Fe)

(d) Platinum (Pt) belongs to d⁸ third transition series.