Discussion Forum



1)

Among the following complexes

$(K-P),K_{3}[Fe(CN)_{6}](K),$

$[Co(NH_{3})_{6}Cl_{3}(L),Na_{3}[Co(ox)_{3}](M)$

$[Ni(H_{2}O)_{6}]Cl_{2}(N); K_{2}[Pt(CN)_{4}](O)$ and 

$[Zn(H_{2}O)_{6}](NO_{3})_{2}$(P)

 the diamagnetic complexes are


A) K, L, M, N

B) K, M, O, P

C) L, M, O, P

D) L, M, N, O

Answer:

Option C

Explanation:

For a diamagnetic complex,  there should not be any unpaired electron in the valence shell of central metral.In $K_{3}[Fe(CN)_{6}],Fe(III)$ has $d^{5}$- configration (odd electrons). Hence it is  paramagnetic. In $[Co(NH_{3})_{6}]Cl_{3},Co(III)$ has d6 configuration in a strong ligand field, hence  all the electrons are paired and the complex is diamagnetic  .In $Na_{3}[Co(ox)_{3}],Co(III)$ has $d^{6}$- configuration  and oxalate being a chelating  ligand, very strong ligand and all the six electrons remains paired in lower  $t_{2g}$  level. diamagnetic.In $[Ni(H_{2}O)_{6}]Cl_{2},Ni(II)$ has $3d^{8}$ -configuration and $H_{2}O$ ia a weak ligand hence 

 8122021689_d5.PNG

In $K_{2}[Pt (CN)_{4}]Pt(II)$ has $d^{8}$- configuration and $CN^{-}$  is a strong ligand, hence all the eight electrons are spin paired. Therefore, complex is diamagnetic

In $[Zn(H_{2}O)_{6}](NO_{3})_{2},Zn(II)$  has ${_{3}}d^{10}$ configuration spin paired , hence diamagnetic