Coordination Compounds 2 Question 32
34. Among the following complexes $(K-P)$,
(2011)
$K _3[Fe(CN) _6]$ $(K),[Co(NH _3) _6] $ $Cl _3(L)$,
$Na _3[Co(\text { ox }) _3]$ $(M),[Ni(H _2 O) _6]$ $Cl _2(N)$,
$K _2[Pt(CN) _4]$ $(O),[Zn(H _2 O) _6]$ $(NO _3) _2(P)$
the diamagnetic complexes are
(a) $K, L, M, N$
(b) $K, M, O, P$
(c) $L, M, O, P$
(d) $L, M, N, O$
Show Answer
Answer:
Correct Answer: 34. (c)
Solution:
- For a diamagnetic complex, there should not be any unpaired electron in the valence shell of central metal.
In $K _3[Fe(CN) _6], Fe$ (III) has $d^{5}$-configuration (odd electrons), hence it is paramagnetic.
In $[Co(NH _3) _6] Cl _3$, Co (III) has $d^{6}$-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 $t _{2 g}$ level, diamagnetic.
In $[Ni(H _2 O) _6] Cl _2, Ni$ (II) has $3 d^{8}$-configuration and $H _2 O$ is a weak ligand, hence
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 with all the ten electrons spin paired, hence diamagnetic.
BETA
