Coordination Compounds 2 Question 32
####34. Among the following complexes $(K-P)$,
(2011)
$K _3\leftFe(CN) _6\right,\left[Co\left(NH _3\right) _6\right] Cl _3(L)$,
$Na _3\leftCo(\text { ox }) _3\right,\left[Ni\left(H _2 O\right) _6\right] Cl _2(N)$,
$K _2\leftPt(CN) _4\right,\left[Zn\left(H _2 O\right) _6\right]\left(NO _3\right) _2(P)$
the diamagnetic complexes are
(a) $K, L, M, N$
(b) $K, M, O, P$
(c) $L, M, O, P$
(d) $L, M, N, O$
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Answer:
Correct Answer: 34. (b)
cobalt (III) chlorid
Solution:
- For a diamagnetic complex, there should not be any unpaired electron in the valence shell of central metal.
In $K _3\left[Fe(CN) _6\right], Fe$ (III) has $d^{5}$-configuration (odd electrons), hence it is paramagnetic.
In $\left[Co\left(NH _3\right) _6\right] Cl _3$, Co (III) has $d^{6}$-configuration in a strong ligand field, hence all the electrons are paired and the complex is diamagnetic.
In $Na _3\left[Co(\text { ox }) _3\right]$, 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 $\left[Ni\left(H _2 O\right) _6\right] Cl _2, Ni$ (II) has $3 d^{8}$-configuration and $H _2 O$ is a weak ligand, hence
In $K _2\left[Pt(CN) _4\right]$, $Pt(II)$ has $d^{8}$-configuration and $CN^{-}$is a strong ligand, hence all the eight electrons are spin paired. Therefore, complex is diamagnetic.
In $\left[Zn\left(H _2 O\right) _6\right]\left(NO _3\right) _2$, $Zn$ (II) has $3 d^{10}$ configuration with all the ten electrons spin paired, hence diamagnetic.
