Choose the complex which is paramagnetic:
(i) [Fe(H2O)6]2+
(ii) K3[Cr(CN)6]
(iii) K3[Fe(CN)6]
(iv) K2[Ni(CN)4]
Answer (2)
[Fe(H2O)6]²⁺ is the only paramagnetic complex among the given options due to the presence of unpaired electrons in its electron configuration. The other options are diamagnetic as they have all paired electrons. This makes [Fe(H2O)6]²⁺ the correct choice.
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To determine which complex is paramagnetic, we need to consider the electronic configuration and the presence of unpaired electrons in each complex.
[Fe(H2O)_6]^2+:
Iron in this complex is in the +2 oxidation state, so its electronic configuration is [ A r ] 3 d 6 .
When these electrons are placed in the field created by the water ligands, which are weak field ligands, they do not pair up completely.
This results in unpaired electrons, making [ F e ( H 2 O ) 6 ] 2 + paramagnetic.
K_3[Cr(CN)_6]:
Chromium here is in the +3 oxidation state, so the electronic configuration is [ A r ] 3 d 3 with CN^- as a strong field ligand.
Strong field ligands tend to pair up electrons, but [ C r ( CN ) 6 ] 3 − has unpaired electrons due to its d^3 configuration.
Therefore, it is paramagnetic.
K_3[Fe(CN)_6]:
Iron in this complex is in the +3 oxidation state, giving an electronic configuration of [ A r ] 3 d 5 with CN^- as a strong field ligand.
All electrons pair up, so this complex is diamagnetic (not paramagnetic).
K_2[Ni(CN)_4]:
Nickel is in the +2 oxidation state with an electronic configuration of [ A r ] 3 d 8 in a square planar arrangement.
All electrons pair up in this configuration, so it is diamagnetic.
Therefore, the paramagnetic complexes are (i) [ F e ( H 2 O ) 6 ] 2 + and (ii) K 3 [ C r ( CN ) 6 ] . However, if we are to choose a single paramagnetic complex based on the given options, (i) [ F e ( H 2 O ) 6 ] 2 + is a clear choice given its weak field ligands and unpaired electrons structure.
