A) \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}\]
B) \[[Fe{{(CO)}_{5}}]\]
C) \[{{[Fe{{(CN)}_{6}}]}^{4-}}\]
D) \[[Cr{{(CO)}_{6}}]\]
Correct Answer: A
Solution :
\[{{[Fe{{(CN)}_{6}}]}^{4}}^{-}\] is diametric (\[F{{e}^{2+}}\]has\[3{{d}^{6}}\]configuration and the 6 electro airs up in three d-orbitals followed by (\[{{d}^{2}}s{{p}^{3-}}\]hybridization). \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}\]is paramagnetic (\[C{{r}^{3+}}\]has \[3{{d}^{3}}\] con-figuration. Hybridization is \[{{d}^{2}}s{{p}^{3}}.\]Due to 3 unpaired electrons it is paramagnetic) \[[Cr{{(CO)}_{6}}:Cr(Z=25):{{[Ar]}^{18}}4{{s}^{1}},3{{d}^{5}}.\] The one 4s-electron pairs up with five 3d-electrons in three (d-orbitals. This is followed by \[{{d}^{2}}s{{p}^{3}}-\]hybridization to give octahedral complex. No unpaired electron and hence complex is diamagnetic. \[Fe{{(CO)}_{5}}:Fe(Z=26):{{[Ar]}^{18}}4{{s}^{2}},3{{d}^{6}}.\] The six electrons in o'-subshell pairs up in three d-orbitals. This is followed by \[{{d}^{2}}s{{p}^{3}}-\]hybridization to give octahedral geometry with one vacant hybridized orbital. The resulting shape of the complex is square-based pyramid. As there is no unpaired. electron, the complex is diamagnetic.
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