A) \[[Cr{{(CO)}_{6}}]\]
B) \[[Fe{{(CO)}_{5}}]\]
C) \[{{[Fe{{(CO)}_{6}}]}^{4-}}\]
D) \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}\]
Correct Answer: D
Solution :
\[C{{r}^{3+}}\] has \[4s{}^\circ 3{{d}^{3}}\] electronic configuration with 3 unpaired electrons, hence paramagnetic. In other cases pairing of d-electrons take place in presence of strong field ligands such as \[CO\]or\[C{{N}^{-}}\]. In \[Cr{{(CO)}_{6}}\] molecule 12 electrons are contributed by \[CO\]group and it contain no odd electron
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