CBSE Class 12 · 2009 · All India · Coordination CompoundsShort Answer

Compare the following complexes with respect to their structural shapes, magnetic behaviour and the hybrid orbitals involved:
[Co(NH

_{3}

)

_{6}

]

^{3 +}

, [Cr(NH

_{3}

)

_{6}

]

^{3 +}

, Ni(CO)

_{4}

(Atomic numbers: Co = 27, Cr = 24, Ni = 28)

Solution

[Co(NH

_{3}

)

_{6}

]

^{3 +}

— Co

^{3 +}

^{6}

, strong-field NH

_{3}

pairs all: shape \textbf{octahedral}, hybridisation \textbf{d

^{2}

^{3}

}, \textbf{diamagnetic}.
[Cr(NH

_{3}

)

_{6}

]

^{3 +}

— Cr

^{3 +}

^{3}

: shape \textbf{octahedral}, hybridisation \textbf{d

^{2}

^{3}

}, \textbf{paramagnetic} (3 unpaired, μ ≈ 3.87 BM).
Ni(CO)

_{4}

— Ni(0) rearranged to 3d

^{10}

under CO: shape \textbf{tetrahedral}, hybridisation \textbf{sp

^{3}

}, \textbf{diamagnetic}.

CBSE Class 12 Board · 2009 · All India · Subjective