Hybridization in [Co(NH3)6] is ?

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Solution

In [Co (NH3)6]3+, Co is in +3 oxidation state with the configuration 3d6. In the presence of NH3 a strong ligand, the 3d electrons pair up leaving two d-orbitals empty. Hence, the hybridization is d2sp3 forming an inner orbital octahedral complex.

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Q. What is the hybridization of

{[C O (N H_{3})_{6}]}_{3}^{3 +}

Q. Assertion :Complex ion

[C o (N H_{3})_{6}]^{2 +}

is readily oxidized to

[C o (N H_{3})_{6}]^{3 +}

Reason: Unpaired electron in complex ion

[C o (N H_{3})_{6}]^{2 +}

is present in 4p orbital.

Q. The complex ion

{[C o {(N H_{3})}_{6}]}_{6}^{+ 2}

is readily oxidized to

{[C o {(N H_{3})}_{6}]}_{6}^{+ 3} .

The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, $[C o (N H_{3})_{6}]^{3 +}, [C o (C N)_{6}]^{3 -}, [C o (H_{2} O)_{6}]^{3 +}$
$[C o (N H_{3})_{6}]^{3 +}, [C o (C N)_{6}]^{3 -}, [C o (H_{2} O)_{6}]^{3 +}$
(a) $[C o (C N)_{6}]^{3 -} > [C o (N H_{3})_{6}]^{3 +} > [C o (H_{2} O)_{6}]^{3 +}$
(b) $[C o (N H_{3})_{6}]^{3 +} > [C o (H_{2} O)_{6}]^{3 +} > [C o (N H_{3})_{6}]^{3 +}$
(c) $[C o (H_{2} O)_{6}]^{3 +} > [C o (N H_{3})_{6}]^{3 +} > [C o (C N)_{6}]^{3 -}$
(d) $[C o (C N)_{6}]^{3 -} > [C o (N H_{3})_{6}]^{3 +} > [C o (H_{2} O)_{6}]^{3 +}$

Oxidation number of cobalt in $[{C o ({N H}_{3})}_{6}] {C l}_{2} B r$ is:

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