Coordination Compounds 2 Question 12
####13. The metal $d$-orbitals that are directly facing the ligands in $K _3\left[Co(CN) _6\right]$ are
(2019 Main, 12 Jan I)
(a) $d _{x z}, d _{y z}$ and $d _{z^{2}}$
(c) $d _{x y}, d _{x z}$ and $d _{y z}$
(b) $d _{x^{2}-y^{2}}$ and $d _{z^{2}}$
(d) $d _{x z}$ and $d _{x^{2}-y^{2}}$
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Solution:
- In $K _3\left[Co(CN) _6\right]$, Co have +3 oxidation state and electronic configuration of $Co^{3+}$ is $[Ar] _{18} 3 d^{6}$.
As, $CN^{-}$is a strong field ligands so it pairs up the $d e^{-} s$
$$ \begin{aligned} & \text { (6e- pairs donated } \ & \text { by } 6 CN^{-} \text {ligands) } \end{aligned} $$
In an octahedral complex, the metal is at the centre of the octahedron and the ligands are at the six corners. The lobes of the $e _g$ orbitals $\left(d _{x^{2}-y^{2}}\right.$ and $\left.d _{z^{2}}\right)$ point along the axes $x, y$ and $z$ under the influence of an octahedral field, the $d$-orbitals split as follow.
As the $d$-orbitals, i.e. $d _{x^{2}-y^{2}}$ and $d _{z^{2}}$ are vacant. Hence, these both orbitals are directly facing the ligands in $K _3\left[Co(CN) _6\right]$.
