SATHEE: Coordination Compounds 2 Question 12

Coordination Compounds 2 Question 12

####13. The metal $d$ -orbitals that are directly facing the ligands in $K_{3} [C o (C N)_{6}]$ are

(2019 Main, 12 Jan I)

(a) $d_{x z}, d_{y z}$ and $d_{z^{2}}$

(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 $K_{3} [C o (C N)_{6}]$ , Co have +3 oxidation state and electronic configuration of $C o^{3 +}$ is $[A r]_{18} 3 d^{6}$ .

As, $C N^{-}$ is a strong field ligands so it pairs up the $d e^{-} s$

$\begin{aligned} (6e- pairs donated & by 6 C N^{-} ligands) \end{aligned}$

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 $(d_{x^{2} - y^{2}}$ and $d_{z^{2}})$ 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 $K_{3} [C o (C N)_{6}]$ .

Coordination Compounds 2 Question 12

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Coordination Compounds 2 Question 12

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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