A) \[s{{p}^{3}}{{d}^{2}}\]
B) \[s{{p}^{3}}\]
C) \[{{d}^{2}}s{{p}^{3}}\]
D) \[s{{p}^{3}}d\]
Correct Answer: A
Solution :
In \[{{K}_{4}}Co{{F}_{a}}\] the hybrid state of \[Co\] is \[s{{p}^{3}}{{d}^{2}}\] giving a high spin complex.
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