A) 4 and 4
B) 0 and 2
C) 2 and 4
D) 0 and 4
Correct Answer: D
Solution :
In both \[{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}\] and \[{{[Co{{F}_{6}}]}^{3-}}\], \[Co\] is present as \[C{{o}^{3+}}\] Thus, the electronic configuration of \[Co\] is \[_{27}Co=[Ar]\,\,3{{d}^{7}},4{{s}^{2}}\] \[_{27}C{{o}^{3+}}=[Ar]\,\,3{{d}^{6}},4{{s}^{0}}\] In case of \[{{[Co{{(N{{H}_{3}})}_{7}}]}^{3+}}\], \[N{{H}_{3}}\] is a strong field ligand, so pairing of electrons in 3d-orbital takes place. \[_{27}C{{o}^{3+}}=[Ar]3{{d}^{6}},4{{s}^{0}}\] In \[{{[Co{{F}_{6}}]}^{3-}}\], F is a weak field ligand. Thus not cause pairing. Hence, \[_{27}C{{o}^{3+}}=[Ar]\,\,3{{d}^{6}},4{{s}^{0}}\]You need to login to perform this action.
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