question_answer

The number of unpaired electrons calculated in  \[{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}\]and \[{{[Co({{F}_{6}})]}^{3-}}\] are

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, \[_{27}C{{o}^{3+}}=[Ar]\,\,3{{d}^{6}},4{{s}^{0}}\]