Using valence bond theory, predict the hybridization and magnetic charater of the following:

(a) [Co(NH_{3})_{6}]^{3+}

(b) [Ni(CO)_{4}]

[ At.no: Co = 27, Ni = 28]

 

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

(a) [Co(NH_{3})_{6}]^{3+}

Oxidation nimber of Co is 3+

 Electronic configuration of Co^{3+} is [Ar]3d^{6}4s^{0}

NH3 is a strong field ligand and hence forms inner spin complex.

            excited state

Hence it under goes d^{2}sp^{3} hybridization for coordination number 6. with octahedral geometry. Since there are no unpaired electrons [Co(NH_{3})_{6}]^{3+} is dimagnetic.

(b) [Ni(CO)_{4}]

Oxidation state of Ni in [Ni(CO)_{4}] is zero.

Electronic configuration of Ni : [Ar]3d^{8}4s^{2}.

         

(CO) ligand forms inner complex, since it is strongly field ligand.

Hence hybridization of Ni is sp^{3} with coordination number 4 and has tertahedral geometry. Since there are no unpaired electron, [Ni(CO)_{4}] is diamagnetic.

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