# 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]

(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.

## Related Chapters

### Preparation Products

##### Knockout NEET July 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
##### Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
##### Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
##### Test Series NEET July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-