The tetrahedral [CoI_4]^{2-} and square plana | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. In $[CoI_4]^{2-}$,the oxidation state of $Co$ is $+2$ ($d^7$ configuration). Since $I^-$ is a weak field ligand and the geometry is tetrahedr

Vedclass · Accepted Answer

high spin,low spin

Vedclass · Answer

(B) $1$. In $[CoI_4]^{2-}$,the oxidation state of $Co$ is $+2$ ($d^7$ configuration). Since $I^-$ is a weak field ligand and the geometry is tetrahedral,the crystal field splitting energy $(\Delta_t)$ is small. Thus,it is a high spin complex.$2$. In $[PdBr_4]^{2-}$,the oxidation state of $Pd$ is $+2$ ($d^8$ configuration). $Pd^{2+}$ is a $4d$ series metal ion. For $4d$ and $5d$ series metal ions,the crystal field splitting energy $(\Delta_o)$ is always large enough to cause pairing of electrons,regardless of the ligand strength. Therefore,square planar complexes of $Pd^{2+}$ are always low spin.

The tetrahedral $[CoI_4]^{2-}$ and square planar $[PdBr_4]^{2-}$ complex ions are respectively:

Similar Questions

In the coordination complex $[CoF_6]^{3-}$,the number of unpaired electrons is equal to ......

Which of the following complex compounds exhibits cis-trans isomerism?

Number of stereoisomers possible for the complexes, $[CrCl_3(py)_3]$ and $[CrCl_2(ox)_2]^{3-}$ are respectively $(py = \text{pyridine}, ox = \text{oxalate})$

In the nitroprusside ion,iron and $NO$ exist as $Fe^{II}$ and $NO^{+}$ instead of $Fe^{III}$ and $NO$. Which method is used to confirm this?

The complex with a maximum number of stereoisomers is

Vedclass Test Series

Exam Paper Generator

Online Exam Module