The hybridization of the central metal ion in | vedclass

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

Question

(D) In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $+3$. The atomic number of $Co$ is $27$,so its electronic configuration is $[Ar] 3d^7 4s^2$. For

Vedclass · Accepted Answer

$sp^3d^2$

Vedclass · Answer

(D) In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $+3$. The atomic number of $Co$ is $27$,so its electronic configuration is $[Ar] 3d^7 4s^2$. For $Co^{3+}$,the configuration is $[Ar] 3d^6$. $F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals. Therefore,the $3d$ electrons remain unpaired,and the hybridization involves the outer $4d$ orbitals. The hybridization is $sp^3d^2$.

The hybridization of the central metal ion in $[CoF_6]^{3-}$ is ....

Similar Questions

Which of the following is an incorrect pair?

Which of the following complexes has its hybridisation correctly matched?

The hybridization states of the central atom in the complexes $[Fe(CN)_6]^{3-}$,$[Fe(CN)_6]^{4-}$ and $[Co(NO_2)_6]^{3-}$ are

In $[NiCl_4]^{2-}$,the number of unpaired electrons is

Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex $[NiX_4]^{-2}.$ The number of unpaired electron$(s)$ in the nickel and geometry of this complex ion are,respectively $:$

Vedclass Test Series

Exam Paper Generator

Online Exam Module