The number of unpaired electrons in the compl | vedclass

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

Question

(D) $1$. The atomic number of $Co$ is $27$. The electronic configuration is $[Ar] 3d^7 4s^2$.$2$. In the complex $[CoF_6]^{3-}$,let the oxidation stat

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) $1$. The atomic number of $Co$ is $27$. The electronic configuration is $[Ar] 3d^7 4s^2$.$2$. In the complex $[CoF_6]^{3-}$,let the oxidation state of $Co$ be $x$. Then $x + 6(-1) = -3$,which gives $x = +3$.$3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.$4$. $F^-$ is a weak field ligand,so it does not cause pairing of electrons.$5$. In the $3d^6$ configuration,the electrons are arranged as $t_{2g}^4 e_g^2$. This results in $4$ unpaired electrons.

Hybridisation	Geometry	Complex structure
$(A) \ sp^3$	$(i) \ \text{Square planar}$	$(p) \ [Fe(CN)_6]^{3-}$
$(B) \ d^2sp^3$	$(ii) \ \text{Tetrahedral}$	$(q) \ [ZnCl_4]^{2-}$
$(C) \ dsp^2$	$(iii) \ \text{Octahedral}$	$(r) \ [Ni(NH_3)_4]^{2+}$
-	$(iv) \ \text{Linear}$	$(s) \ [Ag(CN)_2]^-$

The number of unpaired electrons in the complex $[CoF_6]^{3-}$ is ...... . (Atomic number of $Co = 27$)

Similar Questions

Which of the following complexes is tetrahedral as well as diamagnetic?

The hybridisation involved in $[CoF_6]^{3-}$ is :

Which of the following complexes is paramagnetic and an inner orbital complex?

In Wilkinson's catalyst,the hybridization of the central metal ion and its shape are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module