In the complex cis-[Cr(en)_2Cl_2]^+,the oxida | vedclass

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

Question

(C) The complex is $[Cr(en)_2Cl_2]^+$. Let the oxidation state of $Cr$ be $x$.$x + 2(0) + 2(-1) = +1$$x - 2 = +1 \Rightarrow x = +3$.$Cr$ is a $3d$ tr

Vedclass · Accepted Answer

$+3, 3d$ and $6$

Vedclass · Answer

(C) The complex is $[Cr(en)_2Cl_2]^+$. Let the oxidation state of $Cr$ be $x$.$x + 2(0) + 2(-1) = +1$$x - 2 = +1 \Rightarrow x = +3$.$Cr$ is a $3d$ transition metal,so its valence shell involves $3d$ orbitals.The ligand $en$ (ethylenediamine) is bidentate (coordination number $2$) and $Cl^-$ is monodentate (coordination number $1$).Coordination number $= 2(2) + 2(1) = 4 + 2 = 6$.Thus,the values are $+3, 3d$ and $6$.

In the complex $cis-[Cr(en)_2Cl_2]^+$,the oxidation state of $Cr$,the type of $d$-orbital involved in the valence shell,and the coordination number are respectively:

Similar Questions

Among the following,the species that is both tetrahedral and diamagnetic is:

Considering $H_2O$ as a weak field ligand,the number of unpaired electrons in $[Mn(H_2O)_6]^{2+}$ will be (Atomic number of $Mn = 25$).

When $(n-1)d$ orbitals participate in the hybridization of a coordination complex,the resulting complex is known as:

Which of the following is not a square planar complex?

The metal $Ni$ $(Z = 28)$ combines with a unidentate ligand $X^-$ to form a paramagnetic complex $[NiX_4]^{2-}$. What are its geometry and the number of unpaired electrons in $Ni$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module