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

Which of the following statements is correct?

Which $d-$orbital of metal in $[Fe(CO)_5]$ is involved in hybridisation?

The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively $:$
$[Fe(CN)_6]^{3-}, [FeF_6]^{3-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$

Consider the reaction $CuSO_4 + KCN \to CuCN + Y \xrightarrow{\text{excess } K^{+}CN^{-}} Z$. Identify the correct statement regarding the products $Y$ and $Z$.

Which type of atomic orbitals hybridize to form a square planar complex?

Vedclass Test Series

Exam Paper Generator

Online Exam Module