The atomic numbers of Cr and Fe are 24 and 26 | vedclass

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

Question

(D) $1$. $[Cr(CO)_6]$: $Cr$ is in $0$ oxidation state. Configuration is $3d^5 4s^1$. $CO$ is a strong field ligand,causing pairing. All electrons are

Vedclass · Accepted Answer

$[Cr(NH_3)_6]^{3+}$

Vedclass · Answer

(D) $1$. $[Cr(CO)_6]$: $Cr$ is in $0$ oxidation state. Configuration is $3d^5 4s^1$. $CO$ is a strong field ligand,causing pairing. All electrons are paired. Diamagnetic.$2$. $[Fe(CO)_5]$: $Fe$ is in $0$ oxidation state. Configuration is $3d^6 4s^2$. $CO$ is a strong field ligand,causing pairing. All electrons are paired. Diamagnetic.$3$. $[Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing. All electrons are paired. Diamagnetic.$4$. $[Cr(NH_3)_6]^{3+}$: $Cr$ is in $+3$ oxidation state. Configuration is $3d^3$. There are $3$ unpaired electrons in the $t_{2g}$ orbitals. Paramagnetic.

The atomic numbers of $Cr$ and $Fe$ are $24$ and $26$ respectively. Which of the following is paramagnetic in terms of electron spin?

Similar Questions

How many geometrical isomers will $[Pt(NH_3)(NH_2OH)(H_2O)(py)]^+$ form?

The number of geometric isomers possible for the complex $[CoL_2 Cl_2]^{-}$ where $L = H_2NCH_2CH_2O^{-}$ is

Other than $X$-ray diffraction,how can the following pairs of isomers be distinguished from one another?
$[Cr(NH_3)_6][Cr(NO_2)_6]$ and $[Cr(NH_3)_4(NO_2)_2][Cr(NH_3)_2(NO_2)_4]$

The $d-$electron configurations of $Cr^{2+}, Mn^{2+}, Fe^{2+}$ and $Co^{2+}$ are $d^4, d^5, d^6$ and $d^7$ respectively. Which one of the following will exhibit minimum paramagnetic behaviour?
(At. nos. $Cr = 24, Mn = 25, Fe = 26, Co = 27$)

Total number of relatively more stable isomer$(s)$ possible for octahedral complex $[Cu(en)_2(SCN)_2]$ will be $.........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module