The spin only magnetic moment value (in Bohr magneton units) of Cr ( CO )6 is

Question

The spin only magnetic moment value (in Bohr magneton units) of Cr ( CO )6 is (A) 0 (B) 2.84 (C) 4.90 (D) 5.92

Tardigrade Admin · Accepted Answer

In Cr ( CO )6, CO is a neutral ligand, so Cr is in zero oxidation state, but since it is a strong field ligand, it causes pairing of all the electrons. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/c53c4ae1c4ae2099f153b8c1ce383aee-.png" /> ∴ n(. No. of unpaired .e-)=0 ⇒ μ=√n(n+2)=0

Q. The spin only magnetic moment value (in Bohr magneton units) of $C r (CO)_{6}$ is

0

2.84

4.90

5.92

In $C r (CO)_{6}, CO$ is a neutral ligand, so $C r$ is in zero oxidation state, but since it is a strong field ligand, it causes pairing of all the electrons.

$∴ n ($ No. of unpaired $e^{-}) = 0 \Rightarrow μ = n (n + 2) = 0$

Q. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6​ is

0

2.84

4.90

5.92

In Cr(CO)6​,CO is a neutral ligand, so Cr is in zero oxidation state, but since it is a strong field ligand, it causes pairing of all the electrons. ∴n( No. of unpaired e−)=0⇒μ=n(n+2)​=0

Q. The spin only magnetic moment value (in Bohr magneton units) of $C r (CO)_{6}$ is

In $C r (CO)_{6}, CO$ is a neutral ligand, so $C r$ is in zero oxidation state, but since it is a strong field ligand, it causes pairing of all the electrons.

$∴ n ($ No. of unpaired $e^{-}) = 0 \Rightarrow μ = n (n + 2) = 0$