Question:

The spin only magnetic moment value (in $ $B.M.$ $ unit) of $ Cr(CO)_{6} $ is

Updated On: Aug 15, 2022
  • $0$
  • $ 2.84 $
  • $ 4.90 $
  • $ 5.92 $
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The Correct Option is A

Solution and Explanation

In $Cr(CO)_{B}$, $Cr$ is present as $Cr^{0}$ $Cr$ or $Cr^{0}=[Ar] 3d^{5} \, 4S^{-1}$ $CO$ being strong field ligand pair up the unpaired electrons of $Cr$
In this condition, number of unpaired electrons, $n=0$ Magnetic moment, $\mu= \sqrt{n\left(n+2\right)}$ $=\sqrt{0\left(0+2\right)}$ $=0 BM$
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Concepts Used:

Coordination Compounds

A coordination compound holds a central metal atom or ion surrounded by various oppositely charged ions or neutral molecules. These molecules or ions are re-bonded to the metal atom or ion by a coordinate bond.

Coordination entity:

A coordination entity composes of a central metal atom or ion bonded to a fixed number of ions or molecules.

Ligands:

A molecule, ion, or group which is bonded to the metal atom or ion in a complex or coordination compound by a coordinate bond is commonly called a ligand. It may be either neutral, positively, or negatively charged.