Find the spin magnetic moment ratio for complexes \([Cr(Cn)_{6}]^{-3}\) & \([Cr(H_{2}O)_{6}]^{+3}\)
Find the spin magnetic moment ratio for complexes \([Cr(Cn)_{6}]^{-3}\) & \([Cr(H_{2}O)_{6}]^{+3}\)
Solution and Explanation
The correct answer is 1:1
Spin only magnet magnetic for \([Cr(CN)_6]^{-3} (d^3)\) \(= \sqrt{15}\) B.M.
For \([Cr(H_2O)_6]^{+3} (d^3) = \sqrt{15}\) B.M.
\(\therefore \) the ratio is 1:1
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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.