Question

Which of the following is the high spin complex?

• A. $\left[ Cr ( gly )_{3}\right]$
• B. $\left[ CoBr _{2} Cl _{2}( SCN )_{2}\right]^{3-}$
• C. $\left[ Fe ( CN )_{6}\right]^{3-}$
• D. $Na \left[ PtBrCl \left( NO _{2}\right)_{2}\left( NH _{3}\right)_{2}\right]$

Answer 1

Answer: (B)

$\left[ CoBr _{2} Cl _{2}( SCN )_{2}\right]^{3-}$ will form spin complex because $Br^-$ and $Cl^-$ are weak field ligands and they will not compell the d-electrons to be paired.