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The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion \mathrm{(M^{z+})} are \mathrm{-0.8 \Delta_0} and \mathrm{3.87\ BM} respectively. Identify \mathrm{(M^{z+})} :
 
Option: 1 \mathrm{V^{3+}}
Option: 2 \mathrm{Co}^{2+}
Option: 3 \mathrm{Cr}^{3+}
Option: 4 \operatorname{Mn}^{4+}

Answers (1)

best_answer

Given,

\mathrm{CFSE= -0. 8\Delta _{0}}

\mathrm{\mu = 3.87\, BM}.

This means there are 3 unpaired elections and the configuration is \mathrm{d^{7}}.


Hence, the ion is \mathrm{Co^{2+}}

Thus , the correct answer is Option (2)

Posted by

sudhir.kumar

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