In the complex ion ML_6^{n+},M^{n+} has five | vedclass

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(B) In an octahedral complex $ML_6^{n+}$,the $d$-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).
Given that $M^{n+}$

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$5$

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(B) In an octahedral complex $ML_6^{n+}$,the $d$-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).
Given that $M^{n+}$ has $5$ $d$-electrons and $L$ is a weak field ligand,the pairing energy is greater than the crystal field splitting energy $(\Delta_o)$.
According to Hund's rule and the weak field nature,the electrons will occupy the orbitals singly before pairing starts.
Thus,the configuration is $t_{2g}^3 e_g^2$.
This results in $5$ unpaired electrons.

In the complex ion $ML_6^{n+}$,$M^{n+}$ has five $d$-electrons and $L$ is a weak field ligand. According to crystal field theory,the magnetic property of the complex ion is due to how many unpaired electrons?

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