The number of unpaired d-electrons in [Co(H_2 | vedclass

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(C) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.In $[Co(H_2O)_6]^{3+}$,the oxidation state of $Co$ is

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(C) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.In $[Co(H_2O)_6]^{3+}$,the oxidation state of $Co$ is $+3$. Thus,the configuration of $Co^{3+}$ is $[Ar] 3d^6$.$H_2O$ is a weak field ligand,but for $Co^{3+}$ ($d^6$ system),the crystal field splitting energy is high enough that it forces pairing of electrons in the $t_{2g}$ orbitals.According to the crystal field splitting diagram,all $6$ electrons occupy the $t_{2g}$ orbitals in pairs.Therefore,the number of unpaired electrons is $0$.

The number of unpaired $d$-electrons in $[Co(H_2O)_6]^{3+}$ is $\qquad$

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