Reaction of [Co(H_2O)_6]^{2+} with excess amm | vedclass

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(B) The reaction of $[Co(H_2O)_6]^{2+}$ with excess $NH_3$ in the presence of oxygen leads to the oxidation of $Co^{2+}$ to $Co^{3+}$,forming the comp

Vedclass · Accepted Answer

$6$

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(B) The reaction of $[Co(H_2O)_6]^{2+}$ with excess $NH_3$ in the presence of oxygen leads to the oxidation of $Co^{2+}$ to $Co^{3+}$,forming the complex $[Co(NH_3)_6]^{3+}$.$NH_3$ is a strong field ligand,which causes pairing of electrons,resulting in a low-spin octahedral complex.The electronic configuration of $Co^{3+}$ is $3d^6$.In an octahedral field,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.For a $d^6$ low-spin complex,all $6$ electrons occupy the $t_{2g}$ orbitals,giving the configuration $t_{2g}^6 e_g^0$.Thus,the number of electrons in the $t_{2g}$ orbitals is $6$.

Reaction of $[Co(H_2O)_6]^{2+}$ with excess ammonia and in the presence of oxygen results into a diamagnetic product. Number of electrons present in $t_{2g}$-orbitals of the product is $.....$ .

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