Question

In which of the following octahedral complexes of Co(At No. 27) will the magnitude of ${\Delta }_0$ (CFSE in octahedral field) be the highest?

• A. $\left[Co\right(N{H}_3{)}_6{]}^{3+}$
• B. $\left[Cr\right(CN{)}_6{]}^{3-}$
• C. $\left[Co\right(C_2{O}_4{)}_3{]}^{3-}$
• D. $\left[Co\right(H_{2}O{)}_6{]}^{3+}$

Answer 1

Answer: (A)

$\left[Co\right(N{H}_3{)}_6{]}^{3+}$

In the given complexes the one which will have higher CFSE value will be $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ because the ground oxidation state of Co is $3d^{7}4s^2$ and in the complex Co(3+) will have an electronic configuration of $3d^{6}4s^0$. Now, since the $N{H}_3$ is a strong field ligand which will cause the pairing of electrons and the configuration will be $t_2{g}^{6}e{g}^0$ and the CFSE = -0.4${\Delta }_o$x6= -2.4${\Delta }_o$ which is maximum among the given complexes.