Question

Electronic configurations of $\lbrack Cu(NH_{3})_{6}\rbrack^{2 +}$

On the basis of crystal field splitting theory is

• A. $t_{2g}^{4}e_{g}^{5}$
• B. $t_{2g}^{6}e_{g}^{3}$
• C. $t_{2g}^{9}g_{g}^{0}$
• D. $t_{2g}^{5}e_{g}^{4}$

Answer 1

Answer: (B)

$t_{2g}^{6}e_{g}^{3}$

Answer 2

In $\lbrack Cu(NH_{3})_{6}\rbrack^{2 +},$ oxidation state of Cu = +2,

$Cu^{2 +} = 3d^{9}$

$3d^{9} = t_{2g}^{6}e_{g}^{3}$