A) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{1}}\]
B) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{5}}3{{d}^{11}},4{{s}^{1}}\]
C) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{9}},4{{s}^{1}}\]
D) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{5}}3{{d}^{10}},4{{s}^{2}}\]
Correct Answer: A
Solution :
The expected electronic configuration of Cu(29) is as Cu(29) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},3{{d}^{9}},4{{s}^{2}}\] But the actual configuration is as Cu(29) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},3{{d}^{10}},4{{s}^{1}}\] This is attributed to the fact that completely filled electronic configuration (ie, \[{{d}^{10}}\]) has lower energy. Thus, to acquire increased stability one of the 45-electron goes into the nearby 3d-orbital. So, that 3d-orbital get completely filled in Cu configuration. The extra stability of completely filled electronic configuration is due to their symmetrical arrangement and large exchange energy.
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