A) \[Cu>Ag<Au\]
B) \[Cu>Ag>Au\]
C) \[Cu<Ag<Au\]
D) \[Au>Ag<Cu\]
Correct Answer: A
Solution :
| [a]\[Cu=3{{d}^{10}}4{{s}^{1}},\,\,Ag=4{{d}^{10}}5{{s}^{1}},\,\,Au=4{{f}^{14}}5{{d}^{10}}6{{s}^{1}}\] |
| In all the above given cases, unpaired s-electron has to be removed. In the case of Cu, a 4s electron is to be removed which is closer to the nucleus than the 5s electron of Ag. |
| So, IE, of \[Cu>I{{E}_{1}}\]of Ag. |
| However, in case of Au, due to imperfect screening effect of \[14{{e}^{-}}s\]of 4f orbitals, the nuclear charge increases and therefore \[5s\text{ }{{e}^{-}}\] of Au is more tightly held. |
| Thus, the order of \[I{{E}_{1}}\]is\[Cu>Ag<Au.\] |
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