What is the electron configuration of Au+?

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What is the electron configuration of Au+?

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Answer 1 · 2016-01-03T13:50:42

$[X e] 4 f^{14} 5 d^{10}$ $[Xe] 4f^14 5d^10$

Explanation:

The atomic number of Au is 79.

Therefore, its configuration is:

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} 4 f^{14} 5 d^{10} 6 s^{1}$ $1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 4f^14 5d^10 6s^1$

or, $[X e] 4 f^{14} 5 d^{10} 6 s^{1}$ $[Xe] 4f^14 5d^10 6s^1$

For $A u^{+}$ $Au^+$ , one electron is removed from the outermost $6 s$ $6s$ orbital, making the configuration,

$[X e] 4 f^{14} 5 d^{10}$ $[Xe] 4f^14 5d^10$

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What is the electron configuration of Au+?

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