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What is the last sequence for Silver

4d7
4d7

4d9
4d9

4p2
4p2

3d7

1 Answer

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Answer:

4d₉

Step-by-step explanation:

Silver is atomic number 47, Ag. So, count to 47 in orbitals:

1s₂ 2s₂ 2p₆ 3s₂ 3p₆ 4s₂ 3d₁₀ 4p₆ 5s₂ 4d₉

2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 9 = 47

This sequence ends in 4d₉, therefore it is the last sequence.

User Vasiliy Kevroletin
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