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.
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