Respuesta :

addybarlow20

Answer:

4d₉

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.

Otras preguntas