Question

The electrons present in the penultimate orbit of coinage metal atoms are

• A. 8
• B. 2
• C. 18
• D. 32

Answer 1

Answer: (C)

18

Answer 2

Coinage metals (Cu, Ag, Au) have an electronic configuration with 1 electron in the outermost s-orbital ($ns^1$) and a completely filled d-orbital in the penultimate shell ($(n-1)d^{10}$). The penultimate shell therefore contains $s^2p^6d^{10}$ electrons, totaling $2+6+10=18$ electrons.