This question is about a well known exception in transition metal electron filling: a few metals near the end of the 3d series are more stable with a full $d^{10}$ subshell than with the "expected" configuration, and this stability can carry over into their ions too. The task is to find which metal keeps a full $d^{10}$ arrangement both as a neutral atom and after losing one electron.
- Co: Cobalt's configuration is $[Ar]\,3d^7 4s^2$. The d subshell is not full to begin with, and removing an electron to form $Co^{+}$ still leaves it well short of $d^{10}$.
- Cu: Copper breaks the normal filling order and is written as $[Ar]\,3d^{10}4s^1$, so the d subshell is already completely filled in the neutral atom. Removing the single, loosely held $4s$ electron to form $Cu^{+}$ leaves $[Ar]\,3d^{10}$ untouched, so the full d subshell survives in the $+1$ ion as well.
- Ni: Nickel is $[Ar]\,3d^8 4s^2$. Even after losing an electron, the d count stays at 8, never reaching 10.
- Ir: Iridium's ground state configuration is $[Xe]\,4f^{14}5d^7 6s^2$, and its $+1$ ion does not settle into a filled $d^{10}$ arrangement the way copper does.
Copper is the one metal here whose 4s electron sits outside a completely filled d shell in the neutral atom, so losing it leaves the full d shell intact in the $+1$ ion.
Let's summarize:
- Copper's neutral configuration already has a complete $3d^{10}$, with a single extra $4s^1$ electron on top.
- Losing that one outer electron to form $Cu^{+}$ does not disturb the filled d subshell, unlike Co, Ni, or Ir.
The element with completely filled d-orbitals in both its atomic and $+1$ oxidation state is copper (Cu).