Question:
The element that has completely filled d-orbitals in its atomic and +1 oxidation state is:
The element that has completely filled d-orbitals in its atomic and +1 oxidation state is:
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Elements with configuration ending in \(d^{10}s^1\) (like Cu, Ag, Au) often form stable +1 ions with filled d-subshells.
Updated On: Jul 18, 2026
- Co
- Cu
- Ni
- Ir
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The Correct Option is B
Solution and Explanation
Step 1: Understand the condition given in the question.
We need an element whose d-subshell is completely filled (i.e., \(d^{10}\)) in its atomic state as well as in its \(+1\) oxidation state. This means after losing one electron, the remaining configuration should still have a fully filled d-subshell.
Step 2: Write electronic configuration of Cu.
Copper (Cu, Z = 29): \[ [Ar]\,3d^{10}\,4s^1 \] In atomic state, d-orbitals are already completely filled (\(3d^{10}\)).
Step 3: Form Cu\(^+\) ion.
Cu loses one electron from 4s: \[ Cu^+ : [Ar]\,3d^{10} \] Now the d-subshell remains completely filled even in +1 oxidation state.
Step 4: Check other options.
Co: \(3d^7\), Co\(^+\): not \(d^{10}\) Ni: \(3d^8\), Ni\(^+\): not \(d^{10}\) Ir: does not maintain fully filled d in +1 state consistently
Step 5: Key reasoning.
Only Cu has stable \(d^{10}\) configuration in both atomic and +1 oxidation state due to fully filled d-subshell stability.
Step 6: Final conclusion.
\[ \boxed{\text{Cu}} \]
We need an element whose d-subshell is completely filled (i.e., \(d^{10}\)) in its atomic state as well as in its \(+1\) oxidation state. This means after losing one electron, the remaining configuration should still have a fully filled d-subshell.
Step 2: Write electronic configuration of Cu.
Copper (Cu, Z = 29): \[ [Ar]\,3d^{10}\,4s^1 \] In atomic state, d-orbitals are already completely filled (\(3d^{10}\)).
Step 3: Form Cu\(^+\) ion.
Cu loses one electron from 4s: \[ Cu^+ : [Ar]\,3d^{10} \] Now the d-subshell remains completely filled even in +1 oxidation state.
Step 4: Check other options.
Co: \(3d^7\), Co\(^+\): not \(d^{10}\) Ni: \(3d^8\), Ni\(^+\): not \(d^{10}\) Ir: does not maintain fully filled d in +1 state consistently
Step 5: Key reasoning.
Only Cu has stable \(d^{10}\) configuration in both atomic and +1 oxidation state due to fully filled d-subshell stability.
Step 6: Final conclusion.
\[ \boxed{\text{Cu}} \]
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