Question

The outer electronic configuration of \(\text{Pd}^{2+}\) is

• A. \([Kr]4d^8 5s^2\)
• B. \([Kr]4d^{10} 5s^0\)
• C. \([Kr]4d^8 5s^0\)
• D. \([Kr]4d^{10} 5s^2\)

Hint:

Transition metals often lose s-electrons first, but in some cases like Pd, the d-electrons are directly involved due to stability.

Answer 1

The Correct Option is C

Palladium (Pd) has atomic number 46. Its ground state electronic configuration is: \[ \text{Pd} = [Kr]4d^{10} \] On removing two electrons to form \(\text{Pd}^{2+}\), both electrons are removed from the 4d orbital because the 5s orbital is already empty: \[ \text{Pd}^{2+} = [Kr]4d^8 5s^0 \]