The outer electronic configuration of " $\mathrm{Pd}$ " is
The outer electronic configuration of " $\mathrm{Pd}$ " is
$4 d^8 5 s^2$
$4 d^9 5 s^1$
$4 d^{10} 5 s^0$
$4 d^{10} 5 s^1$
Solution
The atomic number of polladium is 46 . Hence, its electronic configuration of palladium (Pd) is $[\mathrm{Kr}] 4 d^{10}, 5 s^0$
So, its outer electronic configuration is $4 d^{10} 5 s^0$.