Pd has exceptional valence shell electronic cofiguration of $4d^{10}5s^0$ It is a member of -

The correct option is A $5^{th}$ period, Group 10
$4d^{10}5s^0$ is the exceptional configuration of $Pd$ becuase at higher atomic number energies are closer and hence 4d is filled over 5s. Its electronic configuration according to Aufbau principle should be ${[}_{36}Kr]4d^8,5s^2$ thus its
period = $5^{th}$ as $n=5$
Group = $ns+(n-1)d$ electrons
= 2 + 8 = 10