The electronic configuration of palladium is

[A r] 4 d^{9} 5 s^{1}

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[K r] 4 d^{8} 5 s^{2}

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[K r] 5 d^{8} 6 s^{2}

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[K r] 4 d^{10} 5 s^{0}

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Solution

The correct option is D $[K r] 4 d^{10} 5 s^{0}$
Palladium is a transition element. It is in 4d series and 10th group. Atomic number of palladium is 46.
General electronic configuration of 4d series is $[K r] 4 d^{(1 - 10)} 5 s^{(1 - 2)}$
Atomic number of Kr is 36.
Thus, electronic configuration of palladium is $[K r] 4 d^{10} 5 s^{0}$
Note: Single element in d-block which has zero electrons in s-subshell.

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