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Question
Electronic configuration of $$Pd$$ is $$[Kr]4d^{10}$$. It belongs to
A
$$4^{th}$$ period and $$10^{th}$$ group
B
$$5^{th}$$ period and $$10^{th}$$ group
C
$$6^{th}$$ period and $$10^{th}$$ group
D
$$4^{th}$$ period and $$8^{th}$$ group
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Solution
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Correct option is A. $$4^{th}$$ period and $$10^{th}$$ group
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