A) 5th Period, Group 10
B) 4th Period, Group 12
C) 6th Period, Group 10
D) 5th Period, Group 14
Correct Answer: A
Solution :
[a] \[4{{d}^{10}}5{{s}^{0}}\] is the exceptional configuration of Pd. Its electronic configuration should be \[\left[ _{36}Kr \right]4{{d}^{8}},5{{s}^{2}}.\] Thus its Period = 5th Group \[=ns+(n-1l)d\] electrons \[=2+8=10\]
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