JEE MAIN - Chemistry (2021 - 25th July Evening Shift - No. 18)
Number of electrons present in 4f orbital of Ho3+ ion is ____________. (Given Atomic No. of Ho = 67)
Answer
10
Explanation
Ho = [Xe]4f116s2
Ho3+ = [Xe] 4f10
So number of e$$-$$ present in 4f is 10.
Ho3+ = [Xe] 4f10
So number of e$$-$$ present in 4f is 10.
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