Q1 / 90 · JEE Main 2019 (10 Jan Shift 2) · Chemistry

The $71^{st}$ electron of an element $X$ with an atomic number of $71$ enters the orbital:

Correct answer
A
$5 d$
B
$4 f$
C
$6 p$
D
$6 s$

Solution

_{71} Lu : [Xe] 4 f^{14} 6 s^{2} 5 d^{1}

(The last electron enters

5 d

orbital)

Hence, the correct answer is option: 'A'.

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