MCQ
$Be's$ $4^{th}$ electron will have four quantum numbers
$n$ $l$ $m$ $s$
- A$1$ $0$ $0$ $+1/2$
- B$1$ $1$ $+1$ $+1/2$
- ✓$2$ $0$ $0$ $-1/2$
- D$2$ $1$ $0$ $+1/2$
✓
Answer
Correct option: C.
$2$ $0$ $0$ $-1/2$
c
Electronic config for $Be$ is $1 s ^2 2 s ^2$. Fourth electron is in $s$ orbital so $n =2,l=0, m =0$ and $s =-1 / 2$
Electronic config for $Be$ is $1 s ^2 2 s ^2$. Fourth electron is in $s$ orbital so $n =2,l=0, m =0$ and $s =-1 / 2$
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