Question

The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is

• A. $+5,+1,+1/2$
• B. $6,0,0,+1/2$
• C. $5,0,0,+1/2$
• D. None of the above

Answer 1

The correct option is C $5,0,0,+1/2$

Electronic configuration of Rubidium (Rb) is $1s^2$ $2s^2$ $2p^6$ $3s^2$ $3p^6$ $4s^2$ $3d^1$${}^0$ $4p^6$ $5s^1$.

Therefore,

Principle Quantum Number $n=5$

Azimuthal Quantum Number $l=0$ (since for s - orbital $l=0$)

Magnetic Quantum Number $m=0$ (since for $l=0$ there is only one value of $m$ i.e. $m=0$)

Spin Quantum Number $s=+1/2$ or $-1/2$ (there is only one electron and it can have clockwise or anti clockwise spins).

Therefore the correct set of Quantum numbers will be $5,0,0,+1/2$ or $-1/2$.

Hence the correct option is 'C'.