1. Find the electronic configuration of rubidium

Rubidium has atomic number $Z=37$.

Its electronic configuration is:

$1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^1$

So, the valence electron is in the $5s$ orbital.

2. Assign the four quantum numbers

For the valence electron in $5s^1$:

• Principal quantum number: $n=5$
• Azimuthal quantum number for an $s$-orbital: $l=0$
• Magnetic quantum number: for $l=0$, only possible value is $m_l=0$
• Spin quantum number: for a single electron, one allowed value is $m_s=+\frac{1}{2}$

Thus, the set is:

$n=5,\quad l=0,\quad m_l=0,\quad m_s=+\frac{1}{2}$

3. Check the options

• A: $5,1,1,+\frac{1}{2}$  incorrect, because $l=1$ means $p$-orbital
• B: $5,1,0,+\frac{1}{2}$  incorrect, because $l=1$ means $p$-orbital
• C: $5,0,0,+\frac{1}{2}$  correct
• D: $5,0,1,+\frac{1}{2}$  incorrect, because for $l=0$, $m_l$ must be $0$, not $1$

Therefore, the correct option is C.