How many orbitals in a $Cr$ atom have $l = 0$?

How many electrons in an atom can have the following quantum numbers?
$(a)$ $n = 4, m_s = -\frac{1}{2}$
$(b)$ $n = 3, l = 0$

In any atom,which sub-shell will have the highest energy among the following?

Identify which of the following sets of quantum numbers are not possible and explain why.
$(a) n = 0, l = 0, m_l = 0, m_s = +1/2$
$(b) n = 1, l = 0, m_l = 0, m_s = -1/2$
$(c) n = 1, l = 1, m_l = 0, m_s = +1/2$
$(d) n = 2, l = 1, m_l = 0, m_s = -1/2$
$(e) n = 3, l = 3, m_l = 3, m_s = +1/2$
$(f) n = 3, l = 1, m_l = 0, m_s = +1/2$

The orbital angular momentum of an electron in $3s$ orbital is $\frac{xh}{2\pi}$. The value of $x$ is . . . . . . .

What are the values of $n$ and $l$ for the $f$-subshell with the lowest energy?