Which one of the following represents an impossible arrangement of quantum numbers?

Describe the orbitals with the following quantum numbers using $s, p, d, f$ notation:
$(a) n = 1, l = 0$
$(b) n = 3, l = 1$
$(c) n = 4, l = 2$
$(d) n = 4, l = 3$

An orbital is defined as:

Arrange the electrons represented by the following sets of quantum numbers in the decreasing order of energy.
$(i) \ n = 4, l = 0, m_l = 0, m_s = +\frac{1}{2}$
$(ii) \ n = 3, l = 1, m_l = 1, m_s = -\frac{1}{2}$
$(iii) \ n = 3, l = 2, m_l = 0, m_s = +\frac{1}{2}$
$(iv) \ n = 3, l = 0, m_l = 0, m_s = -\frac{1}{2}$

In a multi-electron atom,which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic and electric fields?
$1. n = 1, l = 0, m = 0$
$2. n = 2, l = 0, m = 0$
$3. n = 2, l = 1, m = 1$
$4. n = 3, l = 2, m = 0$
$5. n = 3, l = 2, m = 0$

The total value of magnetic quantum number for $n = 4$ is: