How many unpaired electrons will be present in the ground state of an atom which has valence electronic configuration $3d^6$ in its $+3$ oxidation state?

The plot of $r^2 \Psi^2$ versus $r$ for $2s$ orbital of hydrogen atom exhibits :-

Arrange the following in the increasing order of their energy:
$(i)$ $n = 2, l = 0, m_l = 0, m_s = +\frac{1}{2}$
$(ii)$ $n = 2, l = 1, m_l = +1, m_s = -\frac{1}{2}$
$(iii)$ $n = 3, l = 2, m_l = +1, m_s = -\frac{1}{2}$
$(iv)$ $n = 3, l = 1, m_l = -1, m_s = +\frac{1}{2}$

$4s^2$ is the configuration of the outermost orbit of an element. Its atomic number would be $:-$

Assertion : Spin quantum number can have two values,$+\frac{1}{2}$ and $-\frac{1}{2}$.
Reason : $+$ and $-$ signs signify the positive and negative wave functions.

Which of the following electronic configurations has the maximum total spin multiplicity?