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The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination$(s)$ has/have the same energy,list them:
$(1).$ $n=4, l=2, m_l=-2, m_s=-1/2$
$(2).$ $n=3, l=2, m_l=1, m_s=+1/2$
$(3).$ $n=4, l=1, m_l=0, m_s=+1/2$
$(4).$ $n=3, l=2, m_l=-2, m_s=-1/2$
$(5).$ $n=3, l=1, m_l=-1, m_s=+1/2$
$(6).$ $n=4, l=1, m_l=0, m_s=+1/2$

The probability of finding electrons in $d_{xy}$ orbital is

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In an atom,the total number of electrons having quantum numbers $n=4, |m_{l}|=1$ and $m_{s}=-\frac{1}{2}$ is:

Total number of angular nodes of orbitals associated with third shell $(n=3)$ of an atom is

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