Question

which of the following quantum number sets describes a 3d orbital?
( \bigcirc n = 3, l = 1, ml=-1 )
( \bigcirc n = 3, l = 2, ml=-1 )
( \bigcirc n = 4, l = 2, ml=+1 )
( \bigcirc n = 4, l = 3, ml=+2 )
( \bigcirc n = 2, l = 0, ml = 0 )

Explanation:

Step1: Analyze the principal quantum number ($n$)

For a $3d$ orbital, the principal quantum number $n = 3$. So we can eliminate the options with $n=4$ (i.e., $n = 4,l = 2,m_l=+1$ and $n = 4,l = 3,m_l=+2$) and $n = 2$ ($n=2,l = 0,m_l = 0$).

Step2: Analyze the azimuthal quantum number ($l$)

For $d$ - orbitals, the azimuthal quantum number $l=2$. So we can eliminate the option with $l = 1$ ($n = 3,l = 1,m_l=-1$).

Answer:

$n = 3,l = 2,m_l=-1$