Question

How many orbital(s) is/are possible with $n=3$ , $l=1$ and $m_l=-1$ value?

• A. 2
• B. 3
• C. 5
• D. 1

Answer 1

The correct option is D 1

$Principlequantumnumber=3$
$l=1\to pand{m}_l=-1$
gives us the first 3p orbital. The other two 3p orbitals would have m values of $0$ and $-1$.