# Which of the following is a 2p orbital?

## A) B) C) D)

Aug 19, 2016

The $2 p$ orbital is $D$.

• $A$ is an $n s$ orbital, probably $3 s$ or $4 s$, though it depends on their size.
• $B$ is a $3 {d}_{{z}^{2}}$ orbital.
• $C$ is a $3 {d}_{x z}$ orbital.

However, do not trust the image. $B$, $C$, and $D$ are wrong because all the lobe signs are presented to be the same whereas they aren't.

The $3 {d}_{{z}^{2}}$ orbital ring is the opposite sign, the $3 {d}_{x z}$ has diagonal lobes of the same sign ($y z$ and $x y$ nodal planes), and the $2 {p}_{z}$ orbital's second lobe is the opposite sign.

The $2 p$ orbital has:

• $n = 2$, for the principal quantum number. $n$ can be one number in the set $\left\{1 , 2 , 3 , . . .\right\}$.
• $l = 1$, for the angular momentum quantum number since $l = \left\{0 , 1 , 2 , 3 , . . .\right\} \leftrightarrow \left\{s , p , d , f , . . .\right\}$.

If you recall:

• The total number of nodes (radial or angular regions of zero electron density) is equal to $n - 1$.
• The total number of angular nodes (nodal planes or conic nodes) is $l$.
• The total number of radial nodes is $n - l - 1$.

Since $l = 1$, there is one angular node, and since $n - 1 = 1$, it is the only node.

On either side of a nodal plane is a lobe of the opposite sign, and thus, there are two lobes on a $2 p$ orbital. That means it has to be either $B$ or $D$.

The $B$ orbital has $2$ conic nodes, which are angular nodes, and corresponds to how $l = 2$ corresponds to a $d$ orbital.

$B$ is actually a $\boldsymbol{3 {d}_{{z}^{2}}}$ orbital, like so:

Both lobes on $B$ are actually the same sign, whereas it's the ring that is the opposite sign.

The $2 p$ orbital is $\textcolor{b l u e}{D}$.