Question

A certain orbital has no angular nodes and two radial nodes. The orbital is:

• A. $2\mathrm{p}$
• B. $3\mathrm{p}$
• C. $2\mathrm{s}$
• D. $3\mathrm{s}$

Answer 1

The correct option is D

$3\mathrm{s}$

Answer: (D)

Explanation for correct option:

$$\begin{gathered} \left(\mathrm{D}\right)\mathrm{No}\mathrm{angular}\mathrm{nodes},\mathrm{L}=0 \\ \mathrm{Radial}\mathrm{nodes}=\mathrm{n}-\mathrm{L}-1 \\ =3-0-1 \\ =2 \end{gathered}$$

Explanation for incorrect options:

$$\begin{gathered} \left(\mathrm{A}\right)\mathrm{No}\mathrm{radial}\mathrm{nodes},\mathrm{L}=0 \\ \mathrm{If}\mathrm{s},\mathrm{L}=0 \\ \mathrm{p},\mathrm{L}=1 \\ \mathrm{d},\mathrm{L}=2 \\ \mathrm{f},\mathrm{L}=3 \end{gathered}$$

Here for $\mathrm{p}$, $\mathrm{L}=1$ because $2\mathrm{p}$ is not applicable.

$$\begin{gathered} \left(\mathrm{B}\right)\mathrm{No}\mathrm{radial}\mathrm{nodes},\mathrm{L}=0 \\ \mathrm{If}\mathrm{s},\mathrm{L}=0 \\ \mathrm{p},\mathrm{L}=1 \\ \mathrm{d},\mathrm{L}=2 \\ \mathrm{f},\mathrm{L}=3 \end{gathered}$$

Here for $\mathrm{p},\mathrm{L}=1$because $3\mathrm{p}$ is not applicable

$$\begin{gathered} \left(\mathrm{C}\right)\mathrm{No}\mathrm{angular}\mathrm{nodes},\mathrm{L}=0 \\ \mathrm{Radial}\mathrm{nodes}=\mathrm{n}-\mathrm{L}-1 \\ =2-0-1 \\ =1 \end{gathered}$$.

Here radial node is $1$ Because option (C) is incorrect.

Hence option (D) is correct.