Question

Which of the following are the correct statements?

• A. $2\mathrm{s}$- orbital has one nodal surface.
• B. For $2{\mathrm{p}}_{\mathrm{x}}$ orbitals $\mathrm{yz}$ is the nodal plane.
• C. Electron density in the $\mathrm{xy}-\mathrm{plane}\mathrm{in}3{\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ orbital is zero.
• D. Both (a) and (b)

Answer 1

The correct option is D

Both (a) and (b)

Explanation for correct option

(a) $2\mathrm{s}$- orbital has one nodal surface.

• To find the number of nodal surfaces, we use the formula
• $\mathrm{Nodes}=\mathrm{n}-1\mathrm{where}\mathrm{n}\mathrm{is}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{shell}$
• Here it is given that $\mathrm{n}=2$
• So, number of nodal surfaces $$\begin{gathered} =2-1 \\ =1 \end{gathered}$$
• Hence, the option (a) $2\mathrm{s}$- orbital has one nodal surface is correct.

(b) For $2{\mathrm{p}}_{\mathrm{x}}$ orbitals $\mathrm{yz}$ is the nodal plane.

• Nodal plane will be perpendicular to the $\mathrm{x}\mathrm{axis}$.
• Hence $2{\mathrm{p}}_{\mathrm{x}}$ orbitals will have $\mathrm{yz}$ as the nodal plane.
• Hence option (b) For $2{\mathrm{p}}_{\mathrm{x}}$ orbitals $\mathrm{yz}$ is the nodal plane is correct.

(d) Both (a) and (b)

• Since both option (a) and (b) is correct
• And option (d) says both (a) and (b) are correct.
• Hence, (d) Both (a) and (b) is the correct option

Explanation for incorrect option

(c) Electron density in the $\mathrm{xy}-\mathrm{plane}\mathrm{in}3{\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ orbital is zero.

• The orbital lobes of $3{\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ will be along x and y axis.
• So electron density is not zero in the $\mathrm{xy}-\mathrm{plane}\mathrm{in}3{\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$.
• Hence option (c) Electron density in the $\mathrm{xy}-\mathrm{plane}\mathrm{in}3{\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ orbital is zero is incorrect.

Hence, option (d) Both (a) and (b) is the correct option.