Question

Consider following Schrodinger wave equation for an orbital of hydrogen atom.
$(a_0$ is first Bohr radius)
$\Psi =\frac{\sqrt{2r}}{81a_0^2\sqrt{\pi a_0}}(6-\frac{r}{a_0})e^{\frac{-r}{3a_0}}cos\theta$

$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \left(I\right)& \text{Which orbital is this?}& \left(P\right)& 1\\ \left(II\right)& \text{Number of total node(s)}& \left(Q\right)& 2\\ \left(III\right)& \text{Nodal plane}& \left(R\right)& 3p_z\\ \left(IV\right)& \text{Number of radial node(s)}& \left(S\right)& 3p_x\\ & & \left(T\right)& \text{XY plane}\\ & & \left(U\right)& \text{YZ plane}\end{array}$

Match the correct combination considering List-I and List-II

• A. (I),(R) and (II),(P)
  
• B. (I),(S) and (II),(P)
• C. (I),(R) and (II),(Q)
• D. (I),(S) and (II),(Q)

Answer 1

The correct option is C (I),(R) and (II),(Q)

(I) $Y\left(p_z\right)=\left(\frac{3}{4\pi }{)}^{\frac{1}{2}}cos\right(\theta )$
$Y\left(p_x\right)=\left(\frac{3}{4\pi }{)}^{\frac{1}{2}}sin\right(\theta \left)cos\right(\varphi )$
$Y\left(p_y\right)=\left(\frac{3}{4\pi }{)}^{\frac{1}{2}}sin\right(\theta \left)sin\right(\varphi )$
Comparing with these we can tell, the wavefunction represents $3p_z$ orbital. A quick way to figure this out is that the $p_z$ orbital would only have one angular dependence i.e. $\theta$.
(II) Total number of nodes in this orbital $=n-1=3-1=2$.