Question

The wave functions of 3s and 3p₂ orbitals are given by

Ψ₃ₛ = $\frac{1}{9\sqrt{3}}(\frac{1}{4\pi})^{1/2}(\frac{z}{a₀})^{3/2}(6-\frac{4zr}{a₀}+\frac{4z²r²}{9a₀²})e^{\frac{-zr}{3a₀}}$

Ψ₃p₂ = $\frac{1}{9\sqrt{6}}(\frac{3}{4\pi})^{1/2}(\frac{z}{a₀})^{3/2}(4-\frac{2zr}{3a₀})(\frac{2zr}{3a₀})e^{\frac{-zr}{3a₀}}cos\theta$

from these we can conclude

• A. Total number of nodal surface for 3p₂ & 3s orbitals are equal
• B. The angular nodal surface of 3p₂ orbital has the equation $\theta$= π/2
• C. The radial nodal surfaces of 3s orbital and 3p₂ orbitals are at equal distance from the nucleus
• D. 3s electron have greater penetrating power into the nucleus in comparison to 3p₂ electrons

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

The given wave functions for the 3s and 3p$_z$ orbitals are:

Ψ₃ₛ = $\frac{1}{9\sqrt{3}}(\frac{1}{4\pi})^{1/2}(\frac{z}{a₀})^{3/2}(6-\frac{4zr}{a₀}+\frac{4z²r²}{9a₀²})e^{\frac{-zr}{3a₀}}$

Ψ₃p₂ = $\frac{1}{9\sqrt{6}}(\frac{3}{4\pi})^{1/2}(\frac{z}{a₀})^{3/2}(4-\frac{2zr}{3a₀})(\frac{2zr}{3a₀})e^{\frac{-zr}{3a₀}}cos\theta$

Let's analyze each conclusion:

(A) Total number of nodal surfaces for 3p$_z$ & 3s orbitals are equal.

The total number of nodes for an orbital with principal quantum number $n$ is given by $n-1$. For both 3s and 3p$_z$ orbitals, $n=3$.

Total nodes for 3s = $3 - 1 = 2$.

Total nodes for 3p$_z$ = $3 - 1 = 2$.

Alternatively, the total number of nodes is the sum of radial nodes and angular nodes.

The number of angular nodes is equal to the azimuthal quantum number $l$.

For s orbitals, $l=0$. Angular nodes for 3s = 0.

For p orbitals, $l=1$. Angular nodes for 3p$_z$ = 1.

The number of radial nodes is given by $n - l - 1$.

Radial nodes for 3s = $3 - 0 - 1 = 2$.

Radial nodes for 3p$_z$ = $3 - 1 - 1 = 1$.

Total nodes for 3s = Radial nodes + Angular nodes = $2 + 0 = 2$.

Total nodes for 3p$_z$ = Radial nodes + Angular nodes = $1 + 1 = 2$.

The total number of nodal surfaces is indeed equal for 3s and 3p$_z$ orbitals. This conclusion is correct.

(B) The angular nodal surface of 3p$_z$ orbital has the equation $\theta$= π/2.

Nodal surfaces are regions where the wave function $\Psi$ is zero. The wave function Ψ₃p₂ has an angular part proportional to $\cos\theta$. The angular node occurs when the angular part is zero, i.e., $\cos\theta = 0$. This happens when $\theta = \frac{\pi}{2}$ (or $\theta = \frac{3\pi}{2}$, etc.). The surface described by $\theta = \frac{\pi}{2}$ in spherical coordinates is the xy-plane. This conclusion is correct.

(C) The radial nodal surfaces of 3s orbital and 3p$_z$ orbitals are at equal distance from the nucleus.

Radial nodal surfaces occur when the radial part of the wave function is zero (excluding $r=0$ and $r=\infty$).

For Ψ₃ₛ, the radial part is proportional to $(6-\frac{4zr}{a₀}+\frac{4z²r²}{9a₀²})$. Setting this to zero gives the radial nodes:

$6-\frac{4zr}{a₀}+\frac{4z²r²}{9a₀²} = 0$.

Let $x = \frac{zr}{a₀}$. The equation becomes $6 - 4x + \frac{4}{9}x^2 = 0$. Multiplying by 9/4, we get $\frac{54}{4} - 9x + x^2 = 0$, or $x^2 - 9x + \frac{27}{2} = 0$.

Using the quadratic formula, $x = \frac{9 \pm \sqrt{(-9)^2 - 4(1)(27/2)}}{2} = \frac{9 \pm \sqrt{81 - 54}}{2} = \frac{9 \pm \sqrt{27}}{2} = \frac{9 \pm 3\sqrt{3}}{2}$.

Since $x = \frac{zr}{a₀}$, the radial nodes for 3s are at $r = \frac{a₀}{z} \left(\frac{9 - 3\sqrt{3}}{2}\right)$ and $r = \frac{a₀}{z} \left(\frac{9 + 3\sqrt{3}}{2}\right)$. These are two distinct distances.

For Ψ₃p₂, the radial part is proportional to $(4-\frac{2zr}{3a₀})(\frac{2zr}{3a₀})$. Setting this to zero gives the radial nodes:

$(4-\frac{2zr}{3a₀})(\frac{2zr}{3a₀}) = 0$.

This gives two possibilities:

1. $\frac{2zr}{3a₀} = 0$, which implies $r=0$. This is the nucleus, not a radial node.

2. $4-\frac{2zr}{3a₀} = 0$, which implies $\frac{2zr}{3a₀} = 4$, so $r = \frac{4 \cdot 3a₀}{2z} = \frac{6a₀}{z}$.

There is one radial node for 3p$_z$ at $r = \frac{6a₀}{z}$.

Comparing the radial node distances for 3s ($\frac{a₀}{z} \frac{9 - 3\sqrt{3}}{2}$ and $\frac{a₀}{z} \frac{9 + 3\sqrt{3}}{2}$) and 3p$_z$ ($\frac{6a₀}{z}$), we can see they are different. For example, $\frac{9 - 3\sqrt{3}}{2} \approx \frac{9 - 5.196}{2} = 1.902$ and $\frac{9 + 3\sqrt{3}}{2} \approx \frac{9 + 5.196}{2} = 7.098$, while $6$. None of the 3s radial node distances are equal to the 3p$_z$ radial node distance. This conclusion is incorrect.

(D) 3s electron have greater penetrating power into the nucleus in comparison to 3p$_z$ electrons.

Penetrating power refers to the probability of finding an electron close to the nucleus. For a given principal quantum number $n$, orbitals with lower azimuthal quantum number $l$ have greater penetrating power. This is because the radial probability distribution function $P(r) = 4\pi r^2 |\Psi(r,\theta,\phi)|^2$ is non-zero at $r=0$ only for s orbitals ($l=0$). For $l>0$ (like p orbitals where $l=1$), the wave function is zero at $r=0$, making the probability density at the nucleus zero. The 3s orbital ($l=0$) has a non-zero probability density at the nucleus ($|\Psi_{3s}(0)|^2 \neq 0$), while the 3p$_z$ orbital ($l=1$) has zero probability density at the nucleus ($|\Psi_{3p_z}(0)|^2 = 0$). This indicates that 3s electrons are more likely to be found closer to the nucleus than 3p$_z$ electrons, meaning 3s electrons have greater penetrating power. This conclusion is correct.

Based on the analysis, conclusions (A), (B), and (D) are correct.