Question

question 36
draw the orbital diagram of an atom with 16 electrons. explain how this orbital diagram demonstrates
hund’s rule.

question 37
what is the wavelength of 3.61 × 10⁻¹⁹ j? determine the color of the light.

question 38
what is the frequency of photons with an energy of 1.4 × 10⁻²¹ j?

Explanation:

Question 36

Step1: Determine Electron Configuration

The atom with 16 electrons is sulfur (S). The electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^4\).

Step2: Draw Orbital Diagrams

• \(1s\): \(\uparrow\downarrow\)
• \(2s\): \(\uparrow\downarrow\)
• \(2p\): \(\uparrow\downarrow\) \(\uparrow\downarrow\) \(\uparrow\downarrow\)
• \(3s\): \(\uparrow\downarrow\)
• \(3p\): \(\uparrow\) \(\uparrow\) \(\uparrow\downarrow\) (Here, in the \(3p\) subshell (which has 3 orbitals), the first two electrons occupy separate orbitals with parallel spins, and the third electron pairs with one of them, demonstrating Hund's rule (electrons fill degenerate orbitals singly with parallel spins before pairing).)

Step3: Explain Hund's Rule Demonstration

In the \(3p\) subshell (3 degenerate orbitals), we first place one electron in each orbital (with the same spin, parallel) before pairing. So the three \(3p\) orbitals initially have one \(\uparrow\), one \(\uparrow\), and then the fourth electron (since \(3p\) has 4 electrons in total for S) pairs with one of the singly - occupied orbitals (resulting in \(\uparrow\downarrow\) in one orbital and \(\uparrow\) in the other two initially, then after adding the fourth electron, one has \(\uparrow\downarrow\) and two have \(\uparrow\)). This shows that electrons fill degenerate orbitals singly with parallel spins (Hund's rule) before pairing up.

Step1: Recall the Formula

The formula relating energy (\(E\)), Planck's constant (\(h = 6.626\times10^{-34}\ J\cdot s\)), speed of light (\(c = 3.0\times10^{8}\ m/s\)) and wavelength (\(\lambda\)) is \(E=\frac{hc}{\lambda}\), so \(\lambda=\frac{hc}{E}\).

Step2: Substitute the Values

Given \(E = 3.61\times10^{-19}\ J\), \(h = 6.626\times10^{-34}\ J\cdot s\), \(c = 3.0\times10^{8}\ m/s\).
\(\lambda=\frac{6.626\times10^{-34}\ J\cdot s\times3.0\times10^{8}\ m/s}{3.61\times10^{-19}\ J}\)
\(\lambda=\frac{1.9878\times10^{-25}\ J\cdot m}{3.61\times10^{-19}\ J}\approx5.51\times10^{-7}\ m = 551\ nm\).

Step3: Determine the Color

A wavelength of approximately \(551\ nm\) falls in the green - yellow region, more precisely, it is in the green region (visible light spectrum: 400 - 700 nm, green is around 495 - 570 nm).

Step1: Recall the Formula

The formula relating energy (\(E\)) of a photon, Planck's constant (\(h = 6.626\times10^{-34}\ J\cdot s\)) and frequency (\(
u\)) is \(E = h
u\), so \(
u=\frac{E}{h}\).

Step2: Substitute the Values

Given \(E = 1.4\times10^{-21}\ J\), \(h = 6.626\times10^{-34}\ J\cdot s\).
\(
u=\frac{1.4\times10^{-21}\ J}{6.626\times10^{-34}\ J\cdot s}\approx2.11\times10^{12}\ Hz\)

Answer:

Orbital diagram: \(1s:\uparrow\downarrow\), \(2s:\uparrow\downarrow\), \(2p:\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\), \(3s:\uparrow\downarrow\), \(3p:\uparrow\ \uparrow\ \uparrow\downarrow\). Explanation: In \(3p\) (3 degenerate orbitals), electrons fill singly (parallel spins) first: two \(3p\) orbitals have one electron (\(\uparrow\)) each, then the fourth \(3p\) electron pairs, showing Hund's rule.

Question 37