Question

part a: write the electron configuration for arsenic.
part b: write the orbital diagram for arsenic.
part c: give the set of four quantum numbers for the last electron filling an atom of arsenic.
(n, l, m₁, mₛ) ( , , , )
complete the table.

Explanation:

Part A: Electron Configuration for Arsenic

Step1: Determine the atomic number

Arsenic (As) has an atomic number of 33. This means a neutral As atom has 33 electrons.

Step2: Use the Aufbau principle

The electron configuration is built by filling orbitals in order of increasing energy.
The electron configuration of As is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3}\).

Part B: Orbital Diagram for Arsenic

Step1: Draw the orbitals

Start with \(1s\), then \(2s\), \(2p\), \(3s\), \(3p\), \(4s\), \(3d\), and \(4p\) orbitals.

Step2: Fill the orbitals with electrons

• \(1s\): two electrons with opposite spins (\(\uparrow\downarrow\))
• \(2s\): two electrons with opposite spins (\(\uparrow\downarrow\))
• \(2p\): three pairs of electrons (\(\uparrow\downarrow\) in each of the three \(2p\) orbitals)
• \(3s\): two electrons with opposite spins (\(\uparrow\downarrow\))
• \(3p\): three pairs of electrons (\(\uparrow\downarrow\) in each of the three \(3p\) orbitals)
• \(4s\): two electrons with opposite spins (\(\uparrow\downarrow\))
• \(3d\): five pairs of electrons (\(\uparrow\downarrow\) in each of the five \(3d\) orbitals)
• \(4p\): three electrons, each in separate orbitals with parallel spins (\(\uparrow\) in each of the three \(4p\) orbitals)

Part C: Quantum Numbers for the Last Electron in Arsenic

Step1: Identify the last - filled orbital

The last - filled orbital is \(4p\).

Step2: Determine the quantum numbers

• \(n\) (principal quantum number): \(n = 4\) (since it is in the \(4p\) orbital)
• \(l\) (azimuthal quantum number): For \(p\) orbitals, \(l=1\)
• \(m_{l}\) (magnetic quantum number): For \(p\) orbitals (\(l = 1\)), \(m_{l}=- 1,0,1\). Since the last electron is in the \(4p\) orbital and following Hund's rule (filling orbitals singly first), assume \(m_{l}=1\) (could also be \(-1\) or \(0\) depending on the order of filling, but any of the three values for \(m_{l}\) is correct as long as the spin is considered)
• \(m_{s}\) (spin quantum number): Since it is the last electron and following Hund's rule (parallel spins), \(m_{s}=+\frac{1}{2}\)

The set of quantum numbers is \((4,1,1,+\frac{1}{2})\) (or \((4,1, - 1,+\frac{1}{2})\) or \((4,1,0,+\frac{1}{2})\))

Answer:

Part A: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3}\)
Part B: Orbital diagram as described above (with filled orbitals following Aufbau, Pauli, and Hund's rules)
Part C: \((4,1,1,+\frac{1}{2})\) (or \((4,1, - 1,+\frac{1}{2})\) or \((4,1,0,+\frac{1}{2})\))