Question

Use the periodic table to identify the element with the electron configuration 1s²2s²2p⁴. Write its orbital diagram, and give the quantum numbers of its sixth electron.

Answer 1

Answer:

1. Orbital diagram

2p⁴   ║ ↑↓ ║  "↑"  ║   ↑

2s²    ║ ↑↓ ║

1s²     ║ ↑↓ ║

2. Quantum numbers

• n = 2,
• l = 1,

• $m_{l}$ = 0,

• $m_{s}$ = +1/2

Explanation:

The fill in rule is:

• Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2

• Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
• Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).

So, the orbital diagram of given element is as below and the sixth electron is marked between " "

2p⁴   ║ ↑↓ ║  "↑"  ║   ↑

2s²    ║ ↑↓ ║

1s²     ║ ↑↓ ║

The quantum number of an electron consists of four number:

• n (shell number, - 1, 2, 3...)
• l (subshell number or  orbital number, 0 - orbital s, 1 - orbital p, 2 - orbital d...)
• $m_{l}$ (orbital energy, or "which box the electron is in"). For example, orbital p (l = 1) has 3 "boxes", it was number from -1, 0, 1. Orbital d (l = 2) has 5 "boxes", numbered -2, -1, 0, 1, 2
• $m_{s}$ (spin of electron), either -1/2 or +1/2

In our case, the electron marked with " " has quantum number

• n = 2, shell number 2,
• l = 1, subshell or orbital p,

• $m_{l}$ = 0, 2nd "box" in the range -1, 0, 1

• $m_{s}$ = +1/2, single electron always has +1/2