Question

c. write the full electron configuration, noble gas notation, and fill in the orbital diagram for the following elements.
(11) nitrogen
(12) chlorine
(13) sodium
(14) neon
(15) nickel

Explanation:

Step1: Recall electron - filling rules

Electrons fill orbitals according to the Aufbau principle (low - energy orbitals first), Pauli exclusion principle (each orbital can hold at most 2 electrons with opposite spins), and Hund's rule (electrons fill degenerate orbitals singly first before pairing).

Step2: Determine electron configuration for nitrogen (N, atomic number = 7)

The 1s orbital fills first with 2 electrons, then the 2s orbital with 2 electrons, and then 3 electrons go into the 2p orbitals singly. So the electron configuration is $1s^{2}2s^{2}2p^{3}$. In the orbital diagram: 1s has 1 box with 2 paired electrons (↑↓), 2s has 1 box with 2 paired electrons (↑↓), and 2p has 3 boxes with 1 electron in each (↑ ↑ ↑).

Step3: Determine electron configuration for chlorine (Cl, atomic number = 17)

1s fills with 2 electrons, 2s with 2 electrons, 2p with 6 electrons, 3s with 2 electrons, and 3p with 5 electrons. So it is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$. In the orbital diagram: 1s (↑↓), 2s (↑↓), 2p three boxes with 2 electrons in each (↑↓ ↑↓ ↑↓), 3s (↑↓), 3p three boxes with 2, 2, 1 electrons (↑↓ ↑↓ ↑).

Step4: Determine electron configuration for sodium (Na, atomic number = 11)

1s has 2 electrons, 2s has 2 electrons, 2p has 6 electrons, and 3s has 1 electron. So it is $1s^{2}2s^{2}2p^{6}3s^{1}$. In the orbital diagram: 1s (↑↓), 2s (↑↓), 2p (↑↓ ↑↓ ↑↓), 3s (↑).

Step5: Determine electron configuration for neon (Ne, atomic number = 10)

1s has 2 electrons, 2s has 2 electrons, and 2p has 6 electrons. So it is $1s^{2}2s^{2}2p^{6}$. In the orbital diagram: 1s (↑↓), 2s (↑↓), 2p (↑↓ ↑↓ ↑↓).

Step6: Determine electron configuration for nickel (Ni, atomic number = 28)

1s has 2 electrons, 2s has 2 electrons, 2p has 6 electrons, 3s has 2 electrons, 3p has 6 electrons, 4s has 2 electrons, and 3d has 8 electrons. So it is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{8}$. In the orbital diagram: 1s (↑↓), 2s (↑↓), 2p (↑↓ ↑↓ ↑↓), 3s (↑↓), 3p (↑↓ ↑↓ ↑↓), 4s (↑↓), 3d five boxes with 2, 2, 2, 1, 1 electrons (↑↓ ↑↓ ↑↓ ↑ ↑).

Answer:

Nitrogen: 1s (↑↓), 2s (↑↓), 2p (↑ ↑ ↑)
Chlorine: 1s (↑↓), 2s (↑↓), 2p (↑↓ ↑↓ ↑↓), 3s (↑↓), 3p (↑↓ ↑↓ ↑)
Sodium: 1s (↑↓), 2s (↑↓), 2p (↑↓ ↑↓ ↑↓), 3s (↑)
Neon: 1s (↑↓), 2s (↑↓), 2p (↑↓ ↑↓ ↑↓)
Nickel: 1s (↑↓), 2s (↑↓), 2p (↑↓ ↑↓ ↑↓), 3s (↑↓), 3p (↑↓ ↑↓ ↑↓), 4s (↑↓), 3d (↑↓ ↑↓ ↑↓ ↑ ↑)