Question

1. write the electron configuration for each of the following atoms. how many unpaired electrons are present for each atom? a. nitrogen 1s² 2s² 2p³ b. calcium 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² c. sodium

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 a maximum of 2 electrons with opposite spins), and Hund's rule (electrons in degenerate orbitals occupy them singly first with parallel spins).

Step2: Analyze nitrogen (N)

Nitrogen has an atomic number of 7. Its electron configuration is $1s^{2}2s^{2}2p^{3}$. In the 2p sub - shell (which has 3 degenerate orbitals), according to Hund's rule, the 3 electrons in the 2p sub - shell occupy the 3 orbitals singly. So the number of unpaired electrons is 3.

Step3: Analyze calcium (Ca)

Calcium has an atomic number of 20. Its electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}$. All orbitals are completely filled. So the number of unpaired electrons is 0.

Step4: Analyze sodium (Na)

Sodium has an atomic number of 11. Its electron configuration is $1s^{2}2s^{2}2p^{6}3s^{1}$. The 3s orbital has 1 electron, so the number of unpaired electrons is 1.

Answer:

a. Nitrogen: Electron configuration $1s^{2}2s^{2}2p^{3}$, 3 unpaired electrons
b. Calcium: Electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}$, 0 unpaired electrons
c. Sodium: Electron configuration $1s^{2}2s^{2}2p^{6}3s^{1}$, 1 unpaired electron