Question

1. write electron configurations for the following.

a. calcium

b. zirconium

c. arsenic

d. rubidium

Explanation:

Part a: Calcium (Ca)

Step1: Determine atomic number

Calcium has an atomic number of 20, so it has 20 electrons.

Step2: Fill orbitals

Follow Aufbau principle: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\) (since 4s fills before 3d for Ca, as it's an s - block element).

Step1: Determine atomic number

Zirconium has an atomic number of 40, so 40 electrons.

Step2: Fill orbitals

Using Aufbau principle, after argon (\([Ar] = 1s^2 2s^2 2p^6 3s^2 3p^6\)), we fill 4s, 3d, 4p, 5s, 4d. The electron configuration is \([Kr]4d^2 5s^2\) (where \([Kr]\) is krypton's configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\)).

Step1: Determine atomic number

Arsenic has an atomic number of 33, so 33 electrons.

Step2: Fill orbitals

Using Aufbau principle, after argon (\([Ar]\)), we fill 4s, 3d, 4p. The configuration is \([Ar]3d^{10} 4s^2 4p^3\) (since \([Ar]=1s^2 2s^2 2p^6 3s^2 3p^6\), then \(4s^2\), \(3d^{10}\), and \(4p^3\) to reach 33 electrons).

Answer:

\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\) or \([Ar]4s^2\) (where \([Ar]\) is the electron configuration of argon, \(1s^2 2s^2 2p^6 3s^2 3p^6\))

Part b: Zirconium (Zr)