Question

3 mark for review which of the following best represents the ground - state electron configuration for an atom of selenium? a 1s² 2s²2p⁶ 3s²3p² b 1s² 2s²2p⁶ 3s²3p⁴ c 1s² 2s²2p⁶ 3s²3p⁶ 4s² 3d¹⁰ 4p⁴ d 1s² 2s²2p⁶ 3s²3p⁶ 4s² 3d¹⁰ 4p³

Explanation:

Step1: Recall selenium's atomic number

Selenium has an atomic number of 34, so it has 34 electrons in its ground - state.

Step2: Analyze electron - filling order

Electrons fill orbitals in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. The 1s orbital can hold 2 electrons, 2s can hold 2, 2p can hold 6, 3s can hold 2, 3p can hold 6, 4s can hold 2, 3d can hold 10, and 4p can hold 6.

Step3: Calculate electrons in each option

• Option A: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}$ has $2 + 2+6 + 2+2=14$ electrons.
• Option B: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$ has $2 + 2+6 + 2+4 = 16$ electrons.
• Option C: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}$ has $2+2 + 6+2+6+2+10+4=34$ electrons.
• Option D: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3}$ has $2+2 + 6+2+6+2+10+3=33$ electrons.

Answer:

C. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}$