Question

which electron configuration represents the electrons in an atom of ga in the ground state?
○ a 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
○ b 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4d¹⁰ 4p¹
○ c 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p¹
○ a 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3p⁶ 4s² 4p¹

Explanation:

Step1: Determine the atomic number of Ga

Gallium (Ga) has an atomic number of \(31\).

Step2: Write the electron configuration

The electron configuration is built by filling orbitals in the order \(1s < 2s<2p < 3s<3p <4s<3d <4p\).

• \(1s^2\): \(2\) electrons.
• \(2s^2\): \(2\) electrons (\(2 + 2=4\)).
• \(2p^6\): \(6\) electrons (\(4+6 = 10\)).
• \(3s^2\): \(2\) electrons (\(10 + 2=12\)).
• \(3p^6\): \(6\) electrons (\(12+6 = 18\)).
• \(4s^2\): \(2\) electrons (\(18+2 = 20\)).
• \(3d^{10}\): \(10\) electrons (\(20+10 = 30\)).
• \(4p^1\): \(1\) electron (\(30 + 1=31\)).

So the electron configuration of Ga in the ground state is \(1s^22s^22p^63s^23p^64s^23d^{10}4p^1\)

Answer:

C. \(1s^22s^22p^63s^23p^64s^23d^{10}4p^1\)