What is the electron configuration of a bromine atom (Br) after it loses three electrons?

• A. [Ar] 3d^{10} 4s^{2} 4p^{3}
• B. [Ar] 3d^{10} 4s^{2} 4p^{2}
• C. [Ar] 3d^{10} 4s^{2} 4p^{5}
• D. [Ar] 3d^{10} 4s^{2} 4p^{6}

Answer: (B)

Verified step by step guidance

1. Identify the atomic number of bromine (Br), which is 35, meaning a neutral bromine atom has 35 electrons.
2. Write the ground-state electron configuration of neutral bromine: \([\mathrm{Ar}]\ 3d^{10}\ 4s^{2}\ 4p^{5}\), since the 4p subshell has 5 electrons.
3. Determine how many electrons are lost: bromine loses 3 electrons, so the new electron count is 35 - 3 = 32 electrons.
4. Remove electrons starting from the highest energy level orbitals first, which are the 4p electrons, then 4s if needed. Since 3 electrons are lost, remove 3 electrons from the 4p subshell.
5. Write the new electron configuration after losing 3 electrons: \([\mathrm{Ar}]\ 3d^{10}\ 4s^{2}\ 4p^{2}\), reflecting the removal of 3 electrons from the 4p subshell.