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OpenStudy (anonymous):
What is the electron configuration of Br, Z=35?
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OpenStudy (rogue):
The expanded out form would be...\[Br: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5\]The condensed form would be\[Br: [Ar] 4s^2 3d^{10} 4p^5\]
OpenStudy (anonymous):
yea what Rogue said..just full the orbitals then write the electronic configuration closest to that of a noble gas
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