give-the-first-three-ionization-equation-for-br-and-determine-the-process-with-t

Question

Give the first three ionization equation for Br and determine the process with the least energy

Accepted Answer

IE_{1}: Br → $$Br^{+}$$ + $$e^{-}   IE$$_{2}: $$Br^{+}$$ → $$Br^{2+}$$ + $$e^{-}   IE$$_{3}: $$Br^{2+}$$ →  $$Br^{3+}$$ + $$e^{-}$$   The first ionization energy has the least energy because it is farthest from the nuclues of the atom.

Answer

IE_{1}: Br + $$e^{-}$$ → $$Br^{-}   IE$$_{2}: $$Br^{+}$$ + $$e^{-}$$ → $$Br^{2-}    IE$$_{3}: $$Br^{2+}$$ + $$e^{-}$$ →  $$Br^{3-}$$     The first ionization energy has the least energy because a core electron is removed.

Answer

IE_{1}: Br + $$e^{-}$$ → $$Br^{-}    IE$$_{2}: $$Br^{+}$$ + $$e^{-}$$ → $$Br^{2-}    IE$$_{3}: $$Br^{2+}$$ + $$e^{-}$$ →  $$Br^{3-}$$     The first ionization energy has the least energy because it is the closest to the nucleus of the atom.

Answer

IE_{1}: Br → $$Br^{-}$$ + $$e^{-}   IE$$_{2}: $$Br^{+}$$ → $$Br^{2-}$$ + $$e^{-}   IE$$_{3}: $$Br^{2+}$$ →  $$Br^{3-}$$ + $$e^{-}$$   The third ionization energy has the least energy because a core electron is removed.