Competency #6

Answer and Explanation

The correct answer is D.

To answer this question, you first need to know which ion will be formed by each element. Aluminum is in group 13, and has 3 valence electrons. It is a metal so it loses electrons when it makes an ion, so it becomes an Al+3 ion (13 protons, 10 electrons) to leave behind a filled valence shell.

Br is in group 17, and has 7 valence electrons. It is a non-metal, so it gains electrons when making an ion. It needs one electron to get a filled valence shell, so it becomes a Br-1 ion (17 protons, 18 electrons).

When you make an ionic compound, you need to get the same total amount of negative and positive charge, so that the overall charge on the compound is zero.

  1. AlBr

    This would mean you have +3 charge and –1 charge, for an overall charge of +2. AlBr is not a neutral compound!
  2. Al2Br

    This would mean you have two +3 charges, for a total of +6 charge, and you would have and –1 charge, leaving the overall charge on the molecule of +5 (net charge = 6-1). Al22Br is not a neutral compound!
  3. Al3Br

    This would mean you have three +3 charges, for a total of +9 charge, and you would have and –1 charge, leaving the overall charge on the molecule of +8 (net charge = 9-1). Al 3Br is not a neutral compound!
  4. AlBr3

    This would mean you have one +3 charge and you would have three –1 charges, leaving the overall charge on the molecule of 0 (net charge = 3-3). AlBr3 is a neutral compound!
  5. Br3Al

    This formula is neutral in terms of charge, but when writing ionic compounds we always write the positive ion first, which is the metal (Al +3). It is not correct to write it the other way around.

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