Use Lewis symbols to represent the electron transfer between the following atoms to give ions with noble gas configuations:
A. Mg and S
B. Ba and I
A. To determine the Lewis symbols for the electron transfer between Mg and S, follow these steps:
1. Identify the number of valence electrons for each atom.
- Magnesium (Mg) is in Group 2, so it has 2 valence electrons.
- Sulfur (S) is in Group 6, so it has 6 valence electrons.
2. Represent the valence electrons as dots around the atomic symbol.
- For Mg, draw 2 dots around the atomic symbol.
- For S, draw 6 dots around the atomic symbol.
Mg: ••
S: ••••••
3. Determine the electron transfer that will give both atoms a noble gas configuration.
- Magnesium can lose its 2 valence electrons to achieve a stable noble gas configuration, which is the same as the previous noble gas, neon (Ne).
- Sulfur can gain 2 electrons to achieve a stable noble gas configuration, which is the same as the previous noble gas, argon (Ar).
4. Complete the Lewis symbols by representing the electron transfer.
- Magnesium donates its 2 valence electrons to sulfur, resulting in Mg2+ (magnesium ion) and S2- (sulfide ion).
Mg2+: [Mg]2+ (It lost 2 electrons)
S2-: [S]2- (It gained 2 electrons)
B. Now let's determine the Lewis symbols for the electron transfer between Ba and I:
1. Identify the number of valence electrons for each atom.
- Barium (Ba) is in Group 2, so it has 2 valence electrons.
- Iodine (I) is in Group 7, so it has 7 valence electrons.
2. Represent the valence electrons as dots around the atomic symbol.
- For Ba, draw 2 dots around the atomic symbol.
- For I, draw 7 dots around the atomic symbol.
Ba: ••
I: •••••••
3. Determine the electron transfer that will give both atoms a noble gas configuration.
- Barium can lose its 2 valence electrons to achieve a stable noble gas configuration, which is the same as the previous noble gas, xenon (Xe).
- Iodine can gain 1 electron to achieve a stable noble gas configuration, which is the same as the previous noble gas, xenon (Xe).
4. Complete the Lewis symbols by representing the electron transfer.
- Barium donates its 2 valence electrons to iodine, resulting in Ba2+ (barium ion) and I- (iodide ion).
Ba2+: [Ba]2+ (It lost 2 electrons)
I-: [I]- (It gained 1 electron)
I can't draw these on the computer or these boards.
Mg: is the neutral Mg atom.
..
:S: is the neutral S atom.
Now move the two electrons on Mg to the bottom of S. That leaves Mg with no electrons and a 2+ charge and it leaves S with 8 electrons and a 2- charge.
Ba and I are done the same way except the formula is BaI2 so 1 electron from Ba is transferred to one of the I atoms and the other electron from Ba is transferred to the other I atom. Ba then carries a 2+ charge and each I carries a -1 charge.