Question

1. Use electron dot formulas to predict the formulas of the ionic compounds formed from the

Answer 1

Answer:

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Answer 2

Answer:

a) Bes

b) KF

c) MgO

d) H2S

e) AlCl3

f) NaI

g) SrSe

h) Ga2O3

Explanation:

In the picture attached the electron dot representation of different elements is shown.

a) From the picture it an be seen that Beryllium has 2 valence electrons, to form a ionic compound it will lost them and transforms into $Be^{2+}$

Sulfur has 6 valence electrons, to form a ionic compound it will gain 2 electrons (to complete 8 electrons in its outer shell) and transforms into $S^{2-}$

The ionic compounds are electrically neutral, so the charge combination of its ions must be zero. If we combine 1 $Be^{2+}$ cation with 1 $S^{2-}$ anion the net charge will be: +2 + (-2) = 0. Then the formula is BeS.

The other cases are solved analogously.

b)

potassium cation: $K^{+}$

fluorine anion: $F^{-}$

compound formula: KF (net charge: +1 + (-1) = 0)

c)

magnesium cation: $Mg^{2+}$

oxygen anion: $O^{2-}$

compound formula: MgO (net charge: +2 + (-2) = 0)

d)  hydrogen and sulfur don't make an ionic compound, they form hydrogen sulfide which is a covalent compound

e) aluminum and chlorine form AlCl3, which is covalent.

f)

sodium cation: $Na^{+}$

iodine anion: $I^{-}$

compound formula: NaI (net charge: +1 + (-1) = 0)

g)

strontium cation: $Sr^{2+}$

selenium anion: $Se^{2-}$

compound formula: SrSe (net charge: +2 + (-2) = 0)

h)

gallium cation: $Ga^{3+}$

oxygen anion: $O^{2-}$

compound formula: Ga2O3 (net charge: 2*(+3) + 3*(-2) = 0)