Chemistry, 09.01.2020 10:31 shelovejaylocs

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

following elements:

a) beryllium and sulfur
b) potassium and fluorine
c) magnesium and oxygen
d) hydrogen and sulfur
e) aluminum and chlorine
f) iodine and sodium
g) selenium and strontium
h) gallium and oxygen

Answers: 2

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Answer from: bikkiecuanas13

To write the ionic formula of the compound, the cationic and anionic ions are combined with their charges.

(a) Beryllium and Sulfur:

Cation: $\rm Be^2^+$

Anion = $\rm SO_4^2^-$

The net charge = +2-2 = 0

Formula = $\rm BeSO_4$

(b) Potassium and Fluorine

Cation = $\rm K^+$

Anion = $\rm F^-$

Net charge = 1-1 = 0

Formula = KF

Cation = $\rm Mg^2^+$

Anion = $\rm O^2^-$

Net charge = +2 - 2 = 0

Formula = MgO

(d) Hydrogen and Sulfur :

Cation : $\rm H^+$

Anion : $\rm S^2^-$

To combine with 1 molecule of S and balance the net charge 0. 2 hydrogen ions are used.

Formula = $\rm H_2S$

(e) Aluminium and Chloride:

Cation: $\rm Al^3^+$

Anion : $\rm Cl^-$

To combine with 1 molecule of Al and balance the net charge 0. 3 Cl ions are used.

Formula = $\rm AlCl_3$

(f) Iodine and Sodium:

Cation: $\rm Na^+$

Anion: $\rm I^-$

Net charge: +1 - 1 = 0

Formula = NaI

(g) Selenium and strontium:

Cation: $\rm Sr^2^+$

Anion: $\rm Se^2^-$

Net charge = +2 - 2 = 0

Formual = SrSe

(h) Gallium and Oxygen:

Cation: $\rm Ga ^3^+$

Anion: $\rm O^2^-$

Net charge: 0

Formula : $\rm Ga_2O_3$

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

Answer from: j015

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.

potassium cation: $K^{+}$

fluorine anion: $F^{-}$

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

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.

sodium cation: $Na^{+}$

iodine anion: $I^{-}$

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

strontium cation: $Sr^{2+}$

selenium anion: $Se^{2-}$

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

gallium cation: $Ga^{3+}$

oxygen anion: $O^{2-}$

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

Answer from: Quest

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

Answer from: Quest

2b2t

explanation:

2b2t

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