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Anastasy [175]
3 years ago
11

Explain how fluorine and aluminum would exchange valence electrons to form an ionic compound.

Chemistry
2 answers:
Butoxors [25]3 years ago
6 0

Answer: through their valence electrons

Explanation: It is important to know first that the valence electrons are what is involved in bonding.

Fluorine is group 7(halogens) and have 7 electrons. All members of this groups have 7 valence electron each e.g Chlorine, Bromine, Iodine

Aluminum is in group 3 and has 3 valence electrons.

Elements bonds to attain the duplet (2 electron maximum) and octet (8 electron maximum) electronic configuration which is the stable state.

Fluorine requires one electron to attain the octet state while aluminum need to give out 3 electrons to attain the octet state.

So Aluminum will just give out the 3 electrons to 3 fluorine atoms. One electron to each of them and they will both be stable. Simplified below;

Al + 3F - AlF3

o-na [289]3 years ago
5 0

Answer :

Explanation :

The aluminium and fluorine react to give ionic compound aluminium fluoride.

Aluminium has 1 valence electrons in their shell and fluorine has 7 valence electrons in their shell.

For the complete octet, both aluminium and fluorine exchange valence electrons to form an ionic compound.

The aluminium donates its three valence electrons to three fluorine atoms and they form one Al^{3+} and three F^- ions.

Electron transfer image is shown below.

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shepuryov [24]

Answer:

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Explanation:

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Using the equation:

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We can find the pH of the solution.

The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M

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pOH = -log 5.6x10⁻⁴M

pOH = 3.25

pH = 14-pOH

<h3>pH = 10.75</h3>
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Suppose that you measured out 3.50g of Na2SO4. how many moles would you have? (show work)​
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