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3 years ago

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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The weight of the body in the air is .... the weight of the submerged body

stira [4]

Answer:

the correct answer is option 'b': More than

Explanation:

The 2 situations are represented in the attached figures below

When an object is placed in air it is acted upon by force of gravity of earth which is measured as weight of the object.

While as when any object is submerged partially or completely in any fluid the fluid exerts a force in upward direction and this force is known as force of buoyancy and it's magnitude is given by Archimedes law as equal to the weight of the fluid that the body displaces, hence the effective force in the downward direction direction thus the apparent weight of the object in water decreases.

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3 years ago

Question 14 of 25

algol13

Answer:

describes properties characteristic of no more than two electrons in the vicinity of an atomic nucleus or of a system of nuclei as in a molecule

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2 years ago

Read 2 more answers

II. Ionic Equations

mario62 [17]

Answer:

Complete ionic: $\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Net ionic: $\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

Explanation:

Start by identifying species that exist as ions. In general, such species include:

Soluble salts.
Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ . These three salts will exist as ions:

Each $\rm AgNO_3\, (aq)$ formula unit will exist as one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion.
Each $\rm CaCl_2$ formula unit will exist as one $\rm Ca^{2+}$ ion and two $\rm Cl^{-}$ ions (note the subscript in the formula $\rm CaCl_2\!$ .)
Each $\rm Ca(NO_3)_2$ formula unit will exist as one $\rm Ca^{2+}$ and two $\rm {NO_3}^{-}$ ions.

On the other hand, $\rm AgCl$ is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each $\rm AgNO_3\, (aq)$ formula unit will exist as only one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion. However, because the coefficient of $\rm AgNO_3\, (aq)\!$ in the original equation is two, $\!\rm AgNO_3\, (aq)$ alone should correspond to two $\rm Ag^{+}\!$ ions and two $\rm {NO_3}^{-}\!$ ions.

Do not rewrite the salt $\rm AgCl$ because it is insoluble.

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of $\rm Ca^{2+}$ and two units of $\rm {NO_3}^{-}$ . Doing so will give:

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}$ .