Figure 2 shows what happens to the electrons in the outer shells when a sodium atom reacts with a fluorine atom.

Chemistry

1 answer:

QveST [7]2 years ago

4 0

Sodium reacts with fluorine to make a covalent bond between them. A covalent bond is formed when there is a mutual sharing of electrons between the two adjacent atoms.

Here, Sodium is metal and fluorine is non-metal. So, the bond between them may be an ionic bond, but as the Sodium has one valence electron and fluorine is highly electronegative. Sodium wants to lose electrons and fluorine wants to gain electrons to form a chemical bond. Therefore, the sodium donates its valence electron to fluorine, forms a cation, and fluorine accepts the electron and forms an anion and they distribute the difference in the charge between them.

So, we can say that when sodium and fluorine react to form sodium fluoride with a covalent bond.

Learn more about covalent bonds here:

brainly.com/question/12732708

#SPJ10

You might be interested in

How to convert 12g of H2O to moles

Anna35 [415]

Answer:

moles = given mass/atomic mass

so H2O mass = 2 +16=18

so 12g of h2o= 12/16 = 3/4 moles

8 0

2 years ago

Which energy level has the most energy available to it?<br> 4<br> 3<br> 2<br> 1

aliya0001 [1]

Tropic levels have most energy

8 0

2 years ago

What is the missing word. Please Help Me. Thank you!!

Gnoma [55]

Calcium Floride (Caf2)
Hope this helped =D

7 0

2 years ago

Read 2 more answers

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

8090 [49]

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is $P_A$ and vapour pressure of component B is $P_B$ .

$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

$P_A >P^0_A\cdot x_A$ , $P_B>P^0_B\cdot x_B$