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Part 1: Ionic Bonding 1. Choose Sodium (Na). a. What type of element is it? b. How many valence electrons does it have? 2. Choose Fluorine (F). a. What type of element is it? b. How many valence electrons does it have? 3. Answer the question on the screen, "What type of bond is this combination likely to form?" a. Circle: Ionic or Covalent? b. Choose the appropriate number of atoms to make the bond. Record the number of each atom below: 4. Watch the final animation closely (it will play continuously). a. Describe the change in the number of valence electrons in the atoms as the bond is successfully formed: b. What does the positive (+) charge indicate (mention specific subatomic particles in your answer)? c. What does the negative (-) charge indicate (mention specific subatomic particles in your answer)? d. Record the name and molecular formula for the compound below:

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Answer:

Explanation:

Sodium is a group 1 element with atomic number 1. It has 11 electrons. It is soft reactive metal. It has 1 valence electron.

Fluorine is a group 7 element, a hologen with 7 valence electron. It is a most reactive non metal.

When sodium react with fluorine, ionic bond is formed in the resulting compound sodium fluoride.

One sodium and fluorine each totaling 2 atoms are enough to make the bond.

As the bond is formed, both atoms have octet structure. That is they each have 8 electrons on their outermost shells.

The positive charge on sodium indicates that sodium had lost 1 electron to fluorine atom.

The negative charge on fluorine ion indicates that fluorine atom had gained 1 electron from sodium atom to form negative ion.

The name of the compound is sodium fluoride with formula NaF.

Answer:

The reaction of 1st group element Na and 7th group element F, results in formation of Sodium fluoride (NaF).

Explanation:

1. (a) Sodium (Na) is element belonging to the first group of the periodic table.        

(b) The atomic number of Na is 11 and after distribution it has 1 valence electron.

2. (a) Fluorine (F) is the halogen element and belongs to group 7 of the periodic table.

(b) The atomic number of F is 19 and it has 7 valence electrons.

3. The reaction of Na with F results in an ionic interaction with the formation of NaF (Sodium fluoride).

(a) The reaction and combination of Na and F is likely to be ionic in nature.

(b) The one valence electron of Na and F are involved in formation of bond. Total 2 electrons are involved.

4. (a) The successful formation of bond result in octet completion of NaF by sharing of electrons. Both the elements share 8 electrons, i.e. 7 electrons of F and 1 electron of Na.

(b) The +ve charge is imparted for the loss of electron . Na lost its 1 valence electron to F for bond formation and thus imparts the +ve charge.

(c) The -ve charge is imparted with the gain of electron. In reaction of Na with F, fluorine being more electronegative gains its electrons from Na and imparts a -ve charge.

(d) The reaction of Na with F results in the formation of compound known as sodium fluoride with the molecular formula NaF.

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