Q. 4.89 - An Introduction to General, Organic, and Biological Chemistry

Question

Of the elements Na, Mg, Si, S, Cl, and Ar, identify one that fits each of the following: $\left(4.2, 4.6, 4.7\right)$

a. largest atomic size

b. a halogen

c. electron arrangement $2,8,4$

d. highest ionization energy

e. in group $6A\left(16\right)$

f. most metallic character

g. two valence electrons.

Step-by-Step Solution

Answer

a. In the elements Na, Mg, Si, S, Cl, and Ar. The largest size of Na.

b. In the elements Na, Mg, Si, S, Cl, and Ar.  A halogen is Cl.

c. The electron management $2,8,4$ is element  Si.

d. The highest ionization energy is element Ar.

e. The element S is in the group $6A$.

f. The most metallic character is Na.

g. The two valence electrons is Mg.

Part (a) step 1: Given Information

We need to find the largest atomic size in Na, Mg, Si, S, Cl, and Ar.

Part (a) step 2: Simplify

Here,

Na, Mg, Si, S, Cl, and Ar, all the elements are in period $3$ and different groups. And when move in the period from left to the right side of the element decreases. Therefore, the size order is:

$Na >Mg > Si > S > Cl > Ar$

So, the longest in size is Na.

Part (b) step 1: Given Information

We need to find a halogen from the elements  Na, Mg, Si, S, Cl, and Ar.

Part (b) step 2: Simplify

Now, 

Na, Mg, Si, S, Cl, and Ar, all the elements are in periods $3$ and different groups.

Therefore, Cl (Chlorine) is in the group $7A$ and $7A$ is a halogen group. So, Cl is a halogen.

Part (c) step 1: Given Information

We need to find the electron management $2,8,4$from the elements Na, Mg, Si, S, Cl, and Ar.

Part (c) step 2: Simplify

Here,

Na, Mg, Si, S, Cl, and Ar, all the elements are in periods  $3$and different groups.

The electron management $2,8,4$ 

$4$ electrons in valence electron and here Si is in the group. Then Si(Silicon) has this management.

Part (d) step 1: Given Information

We need to find the highest ionization energy from the elements Na, Mg, Si, S, Cl, and Ar. 

Part (d) step 2: Simplify

Here,

Na, Mg, Si, S, Cl, and Ar, all the elements are in periods $3$ and different groups. And when moving in the period from left to the right side ionization energy of the element increases. Therefore, the ionization energy order is:

$Na <Mg < Si < S < Cl <Ar$

So, the highest ionization energy is Ar.

Part (e) step 1: Given Information

We need to find which element is in the group $6A$from the elements Na, Mg, Si, S, Cl, and Ar.

Part (e) step 2: Simplify

Here,

Na, Mg, Si, S, Cl, and Ar, all the elements are in periods $3$ and different groups.

Among the elements S (sulfur) is in group $6A$.

Part (f) step 1: Given Information

 We need to find the most metallic character from the elements Na, Mg, Si, S, Cl, and Ar. 

Part (f) step 2: Simplify

The period metallic character decreases from left to right meaning elements on the left side of the periodic table (metals) have a more metallic character. The safely that the noble gases have the least metallic character.

In a group, the metallic character rises from top to bottom. This is because elements gain electron levels therefore their valence electrons are less attracted to their nucleus. The valence electrons are more likely to get lost.

The elements are all located in the same period. So we know that the metallic character decreases over a period we determine that sodium has the most metallic character.

Part (g) step 1: Given Information

 We need to find the two valence electrons from the elements Na, Mg, Si, S, Cl, and Ar.  

Part (g) step 2: Simplify

Here,

Na, Mg, Si, S, Cl, and Ar, all the elements are in periods $3$ and different groups.

Mg is in a group $2A$ and it has the atomic arrangement of $(2,8,2)$ Therefore, it has $2$ valence electrons.