Q. 4.90 - An Introduction to General, Organic, and Biological Chemistry

Question

Of the elements $Sn, Xe, Te, Sr, I and Rb$identify one that fits each of the following:

a. smallest atomic size 

b. an alkaline earth metal

c. a metalloid

d. lowest ionization energy

e. in Group $4A\left(14\right)$

f. least metallic character

g. seven valence electrons.

Step-by-Step Solution

Answer

a. In the elements Sn, Xe, Te, Sr, I, and Rb. The smallest atomic size is Xe.

b. The Sr (Strontium) is alkaline earth metal. 

c. In the elements Sn, Xe, Te, Sr, I, and Rb. A metalloid is Te (Tellurium).

d. The lowest ionization energy is element is Rb.

e.  The element Sn is in the group $4A$.

f. The least metallic character is Rb.

g.  The seven valence electrons is I. 

Part (a) step 1: Given Information

We need to find the smallest atomic size in Sn, Xe, Te, Sr, I, and Rb.

Part (a) step 2: Simplify

Here, 

 Sn, Xe, Te, Sr, I, and Rb. all the elements are in period $5$  and different groups.

And when move in the period from left to the right. Therefore, the size order is: 

$Xe<I<Te<Sn<Sr<Rb$

So, the smallest in size is  Xe(Xenon). 

Part (b) step 1: Given Information

We need to find an alkaline earth metal from the elements  Sn, Xe, Te, Sr, I, and Rb. 

Part (b) step 2: Simplify

Here, 

 Sn, Xe, Te, Sr, I, and Rb. all the elements are in period $5$  and different groups.

The elements in the group $2A$ are alkaline earth metals and Sr is in group $2A$. Therefore, Sr (Strontium) is alkaline earth metal.

Part (c) step 1: Given Information

We need to find a metalloid from the elements  Sn, Xe, Te, Sr, I, and Rb.

Part (c) step 2: Simplify

Here, 

 Sn, Xe, Te, Sr, I, and Rb. all the elements are in period $5$ and different groups.

Among these Te (Tellurium) is a metalloid.

Part (d) step 1: Given Information

We need to find the lowest ionization energy from the elements  Sn, Xe, Te, Sr, I, and Rb.

Part (d) step 2: Simplify

Here,

Sn, Xe, Te, Sr, I, and Rb. all the elements are in period  $5$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:

 $Xe>I>Te>Sn>Sr>Rb$

So, the lowest ionization energy is Rb.

Part (e) step 1: Given Information

We need to find  which element is in the group $4A$ from the elements  Sn, Xe, Te, Sr, I, and Rb.  

Part (e) step 2: Simplify

Here, 

 Sn, Xe, Te, Sr, I, and Rb. all the elements are in period $5$ and different groups.

Among these Sn is in group $4A$.

Part (f) step 1: Given Information

We need to find the least metallic character from the elements  Sn, Xe, Te, Sr, I, and Rb.  

Part (f) step 2: Simplify

Here, 

 Sn, Xe, Te, Sr, I, and Rb. all the elements are in period $5$ and different groups. And when moving in the period from left to the right side metallic character of the element decreases. Therefore, the metallic character order is:

$Xe<I<Te<Sn<Sr<Rb$

Thus, the lowest metallic character is Rb.

Part (g) step 1: Given Information

 We need to find the seven valence electrons from the elements  Sn, Xe, Te, Sr, I, and Rb.  

Part (g) step 2: Simplify

Here, 

 Sn, Xe, Te, Sr, I, and Rb. all the elements are in period $5$ and different groups. I (Iodine) is in group $7A$ and it has $7$valence electrons.