Electropositivity of An Element

Electropositivity is the ability of an atom (mainly metals) to lose its valence electron(s) in order to achieve stable electron arrangement in the form of positive ion. The electropositivity of an atom of an element usually associates with its reactivity in a chemical reaction. The more electropositive of an element, the more reactive the element is.

Note: Stable electron arrangement is a state where the outermost shell of an atom is completely filled with electrons, which is commonly 8 electrons [octet configuration] or 2 electrons [duplex configuration] in the case of hydrogen and helium.

As shown in the diagram below, the electron arrangement of sodium atom is 2.8.1. So in order to achieve stable electron arrangement, sodium atom has to lose its one valence electron and transforms itself into a positively charged ion with electron arrangement of 2.8.

Note: The charge of an ion depends on the number of valence electrons an atom lost. If it loses 2 valence electrons, then the charge of ion formed would be 2+.

Factor of Electropositivity
 
The electropositivity of an element is determined by the attraction force between the nucleus of atom and its valence electrons. The farther the distance between the nucleus and the valence electrons, the weaker the attraction force between them, which makes it easier for an atom to lose its valence electrons, thus being more electropositive. 

Since the distance between nucleus of an atom and its valence electrons is related to its atomic size, it also can be concluded that element with bigger atomic size would be more electropositive than element with smaller atomic size.

In the example below, potassium atom has bigger atomic size than sodium atom, thus the distance between its nucleus and valence electron is farther, which causes the attraction force between them is weaker. Therefore, it is easier for potassium atom to lose its valence electron which makes potassium is more electropositive than sodium. 

And talking about atomic size, checks out the diagram below as a reference on how electropositivity of elements changes across the Periods and down the Groups of the Periodic Table of Elements.

That is all for today. I decided to write about this topic after reading my students' answer of a question regarding electropositivity of elements in their recent exam. Of course at the time they learned it, the water was clear as crystal, but now the water has turned very cloudy and very muddy, which makes me a bit angry.

Question 4

A reaction between 6.5 g of zinc and 50 cm³ of 1.0 moldm^-3 hydrochloric acid has produced zinc chloride solution and hydrogen gas. Calculate the maximum mass of zinc chloride formed.

[RAM: Zn, 65; Cl, 35.5]

Answer:

I gave this question to my students few weeks ago in a short quiz. Unfortunately for such a simple question, none of them got the answer right despite of how self-assured they were with it. Apparently they did a mistake by taking the number of moles of zinc as a basis to determine the number of moles of zinc chloride and its mass, instead of using the number of moles of hydrochloric acid to do so. 

Here is why. Below is the chemical equation of the reaction.

                                       Zn    _✛  2HCl  →   ZnCl₂    ✛   H₂  

In the case of number of moles of zinc which is 0.1 mol (6.5 g / 65 gmol^-1 ) used as a basis, it means based on chemical equation, 0.1 mol of zinc reacts with 0.2 mol of hydrochloric acid to produce 0.1 mol of zinc chloride. That is just impossible, because given the volume and the molarity of hydochloric acid, the maximum number of moles of hydrochloric acid involved in the reaction is only 0.05 mol (0.05 dm³ x 1.0 moldm^-3).

Sometimes tricky question like this was designed to test students' understanding on chemical reaction. 

Okay now back to the correct answer.

Step 1: The number of moles of hydrochloric acid

= Molarity x Volume of HCl
= 1.0 moldm^-3 x  0.05 dm³
= 0.05 mol

Step 2: Find the number of moles of zinc chloride

From chemical equation
2 mol of HCl  produces 1 mol of ZnCl₂
Therefore, 0.05 mol of HCl  produces 0.025 mol of  ZnCl₂

Step 3: Find the mass of zinc chloride formed

= Number of moles x RMM of zinc chloride
= 0.025 mol x  [65 + 35.5x2] gmol^-1
= 0.025 mol x  136 gmol^-1
= 3.4 g of zinc chloride
 

Once in a while, I like to throw tricky question in the expectation that my students would make a mistake in their answer. so that they could learn a hard lesson from it. In my experience, it's a very effective teaching method.