Question

What three (3) factors determine the reactivity of elements? Explain and give examples of each.

Answer 1

Explanation:

The three factors which determine the reactivity of elements are as follows.

(1) Number of valence electrons.

(2) Size of an atom.

(3) Electro negativity of an atom.

All these factors are explained as follows.

(1) Number of valence electrons

According to the octet rule, every atom requires to attain stability. Therefore, number of electrons in the outermost shell will decide the reactivity of an atom as every atom needs to fill its shells or sub shells as per the octet rule. As a result, the reactivity increases.

For example, the atomic number of chlorine is 17 and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$

In the 3p sub shell, there is deficiency of one electron. So, in order to attain stability chlorine atom will react readily with another species or atom which can either donate or share one electron.

Whereas an atom with completely fill shell will have low reactivity.

(2) Size of an atom

On moving down the group size of atom increases. As a result, the attraction between the nucleus and electrons decreases due to the shielding effect. Thus, the larger is an atom, the more easily it can give electrons. Therefore, the reactivity increases.

For example, the atomic number of calcium is 20 and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}$

Also it is known that first shell of every atom can have 2 electrons, second shell can have 8 electrons, third shell can have 18 electrons and so on.

Since, calcium is larger in size as compared to beryllium and magnesium so it will readily loose 2 electrons to attain stability. Thus, it will attain a $Ca^{2+}$ charge.

(3) Electro negativity of an atom

An atom with uneven distribution of electrons will acquire a certain amount of charge. Thus, it becomes polar in nature and in order to gain stability, the atom will either loose or gain electrons according to its charge.

For example, atomic number of fluorine is 9 and its electronic configuration is $1s^{2}2s^{2}2p^{5}$.

So, in order to gain stability fluorine will readily accept 1 electron to completely fill its 2p sub shell. As a result, the electronic configuration will become $1s^{2}2s^{2}2p^{6}$

Therefore, fluorine will have a charge of -1 because it has gained one electron.