
Which of the following is the most basic oxide?
A.$S{O_3}$
B.$Se{O_3}$
C.$Po{O_2}$
D.$Te{O_2}$
Answer
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Hint: While answering this question we should keep in mind the rule that in a periodic table the basicity of an element increases down a group. Knowing the atomic number will help us to find the basicity of the element.
Step by step answer:
As the atomic size of the element increases, its electron affinity decreases consequently. It turns out that when moving vertically in the periodic table, the size of the atom outdoes is electronegativity with regard to basicity.
Thus, on the basis of the position of the element in the periodic table we can determine the most basic oxide. Let us examine the position of each of the options.
$S{O_3}$: The atomic number of Sulphur is 16. It is placed in period 3, group 16 of the periodic table.
$Se{O_3}$: The atomic number of Selenium is 34. It is placed in period 4, group 16 of the periodic table.
$Po{O_2}$: The atomic number of Polonium is 84. It is placed in period 6, group 16 of the periodic table.
$Te{O_2}$: The atomic number of Tellurium is 52. It is placed in period 5, group 16 of the periodic table.
The order of the elements in the periodic table is: S, Se, Te, Po. So the basicity of the oxides of these elements will be in the order:
$S{O_3} < {\text{ }}Se{O_3} < {\text{ }}Te{O_2} < {\text{ }}Po{O_2}$
So the most basic oxide is $Po{O_2}$. So Option C is the correct answer.
Note:
1.Basicity is the property of the substance to donate unshared pairs of electrons. This is also known as the Lewis Concept.
2.Basicity is also defined as the ability of a substance to gain ${H^ + }$ ions. This is also known as Bronsted-Lowry Concept.
3.Basicity is also the ability of a substance to give $O{H^ - }$ ions in its aqueous solution.
Step by step answer:
As the atomic size of the element increases, its electron affinity decreases consequently. It turns out that when moving vertically in the periodic table, the size of the atom outdoes is electronegativity with regard to basicity.
Thus, on the basis of the position of the element in the periodic table we can determine the most basic oxide. Let us examine the position of each of the options.
$S{O_3}$: The atomic number of Sulphur is 16. It is placed in period 3, group 16 of the periodic table.
$Se{O_3}$: The atomic number of Selenium is 34. It is placed in period 4, group 16 of the periodic table.
$Po{O_2}$: The atomic number of Polonium is 84. It is placed in period 6, group 16 of the periodic table.
$Te{O_2}$: The atomic number of Tellurium is 52. It is placed in period 5, group 16 of the periodic table.
The order of the elements in the periodic table is: S, Se, Te, Po. So the basicity of the oxides of these elements will be in the order:
$S{O_3} < {\text{ }}Se{O_3} < {\text{ }}Te{O_2} < {\text{ }}Po{O_2}$
So the most basic oxide is $Po{O_2}$. So Option C is the correct answer.
Note:
1.Basicity is the property of the substance to donate unshared pairs of electrons. This is also known as the Lewis Concept.
2.Basicity is also defined as the ability of a substance to gain ${H^ + }$ ions. This is also known as Bronsted-Lowry Concept.
3.Basicity is also the ability of a substance to give $O{H^ - }$ ions in its aqueous solution.
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