Which of the following is the strongest oxidizing agent:
A.\[Cr{O_4}^{2 - }\]
B.\[Mo{O_4}^{2 - }\]
C.\[W{O_4}^{2 - }\]

Hint:The physical and chemical properties of an element depend upon the electronic distribution in the various orbits or shells of an atom. Especially, the distribution of electrons in the valence shell helps in predicting the physical and chemical properties of that element. As the element with similar valence shell electronic configurations recurs at regular intervals in the periodic table, it is the main cause of periodicity. One of the important parts of the long-form periodic table is that several properties vary regularly in moving across a period or down a group which is known as periodic properties.

Step-by-step explanation:
Step 1
In a covalent bond formed between two similar atoms the shared electron pair lies just between the two atoms and contributes equally to both the atoms. So, there is no polarity developed between the combining atoms in this type of bond. The situation is different for a heteroatomic covalent bond in which the electron cloud is shifted towards a particular atom and thereby develops a partial negative charge on it and consequently a partial positive charge develops on the other atom.
Step 2
According to Pauling, the shared pair of electrons shifts towards that atom which exerts a greater pull and named this tendency as electronegativity.
Step 3
Variation across a period: The electronegativity increases on moving from left to right in a period of the periodic table.
Variation down a group: The electronegativity decreases down the group.
Step 4
We know oxidation is the loss of an electron, so the strength of an oxidizing agent depends upon the ease of accepting electrons. So, the element which has a greater value of electronegativity has a higher tendency of accepting electrons and hence will be a stronger oxidizing agent.
The higher oxidation states for heavier elements in the d block are more stable. Cr, Mo, and W are the members of Group 6 the d block elements. So, the order of stability for the given molecules are: \[W{O_4}^{2 - } > Mo{O_4}^{2 - } > Cr{O_4}^{2 - }\].

Hence, \[Cr{O_4}^{2 - }\]having the highest value of electronegativity compared to the rest, and being the least stable compound serves as the strongest oxidizing agent.

Note: Oxidation is the process of loss of electrons and oxidizing agents are good acceptors of electrons. The reduction is the process of gain of electrons and reducing agents are the good donor of electrons.