Question

Which elements exhibit the greatest number of oxidation states in its compounds?
a. Ca
b. V
c. Cu
d. Na

Answer 1

Hint: Try to recall that the elements which show the greatest number of oxidation states occur in or near the middle of the series. Now, by using this you can easily find the correct option from the given ones.

Complete step by step solution:
1. It is known to you that the transition elements exhibit a large number of oxidation states in their compounds. With the exception of a few elements, most of these show variable oxidation states.
2. The different oxidation states of the transition elements are related to their electronic configuration.
The existence of transition metals like V, etc. In different oxidation states means that their atoms can lose different numbers of electrons.
3. Out of all elements given in options, only vanadium (V) and copper (Cu) are transition metals.
4. For vanadium (V), the minimum oxidation state is equal to the number of electrons in the s-orbitals and the other oxidation states are given by the sum of outer s- and some or all d –electrons.
5. The electronic configuration of vanadium (V) is $\[3{d^3}4{s^2}\]$ and the oxidation states it exhibits are +2, +3 and +4.
6. The oxidation state exhibited by copper is +1 and +2.
7. The oxidation state of calcium (Ca) and sodium (Na) is fixed which is +2 and +1 respectively.
Therefore, from above we can conclude that option B is the correct option to the given question.

Note:
1. It should be remembered to you that the highest oxidation states of transition metals are found in compounds of fluorides and oxides because fluorine and oxygen are the most electronegative elements.
2. Also, you should remember that the transition elements show low oxidation states in some compounds or complexes having ligands such as CO.