What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: $3{{d}^{3}},3{{d}^{5}},3{{d}^{8}}$ and $3{{d}^{4}}$ ?
Answer
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Hint: We know that the transition metals exhibit a variable number of oxidation states in their compounds. This is one of the notable features of the transition elements. Few elements show exceptions for this case, most of these show variable oxidation states.
Complete answer:
As we know that the transition elements have low ionization energies therefore, they can have a wide range of oxidation states or positively charged forms. The positive oxidation states allow transition elements to form many different ionic and partially ionic compounds. Transition elements are those which have unpaired electrons either in their ground state or their most stable oxidation state. All d-block elements are not transition elements but all transition elements are d block elements. All the transition elements except the first and last element of the series exhibit variable oxidation state. Now we will find the number of elements having four electrons in their \[+2\] oxidation state up to Mn, the maximum stable oxidation state corresponds to the sum of s and d electrons. After Mn, there is decrease in the stability of higher oxidation states. The stable oxidation states of the transition elements with the following d electron configurations in the ground state of their atoms.
It can be noted from the below tables that the stability of a given oxidation state is dependent upon the nature of the elements with which the metal is combined. The highest oxidation states are shown in the compounds of fluorides and oxides because fluorine and oxygen are the most electronegative elements.
Note:
Remember that the stability of oxidation states for transition metals depends on the electronic configuration of the elements. The elements that show the greatest high number of oxidation states occur in or near the middle of the series. The small number of oxidation states at the extreme left-hand side end is due to the lesser number of electrons to lose or share.
Complete answer:
As we know that the transition elements have low ionization energies therefore, they can have a wide range of oxidation states or positively charged forms. The positive oxidation states allow transition elements to form many different ionic and partially ionic compounds. Transition elements are those which have unpaired electrons either in their ground state or their most stable oxidation state. All d-block elements are not transition elements but all transition elements are d block elements. All the transition elements except the first and last element of the series exhibit variable oxidation state. Now we will find the number of elements having four electrons in their \[+2\] oxidation state up to Mn, the maximum stable oxidation state corresponds to the sum of s and d electrons. After Mn, there is decrease in the stability of higher oxidation states. The stable oxidation states of the transition elements with the following d electron configurations in the ground state of their atoms.
It can be noted from the below tables that the stability of a given oxidation state is dependent upon the nature of the elements with which the metal is combined. The highest oxidation states are shown in the compounds of fluorides and oxides because fluorine and oxygen are the most electronegative elements.
| $3{{d}^{3}}4{{s}^{2}}$ | $+5$ |
| \[3{{d}^{5}}4{{s}^{1}}\] | $+1,+6$ |
| $3{{d}^{5}}4{{s}^{2}}$ | $+2,+7$ |
| $3{{d}^{8}}4{{s}^{2}}$ | $+2$ |
| $3{{d}^{4}}4{{s}^{2}}$ | $+1,+6$ |
Note:
Remember that the stability of oxidation states for transition metals depends on the electronic configuration of the elements. The elements that show the greatest high number of oxidation states occur in or near the middle of the series. The small number of oxidation states at the extreme left-hand side end is due to the lesser number of electrons to lose or share.
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