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Question
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
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Solution
Due to its small size and high electronegativities, oxygen or fluorine elements can oxidize the metal to its higher oxidation state.
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RELATED QUESTIONS
What are interstitial compounds?
Why do the transition elements have higher enthalpies of atomisation?
|
`E_((M^(2+)/M)` |
Cr | Mn | Fe | Co | Ni | Cu |
| -0.91 | -1.18 | -0.44 | -0.28 | -0.25 | -0.34 |
From the given data of E0 values, answer the following questions :
(1) Why is `E_(((Cu^(2+))/(Cu)))` value exceptionally positive
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(3) Which is the stronger reducing agents Cr2+ or Fe2+ ? Give Reason.
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Electronic configurations
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How would you account for the following?
Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
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Why is \[\ce{HCl}\] not used to make the medium acidic in oxidation reactions of \[\ce{KMnO4}\] in acidic medium?
Which of the following will not act as oxidising agents?
(i) \[\ce{CrO3}\]
(ii) \[\ce{MoO3}\]
(iii) \[\ce{WO3}\]
(iv) \[\ce{CrO^{2-}4}\]
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Account for the following:
Copper has an exceptionally positive `"E"_("M"^(2+)//"M")^0` value.
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Give a reason for the following:
Zinc, cadmium and mercury are considered as d-block elements but not regarded as transition elements.
Give a reason for the following:
Transition metals possess a great tendency to form complex compounds.
The E° value for the Mn2+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to the change of:
