
Select the strongest oxidant oxoanion and weakest oxidant oxoanion among the following: $VO_4^{3 - },CrO_4^{2 - },MnO_4^{2 - },FeO_4^{2 - }$
A) Strongest: $FeO_4^{2 - }$ : weakest: $VO_4^{3 - }$
B) Strongest: $VO_4^{3 - }$ : weakest: $FeO_4^{2 - }$
C) Both are equally strong
D) None of these
Answer
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Hint: As we know that the oxidation state of any element is basically the charge present on that element and we also know that higher the value of oxidation state, strongest will be the oxidant oxoanion.
Complete step-by-step answer:
As we know that an oxoanion is basically an ion which possesses the generic formula as ${A_x}O_y^{z - }$. We are also aware of the oxidation number of the central atom which can be calculated using the sum of all the elements and number of atoms present in the compound.
So let us calculate the oxidation number of each compound. The first one is Vanadium oxoanion $VO_4^{3 - }$ and let us consider the vanadium as $x$, so, the oxidation number of this compound will be:
$\Rightarrow x + ( - 8) = - 3$
$\Rightarrow x = + 5$
Similarly, for second compound which is chromate and the oxidation number would be:
$\Rightarrow x + ( - 8) = - 2$
$\Rightarrow x = + 6$
The third one is manganate oxoanion whose oxidation number can be similarly calculated as chromate oxoanion and it would be equivalent to:
$\Rightarrow x + ( - 8) = - 2$
$\Rightarrow x = + 6$
And lastly we have ferrate oxoanion and its oxidation number would be found as:
$\Rightarrow x + ( - 8) = - 2$
$\Rightarrow x = + 6$
So, we can see that the lowest oxidation state is of vanadium oxoanion and the rest possess a similar oxidation state. And we also discussed that lower the value of oxidation number, weaker will be the oxoanion. So, we can conclude that the vanadium oxidant oxoanion is the weakest one because it has the lowest oxidation number among all the given oxoanions.
Now, among manganate, ferrate and chromate, ferrate oxidant oxoanion is the strongest because it possesses the highest effective nuclear charge as compared to the other two oxoanions as well as it has a highest oxidation number among the four.
Therefore the correct answer is option (A).
Note: Always remember that the oxidation number of a central element in case of an oxidant oxoanion is inversely related to its strength. If the value of oxidation number is higher, the oxoanion will be the strongest and if it is lower, the oxoanion will be weakest.
Complete step-by-step answer:
As we know that an oxoanion is basically an ion which possesses the generic formula as ${A_x}O_y^{z - }$. We are also aware of the oxidation number of the central atom which can be calculated using the sum of all the elements and number of atoms present in the compound.
So let us calculate the oxidation number of each compound. The first one is Vanadium oxoanion $VO_4^{3 - }$ and let us consider the vanadium as $x$, so, the oxidation number of this compound will be:
$\Rightarrow x + ( - 8) = - 3$
$\Rightarrow x = + 5$
Similarly, for second compound which is chromate and the oxidation number would be:
$\Rightarrow x + ( - 8) = - 2$
$\Rightarrow x = + 6$
The third one is manganate oxoanion whose oxidation number can be similarly calculated as chromate oxoanion and it would be equivalent to:
$\Rightarrow x + ( - 8) = - 2$
$\Rightarrow x = + 6$
And lastly we have ferrate oxoanion and its oxidation number would be found as:
$\Rightarrow x + ( - 8) = - 2$
$\Rightarrow x = + 6$
So, we can see that the lowest oxidation state is of vanadium oxoanion and the rest possess a similar oxidation state. And we also discussed that lower the value of oxidation number, weaker will be the oxoanion. So, we can conclude that the vanadium oxidant oxoanion is the weakest one because it has the lowest oxidation number among all the given oxoanions.
Now, among manganate, ferrate and chromate, ferrate oxidant oxoanion is the strongest because it possesses the highest effective nuclear charge as compared to the other two oxoanions as well as it has a highest oxidation number among the four.
Therefore the correct answer is option (A).
Note: Always remember that the oxidation number of a central element in case of an oxidant oxoanion is inversely related to its strength. If the value of oxidation number is higher, the oxoanion will be the strongest and if it is lower, the oxoanion will be weakest.
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