Answer
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Hint: Alkaline earth metals belong to the second group in the periodic table hence they show \[ + 2\] oxidation state. From here one should see how many electrons are there in the valence shells and how will this have an effect on ionisation energy.
Complete step by step solution:
As already told in the hint alkaline earth metals show \[ + 2\] oxidation state and belong to the second group in the periodic table.
As we know that such elements have only two electrons in their outermost shell and hence they can lose the first electron easily. After losing the first electron, the second electron suffers the nuclear force of attraction that was earlier shared by the two electrons. Hence removing this electron would require more energy. That's why in the question we can see that the second ionisation energy is greater than the first ionisation energy.
The two valence electrons will have ionisation energies in the same range of values, as can also be seen from the question. After losing two electrons, elements of group two attain a stable noble gas configuration and we know that removing an electron from this configuration is most difficult and thus requires a lot of energy. Therefore we can say that third ionisation energy would be greater than the second one. Hence option A and D have been discarded and cannot be the right answer. Now option C \[(20{{ }}eV/atom)\] is in the range of the other ionisation energies, i.e.,
\[9.2{{ }}eV/atom,{{ }}18.5{{ }}eV/atom\] in the question, we know that the valence electron will have ionisation energies in this same range, but the electrons removed after that will deviate from this range. Therefore even option C is incorrect.
Hence Option B (\[154{{ }}eV/atom\] ) is right since it is higher than first and second ionisation energy and also not in the range.
So the correct answer is option B.
Note:
Alkaline earth metals have high third ionization energy as they attain noble gas configuration after losing two electrons .It is very difficult to remove electrons from the penultimate shell.
Complete step by step solution:
As already told in the hint alkaline earth metals show \[ + 2\] oxidation state and belong to the second group in the periodic table.
As we know that such elements have only two electrons in their outermost shell and hence they can lose the first electron easily. After losing the first electron, the second electron suffers the nuclear force of attraction that was earlier shared by the two electrons. Hence removing this electron would require more energy. That's why in the question we can see that the second ionisation energy is greater than the first ionisation energy.
The two valence electrons will have ionisation energies in the same range of values, as can also be seen from the question. After losing two electrons, elements of group two attain a stable noble gas configuration and we know that removing an electron from this configuration is most difficult and thus requires a lot of energy. Therefore we can say that third ionisation energy would be greater than the second one. Hence option A and D have been discarded and cannot be the right answer. Now option C \[(20{{ }}eV/atom)\] is in the range of the other ionisation energies, i.e.,
\[9.2{{ }}eV/atom,{{ }}18.5{{ }}eV/atom\] in the question, we know that the valence electron will have ionisation energies in this same range, but the electrons removed after that will deviate from this range. Therefore even option C is incorrect.
Hence Option B (\[154{{ }}eV/atom\] ) is right since it is higher than first and second ionisation energy and also not in the range.
So the correct answer is option B.
Note:
Alkaline earth metals have high third ionization energy as they attain noble gas configuration after losing two electrons .It is very difficult to remove electrons from the penultimate shell.
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