Question

5. The first, second and third ionsiation energies (E1, E2 & E3) for an element are 7 eV, 12.5 eV and 42.5 eV respectively. The most stable oxidation state of the element will be:

Answer 1

Answer:

Answer is given below by step by Step explanation

Explanation:

1 The difference between first two ionisation energies is very small, so the element can achieve

Answer 2

Answer:

To remove the first electron, you need to provide 7eV, to remove the second one; you need to access between 12.5eV and 42.5eV that is required to remove the third electron.  

This gives you the most stable oxidation state for this element to be +2.

Here the 3rd and 2nd ionization enthalpy difference is more than 15 eV and evaluate it accordingly.

Explanation:

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