The first four I.E. values of an element are | vedclass

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(C) The ionization energy $(I.E.)$ values provided are $I.E._1 = 284 \ kJ \ mol^{-1}$,$I.E._2 = 412 \ kJ \ mol^{-1}$,$I.E._3 = 656 \ kJ \ mol^{-1}$,an

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$3$

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(C) The ionization energy $(I.E.)$ values provided are $I.E._1 = 284 \ kJ \ mol^{-1}$,$I.E._2 = 412 \ kJ \ mol^{-1}$,$I.E._3 = 656 \ kJ \ mol^{-1}$,and $I.E._4 = 3210 \ kJ \ mol^{-1}$.Calculating the successive differences:$I.E._2 - I.E._1 = 412 - 284 = 128 \ kJ \ mol^{-1}$$I.E._3 - I.E._2 = 656 - 412 = 244 \ kJ \ mol^{-1}$$I.E._4 - I.E._3 = 3210 - 656 = 2554 \ kJ \ mol^{-1}$There is a very large jump in energy between the third and fourth ionization energies $(2554 \ kJ \ mol^{-1})$,which indicates that the fourth electron is being removed from a stable,noble gas-like core configuration.Therefore,the element has $3$ valence electrons.

The first four $I.E.$ values of an element are $284, 412, 656$ and $3210 \ kJ \ mol^{-1}$. The number of valence electrons in the element are

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