The first,second and third ionisation energie | vedclass

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Question

(D) The stability of an oxidation state is determined by the jump in ionization energy.Given: $E_1 = 7 \ eV$,$E_2 = 12.5 \ eV$,and $E_3 = 42.5 \ eV$.T

Vedclass · Accepted Answer

$+2$

Vedclass · Answer

(D) The stability of an oxidation state is determined by the jump in ionization energy.Given: $E_1 = 7 \ eV$,$E_2 = 12.5 \ eV$,and $E_3 = 42.5 \ eV$.The jump from $E_2$ to $E_3$ is $42.5 - 12.5 = 30 \ eV$,which is significantly large compared to the jump from $E_1$ to $E_2$ $(12.5 - 7 = 5.5 \ eV)$.$A$ large jump in ionization energy indicates that the electron is being removed from a stable,noble gas-like configuration after the removal of two electrons.Therefore,the element most easily loses two electrons to achieve a stable configuration,making the $+2$ oxidation state the most stable.Hence,$D$ is the correct answer.

List-$I$ (Electronic configuration)	List-$II$ ($\Delta_{i}H$ in $kJ\ mol^{-1}$)
$(a)$ $1s^{2} 2s^{2}$	$(i)$ $801$
$(b)$ $1s^{2} 2s^{2} 2p^{4}$	$(ii)$ $899$
$(c)$ $1s^{2} 2s^{2} 2p^{3}$	$(iii)$ $1314$
$(d)$ $1s^{2} 2s^{2} 2p^{1}$	$(iv)$ $1402$

The first,second and third ionisation energies ($E_1, E_2$ and $E_3$) for an element are $7 \ eV$,$12.5 \ eV$ and $42.5 \ eV$ respectively. The most stable oxidation state of the element will be

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