The successive ionization energies of an elem | vedclass

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(B) The successive ionization energies are $IE_1 = 20 \ eV, IE_2 = 45 \ eV, IE_3 = 150 \ eV, IE_4 = 900 \ eV, IE_5 = 1800 \ eV$. There is a large jump

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$AX_3$

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(B) The successive ionization energies are $IE_1 = 20 \ eV, IE_2 = 45 \ eV, IE_3 = 150 \ eV, IE_4 = 900 \ eV, IE_5 = 1800 \ eV$. There is a large jump in ionization energy between $IE_3$ and $IE_4$ $(900 - 150 = 750 \ eV)$. This indicates that the fourth electron is being removed from a stable noble gas core,meaning the element $(A)$ has $3$ valence electrons. Therefore,the stable oxidation state of $(A)$ is $+3$. Thus,the formula of the halide of $(A)$ is $AX_3$.

The successive ionization energies of an element $(A)$ are given as $IE_1 = 20 \ eV, IE_2 = 45 \ eV, IE_3 = 150 \ eV, IE_4 = 900 \ eV, IE_5 = 1800 \ eV$. What is the formula of the halide of $(A)$?

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