Find the formula of the halide of a metal who | vedclass

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(C) The successive ionization energies are $10, 16, 25, 400, 500 \ eV \ mol^{-1}$.There is a very large jump in ionization energy between the $3^{rd}$

Vedclass · Accepted Answer

$MX_3$

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(C) The successive ionization energies are $10, 16, 25, 400, 500 \ eV \ mol^{-1}$.There is a very large jump in ionization energy between the $3^{rd}$ $(25 \ eV \ mol^{-1})$ and $4^{th}$ $(400 \ eV \ mol^{-1})$ ionization energies.This indicates that the $4^{th}$ electron is being removed from a stable noble gas core.Therefore,the metal $M$ has $3$ valence electrons and exhibits a stable oxidation state of $+3$.The formula of its halide (where $X$ is a halogen with $-1$ charge) is $MX_3$.

Find the formula of the halide of a metal whose successive ionization energies are $10, 16, 25, 400, 500 \ eV \ mol^{-1}$ respectively $:-$

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