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(C) The tendency to lose electrons is a characteristic property of metals,which increases down a group and decreases across a period.$F$ (Fluorine) is

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$Fe$

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(C) The tendency to lose electrons is a characteristic property of metals,which increases down a group and decreases across a period.$F$ (Fluorine) is a non-metal with the highest electronegativity,so it has the least tendency to lose electrons.$S$ (Sulfur) is a non-metal.Both $Fe$ (Iron) and $Be$ (Beryllium) are metals.$Be$ has a relatively stable electronic configuration $(1s^2 2s^2)$,making it more difficult to remove an electron compared to $Fe$,which is a transition metal with a larger atomic size and lower ionization energy.Therefore,$Fe$ has a greater tendency to lose electrons than $Be$.

Which element has the greatest tendency to lose electrons?

Similar Questions

Among the statements $(I-IV)$,the correct ones are:
$(I)$ $Be$ has smaller atomic radius compared to $Mg$.
$(II)$ $Be$ has higher ionization enthalpy than $Al$.
$(III)$ Charge/radius ratio of $Be$ is greater than that of $Al$.
$(IV)$ Both $Be$ and $Al$ form mainly covalent compounds.

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