Group II metals have a smaller cationic size but one more valence electron than Group I metals. Why would the higher charge density of a metal cation affect the strength of the bonding of it in its elemental state?
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2$\begingroup$ Possibly related: chemistry.stackexchange.com/questions/2883/… $\endgroup$– Todd MinehardtJul 8, 2015 at 15:01
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1 Answer
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A good way to conceptualize metallic bonding is that the valence electrons on metal ions become delocalized, leaving positive ions to be electrostatically attracted to the electrons.
Group II metals contribute 2 valence electrons to the delocalized "sea of electrons", and thus have a +2 charge on the remaining cation. These combined create a stronger attractive force.
It is true that the radii of Group II metal cations are smaller than those of Group I cations (within the same row). This also affects bonding, as the delocalized electrons are closer to the nucleus in the Group II case.
In summary: the increased coulombic force between a higher number of delocalized electrons and higher charges on metallic cations create a stronger metallic bond. Transition metals tend to have even stronger metallic bonding due to the participation of electrons from d orbitals.
