b.p. & m.p. of magnesium

I just quote what I have answered before:
The strength of metallic bonds (金屬鍵)depends on(1) the number of outermost shell e- involved in forming metallic bonds (2)the size of the metal atom. The smaller the size, the closer is the distance between nucleus and the outermost shell e- . The greater attraction between the e- and nucleus, the higher will be the b.p. and m.p. of the metal. The size of the atom is governed by the effective nuclear charge.
However, due to the same group, they have same number of outermost shell e- . Just for Ca,its electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s 2 . After Ca, belongs to transition metals [21 - scandium] in which e-will be filled in 3d orbital. Due to the little energy difference between 3d and 4s orbital, Ca can use 3d orbital to strengthen the metallic bond. Hence it has a higher metallic strength. [由於鈣近過渡金屬,所以能用3d軌道去加強金屬鍵]
From Sr [1s2 2s2 2p6 3s2 3p6 4s 2 3d10 4p6 5s2 ] to Ba [1s2 2s2 2p6 3s2 3p6 4s 2 3d10 4p6 5s2 4d105p6 6s2 ] , their d orbitals can take part in strengthening the metallic bond,making their m.p. and b.p. higher in spite of having a larger atomic radius.
For Be due to its small atomic size, the distance between nucleus and the outermost shell e- is closer. The greater attraction between the e- and nucleus, the higher will be the b.p. and m.p. of the metal. Hence its m.p. and b.p. is very high.
The above reasons explain why Be, Ca, Sr and Ba have higher m.p. and b.p. comparing Mg itself.
m.p. = melting point
b.p. = boiling point
I hope you will find them helpful.