More reactive the metal, lesser the melting it has. Right?

Dec 26, 2016

It sounds like a coincidence to me. You're talking about chemical properties and trying to relate them to physical properties, which is usually not necessarily clear-cut.

Claim:

• Aluminum is more reactive than molybdenum, so it has a lower melting point?

Well, it IS more reactive, since it was more easily oxidized if the reaction is correct, and it DOES have a lower melting point... (${1221}^{\circ} \text{F}$ vs. ${4753}^{\circ} \text{F}$)

(Though you should consider that molybdenum (VI) oxide is a solid and not aqueous, and that the oxidation state of molybdenum in it is particularly large in comparison to aluminum ion.)

Counterexample:

We can see that aluminum has a higher melting point than zinc.
(${1221}^{\circ} \text{F}$ vs. ${787.2}^{\circ} \text{F}$)

Therefore, reactivity and melting point are not related, or there are complicating factors that we're not considering. In other words, there is no clear link between reactivity and melting point.

Instead, I can think of a few physical reasons why aluminum has a lower melting point (a physical property).

Melting point for metals has to do with:

• How many electrons are "free" to move between metal atoms in the metallic crystal structure.
• How well they are delocalized throughout the metal atoms.

The more half-filled orbitals a metal has, the more electrons it can contribute""^([1]) to the conduction band""^([2]), and thus, the more delocalized its valence electrons are.

The following electron configurations for $\text{Al}$ and $\text{Mo}$ are:

$\text{Al} : \left[N e\right] 3 {s}^{2} 3 {p}^{1}$
$\text{Mo} : \left[K r\right] 5 {s}^{1} 4 {d}^{5}$ (compare with $\text{Cr}$ for the electron configuration, though $\text{W}$ does not have the same valence electronic structure.)

Since $\text{Mo}$ has six unpaired electrons, it contributes more electrons to the conduction band, so it has much more electron delocalization than $\text{Al}$, so its metallic bonding is stronger, and it has a much higher melting point than $\text{Al}$ (${4753}^{\circ} \text{F}$ vs. ${1221}^{\circ} \text{F}$).

Furthermore, its significantly higher atomic number ($42$ vs. $13$) means it has a significantly higher effective nuclear charge ($20.25$ vs. $9.5$ from Slater's Rules), so it has more closely-packed metal atoms, and thus stronger metallic bonding and a higher melting point.