Increasing order of melting point by searching the temperature on the web:

$$\ce{NH3} <\ce{I2} <\ce{Zn}<\ce{C_{diamond}}.$$

The idea I have read on "Chemistry: The Central Science" by Theodore L. Brown, ‎H. Eugene LeMay Jr., ‎Bruce E. Bursten on Chapter 11 says something related:

We need more energy to break intramolecular bonds than overcame an intermolecular bond. So, as much stronger is the bond, most energy is required to overcome it.

I want to know why $I_2$ has a higher melting point than $NH_3$. I suspect that $I_2$ has the strongest bond (so a bigger melting point). $I_2$ has intermolecular bonds such as dispersion force (It has big polarizability because of its larger size so his intermolecular bond is strong) and $NH_3$ it has Hydrogen bonding force (the strongest intermolecular force) and also dispersion force. So, I think, but I am not very sure that iodine's force is bigger than NH3 force because of its size (and polarizability).

Is it correct? Must I have into account covalent force in $I_2$?

And why Zinc has an upper melting point in relation with C? Is it because Zn has a covalent bond and Zinc metallic bond or it does not matter?

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    $\begingroup$ Comparing the melting points of things with radically different structures isn't going to reveal useful patterns. Nor is comparing ammonia to iodine on the basis of bond types. An iodine molecule is 15 times heavier than an ammonia molecule; why expect them to be the same based on bonding alone? $\endgroup$ – matt_black Aug 1 '18 at 12:51
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    $\begingroup$ We prefer to not use MathJax in the title field, see here for details. Also @manooooh $\endgroup$ – Martin - マーチン Aug 2 '18 at 18:00
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    $\begingroup$ I am trying to correct the question to be clearer. $\endgroup$ – Geraldine Aug 3 '18 at 15:23
  • $\begingroup$ It is clear actually and should be rather closed as duplicate. It's just that big molecules bind stronger with vdW then small with hydrogen bonds. That's it. $\endgroup$ – Mithoron Aug 3 '18 at 18:42
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    $\begingroup$ @Mithoron Thanks! What about Z and C(Diamond) ? $\endgroup$ – Geraldine Aug 12 '18 at 17:57

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