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What property of the elements make them form different types of carbides like:

$\ce{Be}$ and $\ce{Al}$ - $\ce{Be2C}$ and $\ce{Al4C3}$ (Methanides) contains $\ce{C^4-}$ ion

$\ce{Na}$ and $\ce{Ca}$ - $\ce{Na2C2}$ and $\ce{CaC2}$ (acetylides) contains $\ce{C2^{2-}}$ ion

$\ce{Li}$ and $\ce{Mg}$ - $\ce{Li4C3}$ and $\ce{Mg4C3}$ (sesquicarbides) contains $\ce{C3^4-}$ ion.

Boron - $\ce{B4C}$ (covalent carbides)

Titanium and tungsten - $\ce{TiC}$ and $\ce{WC}$ (interstitial carbides)

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    $\begingroup$ This isn't necessarily a full answer but I note that the most polarising cations ($\ce{Be^{2+}}$ & $\ce{Al^{3+}}$) are associated with the least polarisable anions ($\ce{C^{4-}}$) and vice versa. Additionally, insterstitial carbides only form with large cations, as you might expect if the anion has to fit in the spaces in the cation crystal structure. $\endgroup$ – bon Jul 17 '15 at 10:22
  • $\begingroup$ I think it is due to the high lattice energy released when these compounds are formed. Also what @bon said. $\endgroup$ – ShankRam Feb 3 '16 at 13:33
  • $\begingroup$ @bon With that many negative charges on one atom, I would think $\ce{C^{4-}}$ would be the most polarizable anion of the three, wouldn't it? $\endgroup$ – hBy2Py Feb 12 '16 at 2:58
  • $\begingroup$ You might wanna consider diagonal relationship which accounts for similarities between those particular metals. $\endgroup$ – jatin Mar 18 '16 at 17:18
  • $\begingroup$ @jatin you gave a very important hint. thank you. I am going to write the answer. $\endgroup$ – Nilay Ghosh Mar 22 '16 at 10:50
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According to wikipedia,

Carbides can be generally classified by chemical bonding type as follows:

  1. salt-like carbides
  2. covalent carbides
  3. interstitial carbides
  4. "intermediate" transition metal carbides.

This type of classification is based on the electronegativity of the element to which the carbon is bonded.

According to this site:-

The most electropositive metals form ionic or saltlike carbides, the transition metals in the middle of the periodic table tend to form what are called interstitial carbides, and the nonmetals of electronegativity similar to that of carbon form covalent or molecular carbides.

Actually credits goes to @jatin who gave me an important hint:-

You might wanna consider diagonal relationship which accounts for similarities between those particular metals.

As you can see that Li and Mg, Na and Ca and Be and Al all form diagonal element pairs. So, they are expected to have similar properties. So, considering their electronegativity, the diagonal element pairs have almost same value (though Li and Mg electronegativity is quite different). So they form similar type of carbide.

enter image description here

(original picture)

As for the intestitial carbides, @bon said that

Insterstitial carbides only form with large cations (Ti and W), as you might expect if the anion (very small) to fit in the spaces in the cation crystal structure.

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  • $\begingroup$ There is more to the story than electronegativity of the metal. How does Li form an allene based carbidevand Ca forms an acetylide when their electronegaruvities are only 0.02 apart? Small cation size, thus concentrated charge, has some importance beyond electronegativity, as the end carbons in the allenide ion have more concentrated negative charge than the acetylide ion. $\endgroup$ – Oscar Lanzi Jan 1 '18 at 1:20

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