Which of the following is a covalent bond? 1.Al2O3 2.AlF3 3.AlCl3 4.Al2(SO4)3 The correct answer is AlCl3.But I have a confusion between option c and d.A/c Fajan's rule more the size of anion more is the covalent character. So,SO4^-2 have large size than Cl^-3 .So option d is correct but option c is the right answer.Please explain this question .

  • $\begingroup$ Perhaps because the charge carrying oxygens in sulphate are in resonance? $\endgroup$ Nov 25 '18 at 13:57
  • $\begingroup$ But resonance also mostly happens in covalent compounds $\endgroup$
    – Kaushki
    Nov 25 '18 at 14:05
  • $\begingroup$ So you have to pull the electron density towards the cation. If those electrons are busy oscillating elsewhere, then won't it be difficult to pull such a distributed electron density? That's what I feel.. $\endgroup$ Nov 25 '18 at 14:07

Fajan's rule is correct, but the sulfate ion is complex. We need to ask what is being polarized, a large sulfate ion as a whole, or an oxygen atom specifically. As it turns out, the aluminum acts on a nearest oxygen, so that even if this oxygen is more polarizable because it has a sulfur to draw on, a chlorine is still more polarizable.

Another thing is that aluminum chloride is dimeric in the gas phase, whereas aluminum sulfate is just an ionic solid - although you might still say there is some covalent character in the aluminum-sulfate bonds.

  • $\begingroup$ But since charge is distributed throughout the molecule via resonance, then how can you visualize that only oxygen is being polarized rather than the entire molecule? $\endgroup$ Nov 25 '18 at 15:03
  • $\begingroup$ That's the issue that makes the question complex. If the charge is distributed freely, then the anion is large. But if a single oxygen is polarized first, and then, using its slightly deformed charge distribution, slightly polarizes the sulfur, which then slightly polarizes the other oxygens, then you might as well just concentrate on the first oxygen (small anion). So I suppose that the charge does not distribute equally over the whole molecule if only one oxygen is polarized. Degrees of resonance might be an explanation. $\endgroup$ Nov 25 '18 at 15:18
  • $\begingroup$ What are 'degrees of resonance'? $\endgroup$ Nov 25 '18 at 16:50
  • $\begingroup$ @YUSUFHASAN You don't get what "resonance" is supposed to be. It's about any oscillation. It's about mesomeric structures contributing to resonance hybrid. Also this is a place for discussion about that, since it has negligible importance for this topic. $\endgroup$
    – Mithoron
    Nov 25 '18 at 23:15
  • 1
    $\begingroup$ At the anode, electrons will be removed from something and oxygen will be evolved. The simplest equation doesn't even involve SO4--. The oxidation of water is: H2O - 2e -> 1/2 O2 + 2H+. If you must put SO4-- in there, then: SO4-- - 2e-> SO4 -> SO3 + 1/2 O2. Then SO3 + H2O -> H2SO4, so you get the SO4-- back, along with 2 H+. So neither SO2 nor SO3 is liberated. $\endgroup$ Nov 26 '18 at 19:54

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