My chemical bonding professor says that the $\ce{O-S-O}$ bond angles in $\ce{SO4^2-}$ are ideal (109.5°).

Why? Is this because all bonds are equivalent, and electron distribution is shared among all oxygens?

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    $\begingroup$ Exactly. You may think of it as a superposition of resonance structures with all possible combinations of $\ce{S=O}$ and $\ce{S-O-}$. $\endgroup$ Sep 25 '15 at 5:29

Yes. All the bonds are equivalent as in the 4 resonating structures, each with 1.5 S and O bonds. So the resonant hybrid has four 1.5 double bonds (sigma and pi) with -0.5 formal charge on each O. This is why, the ion has a perfect tetrahedral geometry and shape as it is very symmetric and equivalent. Hope it helps.

The picture below on the right is what is thought to be the most accurate representation nowadays. However, the resonance structure on the left (and all of its equivalent representations) are more than likely acceptable for introductory chemistry classes.

Note that no matter which resonance structure you think is a better representation of the sulfate ion, the bonds are all equivalent and therefore sulfate ion has the ideal tetrahedral bond angle of 109.5 degrees.

enter image description here

  • 1
    $\begingroup$ Yes, all the bonds are equal in $\ce{SO4^2-}$, but thinking of them as bond order 1.5 is no longer accepted as correct. $\endgroup$
    – Jan
    Sep 25 '15 at 13:52
  • $\begingroup$ @Diyanko Bhowmik - also, each O doesn't have a -1.5 formal charge. That would give the overall molecule a -6 charge. $\endgroup$
    – Dissenter
    Dec 24 '15 at 20:12

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