The more thoughtful answer would be to provide examples of the ones that can exist, but that wasn't required. For completeness, let X, Y, and Z be other atoms and : be lone pairs of electrons; then
: Z X / / / (a) X = N (c) X = N (d) : N - Y \ \ \ Y Y Z
2- O S O S O
(-1) .. .. .. .. .. (-1) :O - S - O - S - O: ¨ ¨ ¨ ¨ ¨It is true that sulfur will accept more than an octet, so we could imagine additional resonance structures that would involve using the extra lone pairs on the end oxygens to make a double bond to one or both sulfurs, but that would then put the (-1) formal charge on the sulfur rather than the oxygen! And of the two, oxygen would far rather have it due to its higher electronegativity. So the 3 structuress like the following would contribute rather little to extra stability (see question #11) in the ion.
(-1) .. .. .. .. .. (-1) :O = S - O - S - O: ¨ ¨ ¨ ¨With or without the alternate resonance structures, the electronic structure about each of the central atoms, SOS, involves 4 directions in space whose electrons must avoid one another. So all will be bent, and the entire molecule will zig-zag like /\/\.
.. - .. - .. .. - (a) [ :S : N ::: C: ] (b) [ :S : C ::: N: ] (c) [ :C :: S :: N: ] ¨ ¨ (-1) (+1) (-1) (-1) (0) (0) (-2) (+2) (-1)
(Red is oxygen, gray is carbon, and white is hydrogen. The "ring" is that cycle of atoms at the right end of the molecule comprised of one oxygen and four carbons with other stuff dangling off. So we don't get confused, let's number the ring atoms starting with the oxygen as number 1 and proceeding clockwise around the ring to number 5, the carbon just below #1.)
Atom #2, the topmost carbon, is hybridized sp2 to make three s bonds to the atoms attached to it. But that leaves carbon's extra p electron to double bond with the top-right oxygen. That oxygen must also be hybridized sp2 to accommodate its pair of lone pairs and its double bond to atom #2.
Atom #3, the rightmost carbon, is also hybridized sp2 since he too is bonded to three other atoms. But his bond to his oxygen is only single as we can see since his oxygen is also bonded to a hydrogen (the rightmost). His oxygen is therefore sp3 hybridized to accommodate two single bonds and its pair of lone pairs. That means atom #3 has no choice but to use its extra p electron to double bond to atom #4, the bottomost carbon.
Fortunately, atom #4 is also sp2 hybridized to accept 3's double bond. And 4's oxygen is an OH too. So it too is sp3.
Finally, #4 has only a single bond left over for #5, the carbon atom that closes the ring back to that (#1) oxygen. So atom #5 is single bonded to four other atoms and must be sp3 hybridized.
So a Lewis structure for vitamin C would be sort of
H :O: \ .. H H :O: :O - C \ | | / | .. :O - C - C - C C - O: " | | | \ | H H H C H | :O: / Hand everybody has zero formal charge.
Take CO, carbon monoxide, for example. With what second row homonuclear diatomic, E2, is CO isoelectronic? And what does that tell you about the relative magnetic properties of CO and CO+ ?
Remove one electron to make CO+ and its buddy is unpaired! So CO+ would be paramagnetic.
So NH4+ is a weaker acid than PH4+, and equilibrium lies to the right with the weaker acid.
If you remain unconvinced, the weaker conjugate acid (NH4+) is associated with the stronger conjugate base (NH3). So again the weaker base (PH3) makes equilibrium lie to the right! The weaker conjugate acid and the weaker base are both on the same side.
Molecular fluorine is made by the electrolysis of the molten salt, KHF2. The existence of that salt suggests a bizarre HF2- ion! Lewis would absolutely choke on such a species, but MO makes it simple to understand. That ion would hold together only if there was a bond between hydrogen's 1s orbital and the highest occupied molecular orbital of F2-. Since F2- has an odd number of electrons, that highest MO orbital (s2px* ) has only one electron, and it stands to reason that it can bond with the single electron in hydrogen atom.
Remember the origin and hence the shape of that s2px* orbital to prove that HF2- must be linear. (The key is orbital overlap. The x axis runs between the fluorine atoms.)
There's no hope for triangular at all since the sign of the s2px* MO as we look along the F2- bond direction varies as
for the antibonding s2px*. That | is a node in the MO exactly between the fluorines. Were the 1s orbital of hydrogen to attempt to straddle the ion, that change in sign across the node would give exactly as much positive as negative overlap, meaning no overlap at all! So triangular geometries are nonbonding. Bent geometries are weakly overlapping and so weakly bonding, but linear geometries are strongly overlapping hence strongly bonding.
(-1) (-1) .. .. :O: :O: :O: :O: \ // <----> \\ / N N ¨ ¨