Yes, what I wrote on the topic in the previous question was very, very brief and while writing it I was not thinking of boron — an element that usually calls for exceptions due to it only having three valence electrons.
However nonetheless, the first thing you should do for any compound is to consider all atoms in ground state configuration and see whether you can form the required bonds from occupied orbitals in that state. Let’s take the example compounds $\ce{BF3, NF3, AsH3, C2H4}$ (tricoordinate) $\ce{CO2, H2O, H2Se, BeCl2, CH2}$ (dicoordinate) and $\ce{HCl, N2, CO}$ (monocoordinate) for the discussion. Any structural data referenced herein is taken from the corresponding Wikipedia articles.
Ideal cases
I consider the examples $\ce{AsH3, H2Se}$ and $\ce{HCl}$ to be ideal cases. Taking each atom in its ground state ($\ce{As}:{ [\ce{Ar}]}\,\mathrm{4s^2\,4p^3}; \ce{Se}{: [\ce{Ar}]}\,\mathrm{4s^2\,4p^4}; \ce{Cl}{: [\ce{Ne}]}\,\mathrm{3s^2\,3p^5}; \ce{H}{: \mathrm{1s^1}}$), arsenic has three half-filled p orbitas, selenium has two and chlorine has one. Each of these elements (except hydrogen) has a single lone pair in an s orbital and if there are further lone pairs, they are in p orbitals (arsenic has no more). Thus, we can construct $\sigma(\ce{s-p})$ bonds until all valences are satisfied.
Moving on to bond angles: well, $\ce{HCl}$ has no problems as there are only two atoms. But $\ce{H2Se}$ and $\ce{AsH3}$ also have hardly any problems since the central atoms are so large that the comparatively smaller ligands can be attached at almost perfect $90^\circ$ angles without significant steric repulsion. The bond angles repored ($\ce{AsH3}{: 91.8^\circ}; \ce{H2Se}{: 91^\circ}$) are in adequate agreement.
Ideal cases with steric distortion
The next set of examples along the route you should be going are $\ce{NF3}$ and $\ce{H2O}$. These compounds fulfil the general criteria outlined in the previous section. However, their central atoms are decidedly smaller. Thus, perfect $90^\circ$ angles are unobtainable as there would be a steric clash of the ligand atoms. The ligand atoms move apart to prevent steric repulsion; this is achieved by mixing in s-contribution into the bonding orbitals. $25~\%$ s-contribution would correspond to a tetrahedral bond angle of $109.5^\circ$ and $\mathrm{sp^3}$ hybrid orbials. The table below compares the bond angle and length data of the lighter homologues discussed here and the heavier ones discussed above (consider the fluorines in $\ce{NF3}$ to be pseudohydrogens).
\begin{array}{lcccc}\hline
\text{element} & \ce{N} & \ce{As} & \ce{O} & \ce{Se} \\ \hline
d(\ce{X-H})/\mathrm{pm} & 137 & 151.9 & 95.84 & 146 \\
\angle(\ce{H-X-H})/^\circ & 102.5 & 91.8 & 104.5 & 91 \\ \hline\end{array}
$\ce{CH2}$, carbene, is a dangerous case. As far as I know, no definite structures are reported (although Wikipedia mentions bond angles determined by EPR) since it is too reactive and since it is susceptible to dimerisation to form ethene. It also comes in three distinct forms: a singlet form, a linear triplet and a bent triplet. The most ideal structure of these three is the singlet carbene, which also fits nicely into this subsection.
For a singlet carbene, you would also start off from carbon’s ground state $[\ce{He}]\,\mathrm{2s^2\,2p^2}$. The two p orbitals each form a $\sigma(\ce{s-p})$ bond to the hydrogen atoms. The two hydrogens, however, need to move away from each other to ease steric repulsion; thus a little bit of s-contribution is mixed in for a final bond angle of $102^\circ$. A third p orbial remains unoccupied — note the similariy to $\ce{H2O}$ which has everything carbene has plus two electrons in a p-type lone pair.
Practically ideal cases which naturally lead to multiple bond
Herein, we shall discuss $\ce{N2}$ and $\ce{CO}$. For both, we can apply the method outlined above; I’ll start with the simpler $\ce{N2}$ case. We initially have one s lone pair and three, singly occupied p orbitals on each nitrogen. One of these forms a standard $\sigma(\ce{p-p})$ bond with the other nitrogen. Now, thankfully, the other p orbitals are nicely aligned and fit perfectly for generating two $\pi(\ce{p-p})$ bonds. Together, that gives us three bonds or a triple bond, composed of one σ and two π bonds.
The carbon monoxide case is essentially the same except that carbon would have one empty p orbital and oxygen would have one lone pair in a p orbital; only one π bond can be formed in the way outlined above. However, the unoccupied carbon orbital and the fully populated oxygen orbital can align likewise and form what could formally be described as a second, dative π bond. That renders nitrogen and carbon monoxide isoelectronic. Note, however, that the simplification introduced herein is not able to explain the big differences between $\ce{CO}$ and $\ce{N2}$ that arise from their actually rather different molecular orbital schemes.
Cases, in which an s type lone pair is impossible
I’ve cleared away all the simple cases. If you go through the list, the only ones that remain are $\ce{BF3, C2H4}$ and $\ce{BeCl2}$ — although we might as well include linear and bent triplet $\ce{CH2}$ in here. These all have in common that assuming an s type lone pair would not allow for sufficient electrons available for bonding.
If that is the case, the simple answer always suggests hybridisation (athough the MO description does not require it and nonlocalised molecular orbitals would not support it). At least one p orbital must be mixed in to allow for the generation of enough σ bonds; yet you should always try and make the p contribution of the hybrid orbitals as small as possible (in simplified terms) as that means the s lone pair ‘loses’ less energy en route. (Beware, the terms and images I used are inaccurate!) Any orbitals, populated or not, that are not required for σ bonds remain as p type ones. For $\ce{CO2}$ and linear triplet $\ce{CH2}$, that gives us an $\mathrm{sp}$ configuration of the central carbon with two unused p orbitals; in $\ce{C2H4}$, bent triplet $\ce{CH2}$ and $\ce{BF3}$, the central atoms are $\mathrm{sp^2}$ configured with one additional p orbital.
If possible — i.e. in $\ce{CO2}$ and $\ce{C2H4}$ — π bonds should then be formed. Each of the oxygens in carbon dioxide has one p-type semipopulated p orbital available for a π bond as does the second carbon of $\ce{C2H4}$. The carbenes cannot form π bonds due to the lack of partners, and thus they remain as diradicals. The linear one has two degenerate p-type radical orbitals, the bent one almost has an $\mathrm{sp^2}$ type radical orbital and one p type. (In fact, bent triplet carbene has a larger bond angle than the $\mathrm{sp^2}$ predicted $120^\circ$. This is in line with removing part of the s contribution of the radicals and supplying that to the bonding orbitals. Nobody expects you to correctly analyse the structures of carbenes, though.)
Beryllium
You may have noticed that that element had gone missing at the end of the previous section. This is because it is the most special of all cases. All other alkaline earth metals behave as metals (form dications and practically ionic structures) while beryllium still features notable non-metal properties.
I think that the best description of $\ce{BeCl2}$ is via a four-electron-three-centre bond akin to $\ce{XeF2}$. That again requires an unhybridised central atom. The lowest of the three relevant MOs of the 4e3c bond would be formed of the fully bonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ combination; the second bonding MO would then be $\mathrm{p}(\ce{Cl}) + \mathrm{p}(\ce{Cl})$ where the nodal plane traverses $\ce{Be}$ — mixing in its 2p orbital probably increases the bonding character — and a final antibonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ orbital wherein the phase of beryllium’s orbital is of opposite sign with respect to the chlorine orbitals’. If that is the case, it should possibly be discussed with the ‘hypervalent’ cases of the other question.
I could be terribly wrong, though. Thankfully, nobody will request you to predict the MO scheme of $\ce{BeCl2}$ until at least the final bachelor’s examination.
