I’m confused as I am unable to find a completely trusted way of finding coordinately bonded compounds from a bunch of other compounds.

  • $\begingroup$ Please don’t use all-caps in titles (or in the body or anywhere else, for all that matters) Thanks @Wild kitty for being quicker in editing ;) $\endgroup$
    – Jan
    Commented Oct 19, 2015 at 11:38
  • $\begingroup$ Thnx @Wild kitty for being quicker in editing ;) $\endgroup$
    – user21891
    Commented Oct 19, 2015 at 12:10
  • $\begingroup$ Coordinate bonds are covalent bonds and don't really differ in practice from non-coordinate ones. Only difference is source of electrons. $\endgroup$
    – Mithoron
    Commented Oct 19, 2015 at 16:43

2 Answers 2


Coordinate bonds can be distinguished from (other or ‘normal’) covalent ones experimentally.

Let’s look at the bond creation first as Newbie suggested. Typically, one would use a Lewis base and a Lewis acid to create a coordinate bond between them.

$$\ce{NH3 + BH3 -> H3N\bond{->}BH3}\tag{1}$$

In contrast, one would usually consider a normal covalent bond to be created from two radicals:

$$\ce{H3C^{.} + {}^{.}CH3 -> H3C-CH3}\tag{2}$$

But we can also create covalent $\ce{C-C}$ bonds by using a carbanion source and a carbocation source (slightly different example, because both ions are actually known and react with each other):

$$\ce{(H3C)3C+ + {}^{-}CN -> (H3C)3C-CN}\tag{3}$$

Thus, we can’t really tell a difference here. The difference in fact lies in the way a bond dissociates if the bond dissociation enthalpy $D = \Delta_\mathrm{D} H^0$ is provided. Standard covalent bonds will dissociate homolytically by the following equation:

$$\ce{H3C-CH3 -> H3C^. + ^.CH3}\tag{4}$$

A coordinate bond will, however, dissociate heterolytically thus liberation the original Lewis acid and base:

$$\ce{H3N\bond{->}BH3 -> NH3 + BH3}\tag{5}$$

In cases such as the ammonium ion ($\ce{NH4+}$) or say $\ce{[AlCl4]-}$ this means we have a certain number of normal bonds and another certain number of coordinate bonds (three and one, respectively in both examples). Although they are all between the same atoms, we can distinguish if we remember that neither ion can exist by itself, there will always be a corresponding counterion. Thus, if we consider $\ce{NH4Cl}$ or $\ce{Na[AlCl4]}$, we realise how the first bond is broken differently:

$$\begin{align}\ce{NH4Cl &-> NH3 + HCl}\tag{6}\\[0.6em] \ce{Na[AlCl4] &-> NaCl + AlCl3}\tag{7}\end{align}$$

And all further bonds are cleaved homolytically. We gain two different bond dissociation energies, one corresponding to the coordinate bond and one corresponding to the remaining bonds. (Note that salts such as $\ce{AlCl3}$ or $\ce{NaCl}$ are cleaved homolytically by heating in the gas phase — flame-colouring would not happen if we had $\ce{Na+}$ in the flame.)

  • $\begingroup$ Firstly, you seem to be talking about "covalent bond" and "dative covalent bond" as two distinct types of bonds, which is not the case as one is a subset of the other (specifically "dative covalent" is a subset of "covalent"). $\endgroup$ Commented Jun 17, 2017 at 14:58
  • $\begingroup$ Secondly, you also seem to misinterpret "covalent bond" as the pure covalent case (i.e. the bond is not polarised; the electronegativity difference between the two atoms involved is negligble). $\endgroup$ Commented Jun 17, 2017 at 15:01
  • $\begingroup$ Consider the case of the hydrohalic acids. They have been cited in many literature as "covalent substances". By your definition of homolytic fission, the bonds in these chemical species would not be "covalent" because they often dissociate to give ions. By your definition, wouldn't these bonds be considered "coordinate" and not "covalent". $\endgroup$ Commented Jun 17, 2017 at 15:04
  • $\begingroup$ In conclusion, your answer is problematic. These clarify your answer regarding the points I have raised. $\endgroup$ Commented Jun 17, 2017 at 15:05
  • 1
    $\begingroup$ Also, you seem to be seeing covalent bonds as something ‘pure’ or somehow different from polarised covalent bonds (or ionic bonds). They are not; all bonds are some variation of a wave function. All we can do is pick out certain cases and attempt to interpret them with our models. Argument in point: even the ionic NaCl bond dissociates homolytically (as mentioned at the end of the answer) much like any covalent bond, proving it is much closer to a covalent bond than to a dative or coordinate bond. $\endgroup$
    – Jan
    Commented Jun 17, 2017 at 15:06

The main (and probably the only) way is to apply the definition. In other way, you need to understand what the valence structures of the examined compound's ingredients are and based on this information to determine how they have bonded.

If you struggle to implement this analysis directly, try to work step by step, starting with thinking out a synthetic reaction, e.g.:

$$\ce{A + B -> AB}$$

that produces the examined compound $\ce{AB}$. The least you can conclude from knowing $\ce{A}$ and $\ce{B}$ is whether any of them provide elecro-deficient (or ionic) constituents of $\ce{AB}$, which, if so, is a clue that there may be coordinate bonding "in a close proximity" to the bond $\ce{A - B}$ …

Example: In ammonium hydroxide:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

the hydroxide part is an anion, so it has a surplus of electrons. By knowing that $\ce{OH-}$ came from $\ce{H2O}$, it becomes obvious that the additional electron has came from one of the hydrogen atoms in $\ce{NH4+}$, hence, the bond between $\ce{N}$ and this particular $\ce{H+}$ is coordinate (both electrons come from the nitrogen atom $\ce{N}$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.