For the following question

Which molecules react to form a dative/coordinate bond?

A. $\ce{CH4}$ and $\ce{NH3}$
B. $\ce{C2H2}$ and $\ce{Cl2}$
C. $\ce{NH3}$ and $\ce{HF}$
D. $\ce{Cl2}$ and $\ce{HF}$

the answer scheme says C is correct, but I do not see how two already complete molecules can form coordinate bonds. $\ce{NH3}$ has a lone pair of electrons, $\ce{HF}$ having none. How can a coordinate bond between this lone pair and $\ce{HF}$ happen? I am confused.

The others are also equally confusing.

  • 5
    $\begingroup$ I think the question means the H-bonding between HF (H donor) and NH3 (H acceptor) $\endgroup$ – szentsas Aug 12 '19 at 9:57
  • 1
    $\begingroup$ @szentsas I wouldn't call it H-bonding. Wouldn't ammonia, being a lewis base, abstract a proton from HF? In some low-level chemistry courses this happens to be a fundamental example for dative bonds. $\endgroup$ – William R. Ebenezer Aug 12 '19 at 15:01
  • $\begingroup$ I don't think you ever consider this a dative bond. If you transfer the hydrogen, it's definitely not dative. That is a new covalent bond. @WilliamR.Ebenezer $\endgroup$ – Zhe Aug 12 '19 at 17:54
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    $\begingroup$ @Zhe and that's why it's called coordinate covalent bond. It is virtually indistinguishable from a regular coordinate bond. Kindly read: chemguide.co.uk/atoms/bonding/dative.html, and then pass your judgement. $\endgroup$ – William R. Ebenezer Aug 12 '19 at 18:24
  • 1
    $\begingroup$ @WilliamR.Ebenezer no need to be passive aggressive. The point that was being made was that, despite what chemguide says, the majority of chemists would consider something like $\ce{OC:}\rightarrow\ce{BH3}$ to be a dative bond, but not the N–H bond in $\ce{NH4+}$. $\endgroup$ – orthocresol Aug 12 '19 at 20:17

Option C is correct.

You could start by looking up the definition of a coordinate bond.

Here's a tiny excerpt from the above link:

A coordinate bond is a kind of 2-center, 2-electron covalent bond in which the two electrons derive from the same atom.

In your case, as you have correctly noted yourself, the $\ce{NH3}$ molecule will have an "lone pair" of electrons. However, your judgement that $\ce{HF}$ has none is quite incorrect– the fluorine atom has $3$ lone pairs. (not that it's very relevant to the question, but you can look up the structure of $\ce{HF}$ here)

Since the $\ce{NH3}$ is a lewis base, its lone pair of electrons would abstract the $\ce{H}$ from $\ce{HF}$.

NH3 and HF reaction

$\ce{NH4+}$ has four equivalent bonds. You could call one of them the "dative" bond, but there's really no way to distinguish them from each others in general cases.

The concept of dative bond comes here just because the bond between nitrogen and the hydrogen from $\ce{HF}$ involves 2 electrons (like all single bonds), both of which are contributed by the nitrogen atom (unlike most single bonds).

(Image source)

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