Molecules That Break the Octet Rule, and Coordinate Bonds – Formation of Covalent Bonds (Senior 4 Chemistry) | Ganzaa

Formation of Covalent Bonds

Lessons

Senior 4 · Chemistry

Formation of Covalent Bonds

5 lessons

Lesson 2 of 5

Molecules That Break the Octet Rule, and Coordinate Bonds

25 min

White smoke from two invisible gases

Open a bottle of concentrated ammonia solution next to a bottle of concentrated hydrochloric acid, and a white smoke forms in the air between them. The gases ammonia and hydrogen chloride have reacted to make tiny crystals of ammonium chloride. In the ammonium ion, $\mathrm{NH_4^{+}}$ , nitrogen is bonded to four hydrogen atoms, yet a nitrogen atom has only three unpaired electrons to share. The fourth bond is a different kind: both of its electrons come from nitrogen. This lesson looks at bonds like that, and at molecules where the octet rule from Lesson 1 simply does not work.

Three kinds of exception

1. Fewer than eight electrons (electron-deficient). Beryllium has two outer electrons and boron three. In gaseous beryllium chloride, $\mathrm{BeCl_2}$ , beryllium shares two pairs: only 4 electrons around it. In boron trifluoride, $\mathrm{BF_3}$ , boron shares three pairs: 6 electrons. Aluminium chloride, $\mathrm{AlCl_3}$ , is similar. These molecules have an empty orbital and readily accept a lone pair from another molecule.

2. More than eight electrons (expanded octet). From Period 3 onwards, the central atom can have 10 or 12 electrons around it. In phosphorus pentachloride, $\mathrm{PCl_5}$ , phosphorus forms five bonds (10 electrons); in sulfur hexafluoride, $\mathrm{SF_6}$ , sulfur forms six (12 electrons). In the model used at this level, the outer s and p electrons are promoted into the empty 3d subshell, which is close in energy, so more unpaired electrons are available for sharing. Period 2 elements have no 2d subshell, so nitrogen can form $\mathrm{NCl_3}$ but never $\mathrm{NCl_5}$ .

3. An odd number of electrons. Nitrogen monoxide, NO, has $5 + 6 = 11$ outer electrons, and nitrogen dioxide, $\mathrm{NO_2}$ , has $5 + 12 = 17$ . With an odd total, at least one electron must be unpaired, so one atom cannot have an octet. Molecules with an unpaired electron are called free radicals; they are very reactive.

Worked example 1: counting electrons around the central atom

For each molecule, state the number of electrons around the central atom and whether it obeys the octet rule: (a) $\mathrm{BCl_3}$ , (b) $\mathrm{SCl_2}$ , (c) $\mathrm{PF_5}$ , (d) $\mathrm{ClF_3}$ .

Solution. Count the central atom's outer electrons, add one for each bond (each outer atom here supplies one electron), and see what is left as lone pairs.

(a) B has 3; three B–Cl bonds use them all: 3 pairs = 6 electrons, an incomplete octet.
(b) S has 6; two bonds use 2, leaving 4 as 2 lone pairs: 2 bonding + 2 lone pairs = 8, octet obeyed.
(c) P has 5; five P–F bonds: 10 electrons, an expanded octet.
(d) Cl has 7; three bonds use 3, leaving 4 as 2 lone pairs: 3 + 2 pairs = 10 electrons, an expanded octet.

Definition

Coordinate (dative covalent) bond: a covalent bond in which both shared electrons come from the same atom. It forms between a donor, an atom with a lone pair, and an acceptor, an atom with an empty orbital in its outer shell. It is drawn as an arrow from the donor to the acceptor: $\mathrm{H_3N \rightarrow H^{+}}$ .

Coordinate bonds in action

Ammonium ion. $\mathrm{NH_3} + \mathrm{H^{+}} \rightarrow \mathrm{NH_4^{+}}$ . The lone pair on nitrogen is shared with a hydrogen ion, which has an empty 1s orbital. Once formed, the four N–H bonds are identical in length and strength; you cannot tell which was dative, and the positive charge belongs to the whole ion.
Oxonium (hydronium) ion. $\mathrm{H_2O} + \mathrm{H^{+}} \rightarrow \mathrm{H_3O^{+}}$ . Every acid in water forms this ion; it is what makes a solution acidic.
Ammonia–boron trifluoride. $\mathrm{NH_3} + \mathrm{BF_3} \rightarrow \mathrm{H_3N{\rightarrow}BF_3}$ . The electron-deficient boron accepts nitrogen's lone pair and gains an octet.

Aluminium chloride: a molecule that bonds to itself

At about 180 °C aluminium chloride turns to vapour. At moderate temperatures the vapour consists mainly of $\mathrm{Al_2Cl_6}$ molecules rather than $\mathrm{AlCl_3}$ . Each $\mathrm{AlCl_3}$ unit has only six electrons around aluminium. Two units join: a chlorine atom of each unit donates one of its lone pairs to the aluminium of the other, giving two coordinate bonds and an octet on each aluminium. At higher temperatures the dimer splits back into $\mathrm{AlCl_3}$ .

Worked example 2: donor and acceptor

Carbon monoxide reacts with nickel to form $\mathrm{Ni(CO)_4}$ , and phosphine, $\mathrm{PH_3}$ , reacts with hydrogen iodide to form $\mathrm{PH_4^{+}I^{-}}$ . For the new bonds in each, identify the donor and acceptor and explain what each must have.

Solution. A donor needs a lone pair; an acceptor needs an empty outer orbital.

In $\mathrm{Ni(CO)_4}$ , each CO molecule donates the lone pair on its carbon atom to an empty orbital of the nickel atom: donor C (of CO), acceptor Ni.
In $\mathrm{PH_4^{+}}$ , the lone pair on phosphorus (P has 5 outer electrons, 3 used in P–H bonds) is donated to $\mathrm{H^{+}}$ , which has an empty 1s orbital: donor P, acceptor $\mathrm{H^{+}}$ , exactly as in $\mathrm{NH_4^{+}}$ .

Warning

Common mistakes. (1) Thinking a dative bond is weaker or different once formed: in $\mathrm{NH_4^{+}}$ all four N–H bonds are the same. (2) Drawing the arrow the wrong way: it points from the lone pair to the acceptor. (3) Expanding the octet of a Period 2 atom: N, O and F never have more than eight. (4) Forgetting to count the charge of an ion when you add up electrons.

Safety. Concentrated ammonia and concentrated hydrochloric acid both give off choking, corrosive fumes. The white-smoke demonstration is done by a teacher in a fume cupboard; students wear goggles.

Tip

To spot a dative bond, look for an atom that ends up with more bonds than its usual valency (N with 4, O with 3) next to an atom with fewer electrons than an octet ( $\mathrm{H^{+}}$ , B, Al). The lone pair goes from the first to the second.

Chemistry Student: Co-ordinate (dative covalent) Bonding (A-Level Chemistry) (4 min). Watch how the ammonium ion forms and how the arrow is drawn from the donor to the acceptor.

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Practice

Give one example each of a molecule with (a) an incomplete octet, (b) an expanded octet, (c) an odd number of electrons. Draw a dot-and-cross diagram for (a).
Explain why phosphorus can form $\mathrm{PCl_5}$ but nitrogen cannot form $\mathrm{NCl_5}$ .
Define a coordinate bond. Draw the oxonium ion, $\mathrm{H_3O^{+}}$ , showing the coordinate bond and the remaining lone pair.
Explain why all four N–H bonds in $\mathrm{NH_4^{+}}$ have the same length.
Boron trifluoride reacts with fluoride ions to form $\mathrm{BF_4^{-}}$ . Identify the donor and acceptor and give the number of electrons around boron before and after.
Explain why aluminium chloride vapour contains $\mathrm{Al_2Cl_6}$ molecules. Show the bonds in a diagram.
How many outer electrons does $\mathrm{NO_2}$ have? Explain why it must be a free radical.

Show answersHide answers

(a) $\mathrm{BF_3}$ (or $\mathrm{BeCl_2}$ ); (b) $\mathrm{SF_6}$ (or $\mathrm{PCl_5}$ ); (c) NO (or $\mathrm{NO_2}$ ). $\mathrm{BF_3}$ : B in the centre sharing one pair with each F; B has 6 electrons, each F has 8.
Phosphorus (Period 3) has an empty 3d subshell close in energy to 3s and 3p, so electrons can be promoted to give 5 unpaired electrons. Nitrogen (Period 2) has no 2d subshell, so it has at most 3 unpaired electrons and cannot exceed an octet.
A covalent bond in which both electrons come from one atom. $\mathrm{H_3O^{+}}$ : O with two ordinary O–H bonds, one arrow O→H, and one lone pair on O; overall charge +.
Once formed, a dative bond is an ordinary shared pair; the four bonds are equivalent and the positive charge is spread over the whole ion.
Donor: $\mathrm{F^{-}}$ (lone pair). Acceptor: B (empty orbital). Boron has 6 electrons before and 8 after.
Each Al in $\mathrm{AlCl_3}$ has only 6 electrons. A lone pair on a chlorine of one unit is donated to the aluminium of the other, and vice versa, giving two Cl→Al coordinate bonds and an octet on each Al (diagram as in Figure 2.3).
$5 + 2 \times 6 = 17$ : an odd number, so one electron must be unpaired.

Quick check

Which molecule has an expanded octet around its central atom?

Key points

Exceptions to the octet rule: electron-deficient molecules ( $\mathrm{BeCl_2}$ , $\mathrm{BF_3}$ , $\mathrm{AlCl_3}$ ); expanded octets in Period 3 and beyond ( $\mathrm{PCl_5}$ , $\mathrm{SF_6}$ ); odd-electron molecules (NO, $\mathrm{NO_2}$ ).
A coordinate (dative) bond is a shared pair donated by one atom with a lone pair to an atom with an empty orbital; it is drawn as an arrow from donor to acceptor.
Examples: $\mathrm{NH_4^{+}}$ , $\mathrm{H_3O^{+}}$ , $\mathrm{H_3N{\rightarrow}BF_3}$ and the $\mathrm{Al_2Cl_6}$ dimer.
Once formed, a coordinate bond is identical to any other covalent bond between the same atoms.

Sources and credits

Written by Ganzaa.
Video: Co-ordinate (dative covalent) Bonding (A-Level Chemistry), Chemistry Student, embedded from YouTube.
The diagrams, worked examples and Practice items were written and drawn by Ganzaa.

1White smoke from two invisible gases

2Three kinds of exception

3Coordinate bonds in action

4Aluminium chloride: a molecule that bonds to itself

White smoke from two invisible gases

Three kinds of exception

Coordinate bonds in action

Aluminium chloride: a molecule that bonds to itself