AS Chemistry 9701: VSEPR Shapes of BF3 and SF6, and the Expanded Octet

Boron has $3$ outer-shell electrons and forms the compound $\text{BF}_3$ by bonding to three fluorine atoms, with no lone pairs remaining on the boron atom. Sulfur has $6$ outer-shell electrons and forms the compound $\text{SF}_6$ by bonding to six fluorine atoms, also with no lone pairs remaining on the sulfur atom.

(a) State the total number of electron pairs around the central atom in $\text{BF}_3$ , and hence state the shape of $\text{BF}_3$ and its F–B–F bond angle. [2]

(b) State the total number of electron pairs around the central atom in $\text{SF}_6$ , and hence state the shape of $\text{SF}_6$ . State the two different F–S–F bond angles present in the molecule. [3]

(c) Nitrogen, which is in the same period as boron, cannot form an analogous five-fluorine compound $\text{NF}_5$ , whereas sulfur can form $\text{SF}_6$ . Suggest, in terms of available orbitals, why period 3 elements such as sulfur can exceed the "normal" limit of four electron pairs (an octet) around the central atom, while period 2 elements such as nitrogen cannot. [2]

Part (a): Shape and bond angle of BF₃

Boron has $3$ outer-shell electrons, all of which are used to form $3$ single bonds to the $3$ fluorine atoms. No electrons are left over, so boron has no lone pairs.

$\text{Total electron pairs around B} = 3 \text{ (bonding)} + 0 \text{ (lone)} = 3$

By VSEPR theory, $3$ electron pairs (all bonding, none lone) arrange themselves as far apart as possible in a single plane, giving a trigonal planar shape with an F–B–F bond angle of $\boxed{120^\circ}$ . (Boron is an exception to the usual “full octet” pattern seen elsewhere in this topic, since it ends up surrounded by only $6$ electrons rather than $8$ .)

Part (b): Shape and bond angles of SF₆

Sulfur has $6$ outer-shell electrons, all of which are used to form $6$ single bonds to the $6$ fluorine atoms, leaving no lone pairs on sulfur.

$\text{Total electron pairs around S} = 6 \text{ (bonding)} + 0 \text{ (lone)} = 6$

By VSEPR theory, $6$ electron pairs arrange themselves as far apart as possible in an octahedral arrangement: one fluorine atom points to each of the $6$ vertices of an octahedron centred on sulfur (equivalently, along the $+x$ , $-x$ , $+y$ , $-y$ , $+z$ and $-z$ directions).

Because this arrangement has two distinct pairings of fluorine atoms, there are two different bond angles:

$\boxed{90^\circ}$ between any two adjacent fluorine atoms (e.g. $+x$ and $+y$ ).
$\boxed{180^\circ}$ between the two fluorine atoms directly opposite each other through the sulfur atom (e.g. $+x$ and $-x$ ).

Part (c): Why sulfur but not nitrogen can exceed an octet

Nitrogen is a period 2 element, so its outer (valence) shell is the $n=2$ shell, which contains only $2s$ and $2p$ subshells. There are no $2d$ orbitals available at this energy level, so nitrogen’s valence shell has a hard maximum of $4$ orbitals (one $2s$ and three $2p$ ), holding at most $4$ electron pairs, an octet. This is why $\text{NF}_5$ does not exist.

Sulfur is a period 3 element, so its valence shell is the $n=3$ shell, which contains $3s$ , $3p$ and $3d$ subshells. The empty $3d$ orbitals are close enough in energy to be used for additional bonding, allowing sulfur’s valence shell to accommodate more than $4$ electron pairs. This “expanded octet” is why sulfur, unlike nitrogen, can form a compound such as $\text{SF}_6$ with $6$ bonding pairs around it.

Final answers

(a) $3$ electron pairs (all bonding); shape $=$ trigonal planar; bond angle $= \boxed{120^\circ}$ .
(b) $6$ electron pairs (all bonding); shape $=$ octahedral; bond angles $= \boxed{90^\circ}$ (adjacent F atoms) and $\boxed{180^\circ}$ (opposite F atoms).
(c) Sulfur (period 3) has accessible $3d$ orbitals to hold extra electron pairs beyond an octet; nitrogen (period 2) has only $2s$ / $2p$ orbitals available, so it cannot exceed an octet.