1. We need to identify the species in which all bonds are not equivalent, i.e. the bond lengths are unequal.

2. Let us examine each option using shape and symmetry.

Option A: $XeF_4$

• Central atom: Xe
• Total electron pairs around Xe = 6
• Geometry of electron pairs: octahedral
• Lone pairs: 2
• Molecular shape: square planar

In $XeF_4$, the four $Xe-F$ bonds lie in the same square plane and are symmetry-equivalent. So, all bond lengths are equal.

Option B: $BF_4^-$

• Central atom: B
• Hybridisation: $sp^3$
• Molecular shape: tetrahedral

All four $B-F$ bonds in a tetrahedral ion are equivalent by symmetry. So, all bond lengths are equal.

Option C: $SiF_4$

• Central atom: Si
• Hybridisation: $sp^3$
• Molecular shape: tetrahedral

All four $Si-F$ bonds are equivalent in a tetrahedral molecule. So, all bond lengths are equal.

Option D: $SF_4$

• Central atom: S
• Total electron pairs around S = 5
• Electron pair geometry: trigonal bipyramidal
• Lone pairs: 1
• Molecular shape: see-saw

In a trigonal bipyramidal arrangement, the lone pair occupies an equatorial position to minimize repulsion. Thus in $SF_4$:

• There are 2 axial $S-F$ bonds
• There are 2 equatorial $S-F$ bonds

Axial and equatorial bonds are not equivalent. Due to greater repulsion on axial bonds, their bond lengths differ from equatorial bond lengths. Hence, $SF_4$ has unequal bond lengths.

3. Conclusion

The species with unequal bond lengths is: $\boxed{SF_4}$

So, the correct option is D.