In which of the following molecules/ions are all the bonds not equal?

The equality of bond lengths in a polyatomic species is governed by its molecular geometry. Species that place all equivalent ligands in symmetry-equivalent positions give identical bond lengths, whereas geometries having two distinct sets of positions (axial vs. equatorial) give two different bond lengths.

Case A : $$SF_4$$
S has $$6$$ valence electrons. Adding $$4$$ electrons from the four $$F$$ atoms gives $$10$$ electrons, i.e. $$5$$ electron pairs (one of them a lone pair).
 Hybridisation : $$sp^3d$$ (trigonal bipyramidal electron-pair geometry).
 The lone pair occupies an equatorial position (maximum separation), so the geometry of the atoms is see-saw: two axial S-F bonds and two equatorial S-F bonds.
 Because axial-equatorial repulsions weaken the axial bonds (they are longer) while equatorial bonds are shorter, the four S-F bonds are not equal.

Case B : $$SiF_4$$
$$\mathrm{Si}$$ has $$4$$ valence electrons, giving $$8$$ electrons (4 bonding pairs) around the central atom.
 Hybridisation : $$sp^3$$ with a tetrahedral shape. All four $$F$$ atoms occupy symmetry-equivalent corners of a regular tetrahedron, hence all Si-F bond lengths are equal.

Case C : $$XeF_4$$
Xe contributes $$8$$ valence electrons; with $$4$$ electrons from $$F$$ atoms we have $$12$$ electrons (6 pairs: 4 bonding + 2 lone).
 Hybridisation : $$sp^3d^2$$ with an octahedral electron-pair arrangement; the two lone pairs occupy positions trans to each other, giving a square-planar molecular geometry.
 All four $$F$$ atoms lie at the corners of the same square and are symmetry-equivalent, so every Xe-F bond length is identical.

Case D : $$BF_4^-$$
B provides $$3$$ valence electrons, plus $$1$$ extra due to the negative charge: total $$4$$ electrons. With $$4$$ electrons from the four $$F$$ atoms we again have $$8$$ electrons (4 bonding pairs).
 Hybridisation : $$sp^3$$ with a perfect tetrahedral geometry. All $$F$$ ligands are equivalent, giving equal B-F bond lengths.

Therefore, the only species in which there are two sets of distinct bond lengths is $$SF_4$$.

Answer : Option A which is: $$SF_4$$