Which of the species has a permanent dipole moment?

The correct option is SF₄.

To determine which species has a permanent dipole moment, we must analyze the chemical bonding, molecular geometry (using VSEPR theory), and the symmetry of each given molecule. A molecule has a permanent dipole moment if its individual bond dipoles do not cancel each other out due to symmetry.

Let us analyze each option step-by-step:

1. SiF₄ (Silicon Tetrafluoride):
Silicon (Si) belongs to Group 14 and has 4 valence electrons. It forms four single bonds with four fluorine (F) atoms, leaving zero lone pairs on the central silicon atom.
According to VSEPR theory, a molecule with 4 bond pairs and 0 lone pairs adopts a perfect tetrahedral geometry.
Since the geometry is highly symmetrical and all surrounding atoms are identical, the individual polar Si-F bond dipoles point toward the corners of a regular tetrahedron and vectorially cancel each other out completely.
Therefore, the net dipole moment ($\mu$) of SiF₄ is zero ($\mu =0$).

2. SF₄ (Sulfur Tetrafluoride):
Sulfur (S) belongs to Group 16 and has 6 valence electrons. In SF₄, it forms four single bonds with four fluorine atoms, which uses 4 electrons. The remaining 2 valence electrons form 1 lone pair on the sulfur atom.
This gives a steric number of 5 (4 bond pairs + 1 lone pair). According to VSEPR theory, the electron-pair geometry is trigonal bipyramidal. To minimize repulsion, the lone pair occupies an equatorial position, resulting in a seesaw molecular geometry.
Because of this asymmetrical seesaw shape, the polar S-F bond dipoles do not cancel each other out. The presence of the lone pair also contributes to an uneven distribution of charge.
Therefore, SF₄ is polar and has a permanent dipole moment ($\mu \ne 0$).

3. BF₃ (Boron Trifluoride):
Boron (B) belongs to Group 13 and has 3 valence electrons. It forms three single bonds with three fluorine atoms, leaving zero lone pairs on the boron atom.
With 3 bond pairs and 0 lone pairs, the molecular geometry is trigonal planar, with bond angles of 120°.
The three polar B-F bonds lie in a single plane pointing symmetrically outwards. The vector sum of any two B-F bond dipoles is equal and opposite to the third, resulting in complete cancellation.
Therefore, the net dipole moment of BF₃ is zero ($\mu =0$).

4. XeF₄ (Xenon Tetrafluoride):
Xenon (Xe) is a noble gas with 8 valence electrons. In XeF₄, it forms four single bonds with four fluorine atoms, using 4 electrons. The remaining 4 valence electrons form 2 lone pairs on the xenon atom.
This gives a steric number of 6 (4 bond pairs + 2 lone pairs). The electron-pair geometry is octahedral. To minimize lone pair-lone pair repulsion, the two lone pairs position themselves opposite to each other (180° apart), resulting in a square planar molecular geometry.
The four polar Xe-F bonds point towards the corners of a square, cancelling each other out in the horizontal plane. The two lone pairs point in opposite directions vertically, cancelling their dipole contributions as well.
Therefore, the net dipole moment of XeF₄ is zero ($\mu =0$).

In conclusion, only SF₄ has an asymmetrical geometry (seesaw) that prevents its bond dipoles from cancelling out, giving it a permanent dipole moment.