In the 19F NMR spectrum of ClF3, the number of signals and multiplicity at room temperature are (19F; I = \(\tfrac{1}{2}\)) :
Two, a doublet and a triplet
Start from the geometry. ClF3 has a central chlorine with three bond pairs and two lone pairs, so its electron geometry is trigonal bipyramidal. The two lone pairs occupy equatorial positions, giving the familiar T-shaped molecule.
That leaves two chemically distinct fluorine environments: two axial fluorines, which are equivalent to each other, and one equatorial fluorine. Two environments means two signals, in a 2 : 1 intensity ratio.
Now apply the n+1 rule, remembering that coupling is only observed between inequivalent nuclei and that 19F has \(I = \tfrac{1}{2}\).
The two axial fluorines couple to the single equatorial fluorine. With n = 1 neighbour, they appear as a doublet.
The equatorial fluorine couples to the two equivalent axial fluorines. With n = 2 neighbours, it appears as a triplet.
So the spectrum shows two signals, one a doublet and the other a triplet. The single quartet in the last option would require all three fluorines to be equivalent and coupling to some fourth spin, which the T-shaped geometry rules out; and rapid exchange averaging the environments would instead collapse everything to one singlet, which is what happens only at elevated temperature.
Hence the answer is two signals, a doublet and a triplet.
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