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QuestionAnatomy and Physiology

"Write a Lewis structure for PF^3 and then predict the geometry. Tetrahedral Trigonal pyramidal Bent Seesaw"
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Step 1
: Draw the Lewis structure for PF^3.

First,drawthecentralphosphorusatom(P)andsurrounditwiththreeflourineatoms(F),representingthetriplecovalentbondbetweenphosphorusandfluorineinPF3.Then,addallthevalenceelectronsneededtocompletetheoctetsforeachatom.Phosphorushas5valenceelectrons,andeachfluorinehas7valenceelectrons,givingatotalof(5+3×7=)26valenceelectrons.F ∣ P−F−F−F ∣ FFirst, draw the central phosphorus atom (P) and surround it with three flourine atoms (F), representing the triple covalent bond between phosphorus and fluorine in PF^3. Then, add all the valence electrons needed to complete the octets for each atom. Phosphorus has 5 valence electrons, and each fluorine has 7 valence electrons, giving a total of (5 + 3 × 7 =) 26 valence electrons. \begin{array}{cccccc} & & \mathrm{F} & & & \ & & | & & & \ \mathrm{P} & - & \mathrm{F} & - & \mathrm{F} & - & \mathrm{F} \ & & | & & & \ & & \mathrm{F} & & & \end{array}

Step 2
: Predict the geometry of PF^3.

The central phosphorus atom is bonded to three fluorine atoms, forming three electron pairs (two bonding pairs and one lone pair) around the phosphorus atom. The geometry of such an arrangement is trigonal pyramidal, with the lone pair causing slight distortion from the ideal tetrahedral geometry.

Final Answer

The geometry of PF^3 is trigonal pyramidal.

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