U2.5 · Lewis DiagramsPractice · Lewis 结构

P has 5 valence electrons and each O has 6. The 3− charge means 3 extra electrons. Total = 5 + 4 × 6 + 3 = 32. Add electrons for an anion and subtract them for a cation; forgetting the charge is a common slip.

Lone pairs on the central atom = (its valence electrons − electrons it uses in bonds) ÷ 2. $\ce{PBr3}$ P: 1; $\ce{CF4}$ C: 0; $\ce{OCl2}$ O: 2; $\ce{BH3}$ B: 0. P has 5 valence electrons and uses 3 in bonds to Br, leaving 2: one lone pair.

Total valence electrons = 4 + 6 + 2 × 7 = 24. Three single bonds use 6; completing the octets on O and both Cl uses the rest, but C then has only 6 electrons, so one lone pair on O becomes a C=O double bond (all formal charges 0). Result: no lone pairs on C, 2 on O, and 3 on each Cl, for 8 in total.

Electrons around the central atom: $\ce{PCl3}$ P 8, $\ce{SF4}$ S 10, $\ce{NF3}$ N 8, $\ce{BCl3}$ B 6. $\ce{SF4}$ has 34 valence electrons; the four S–F bonds and the lone pairs on F use 32, so the last pair must go on S, giving it 10 electrons. S is in period 3 and can hold more than 8; B in $\ce{BCl3}$ has only 6.

Total valence electrons = 2 × 5 + 4 × 1 = 14. The H₂N–NH₂ skeleton has 5 single bonds using 10, leaving 4: one pair on each N, giving each N exactly 8. An N=N double bond would put 10 electrons around each N, and two lone pairs on each N would need 18 electrons, more than the 14 available.