1 Proton (p) Nguyên – Electron (e) – 21 P = E = Z A=Z+N = P +N+E = 2Z +N = A + Z 31 A Z X 1 (A) proton electron (C) electron (B (D (A) A = Z + N (C) A = Z + P không (B) A = P + N (D) A = E + N (A) Z = N (B) A = Z + E A (C) (A (C Z 2N Z (D) (B (D (A) 24 X (B) 24 16 X (C) 16 X (D) 16 X 2Z N – A Z – – A) B) D) 15 C) 2Z+N=52 A Z N 35 Z 17 N 18 VD2: A) 82 22 X B) 56 26 X C) 2Z N 82 2Z - N=22 56 30 X Z N D) 52 30 X 26 30 56 26 X A) 23 11 Y B) 22 11 Y C) 23 12 Y D) 34 11 Y A) 22 14 X B) 25 11 X C) 24 12 X D) 23 13 X (A) (B) 14 (C) 13 (D) 27 A) 35 17 X B) 36 16 X C) 37 17 X D) 34 18 X N Z 1,52 VD3: Vì ZX < 82 nên ta có N Z 1,52 Z N 1,52Z 2Z + N 3,52Z 18 3,52Z 5,11 Z Vì Z nguyên nên Z = Z=6 Thay ( A A) 23 11 Y (A) 14 X B) 22 11 Y C) 23 12 Y D) 34 11 Y (C) 14 X (D) 13 X B (B) 21 14 X VD4: A ZB 2Z A 2ZB 40 ZA 12 2Z A 2ZB ZB 12 VD5: Chú ý Vì AB2 có nguyên A AB2 ZB 2Z A 4ZB 44 ZA 2ZB 2Z A ZB VD6: 3 ZX M NX M 2ZM NM 3.(2Z X 2ZM 6Z X (NM 2Z X 2ZM NX ) 3NX ) 196 60 hay 2ZM 6Z X (NM 3NX ) 2ZM 6Z X (NM 3NX ) 2Z X 2ZM M 196 (1) 60 (2) (3) + 6ZX NM + 3NX 2ZM + 6ZX 2Z X 6ZM 128 ZX 13 X Al 2ZM 2Z X ZM 17 M Cl AlCl3 C 2B5 B) D) 19 16 A C D A C B) 19 D) 19 16 A C B) 10 D) 11 12 E – M – ne Mn+ M + ne Mn- – – 3+ – Ta có: M3+ M - 3e 2ZM NM 82 2ZM - NM=22 ZM 26 NM 30 M Fe VD8: 2+ X2- 22+ 2+ X2_ ZX M NX M 2ZM 2ZM NM 2Z X 2Z X - (NM NX 60 2ZM NX )=20 NM 2-: 2ZX + 2+: 2ZM - ZM 12 ZX 2- 2Z X NX 40 20 (1) (2) 2ZM - - (2ZX + 2) = 2ZM -2ZX = (3) 2ZX +2 = 18 2+ 2ZM - = 22 F 23 B) 32 16 D) 16 A C G X2A C + B) 23 32 D) 27 28 H + và ion X2- 2X + 2- ion X2A + 2X B C D) 110 u Câu B Câu A A C B C C C C C D B C E B A F B C G B B H A A 10