参考复旦无机课件problems.pptxVIP

Slater规则的应用如估算F原子的第一电离能：F, 1s22s22p5 对(2s, 2p)内的电子而言, σ(F) = 2(0.85) + 6(0.35) = 3.8E(F) (束缚能) = 7(13.6)[(9 – 3.8)/2]2 = 643.6 eVF+, 1s22s22p4 对(2s, 2p)内的电子而言,σ(F+) = 2(0.85) + 5(0.35) = 3.45E(F+) = 6(13.6)[(9 – 3.45)/2]2 = 628.4 eV I1 = E(F) - E(F+) = 643.6 - 628.4 = 15.2 eV 由于Slater规则不考虑外层电子对1s2电子的影响,只需要比较外层电子的束缚能。实验值： 17.4 eV。估算Na和Mg的第二电离能及Ne原子1s电子的束缚能:Na: 1s22s22p63s1 I2(Na) = EB[Na+] –EB[Na2+] = 13.6{8[(11 - 7(0.35) - 2(0.85))/2]2 – 7[(11 – 6(0.35) – 2(0.85))/2]2}) = 42.5 eV (实验值: 47.3 eV)Mg: 1s22s22p63s2 I2(Mg) = EB[Mg+] –EB[Mg2+] = 13.6{[12 – 8(0.85) – 2(1)]/3}2 = 15.5 eV (实验值: 15.0 eV)EB(Ne, 1s) = EB[1s22s22p6] –EB[1s12s22p6] = 13.6{2[(10 - 0.35)/1]2 + 8[(10 – 2(0.85) -7(0.35))/2]2 – (10/1)2 – 8[(10 - 0.85 – 7(0.35))/2]2} = 909 eV将下列离子按半径大小排序：(1) Y3+, Ba2+, Al3+, Co3+, Cs+, La3+, Ir3+, Fe3+答案: Cs+ (181 pm) Ba2+ (149 pm) La3+ (117.2 pm) Y3+ (104 pm) Ir3+ (82 pm) Fe3+ (78.5 pm) Co3+ (75 pm) Al3+ (67.5 pm)(2) Cl-, I-, Te2-, Ar+答案: Te2- (221 pm) I- (216 pm) Cl- (181 pm) Ar+ ( Ar+ Cl Cl-)解释红外振动频率与取代基团电负性总和之间的关系。化合物 P-O(cm-1) ∑χs F3PO 1404 11.70 F2ClPO 1358 10.75 Cl3PO 1295 8.85 Cl2BrPO 1285 8.52 ClBr2PO 1275 8.19 Br3PO 1261 7.86 ?3PO 1190 7.2 Me3PO 1176 6.0[提示：P-O的振动频率越高，P-O键就越强]答案：取代基团电负性总和与P-O强度之间的关系证明P与O之间除P-Oσ单键外还存在P-Oπ键，而不是单纯的σ单键。如果P-O之间只有单键，则随着取代基团电负性的增加，P的电子密度将会下降，导致 P-O键减弱；而从实验结果看正好相反，这就说明O原子会将部分电子反馈给P，形成3d-2pπ键，使P原子上因取代基团诱导而下降的电子密度得以补偿，从而加强了 P=O键。Addition of NO in the H2-Cl2 system leads to explosion. Please suggest a mechanism to explain.AnswerNO + Cl2 → ClNO + Cl.In NO, N atom adopts sp hybrization to form a σ-bond, aπ-bond and a 3e-π-bond , thus N contains a single electron.Estimate the electron hole size of Li, Na, K in liquid ammonia, respectively. (V=4?R3/3, NA=6.022?1023 )Suggestion: only three species of NH3, M+, and e-am exist in the solution which can be taken as balls and can be held without gap.MetalSolubility (mol metal (g metal) /1 kg NH3) (239 K)Molar ratio of NH3/metalA