Question

3.(15 PTS) A textbook of physical chemistry proposes that spʻ hybrid orbitals(corresponding to the geometry shown in the figure) can be constructed from linear combinations of 2s, 2px and 2pz.orbitals: 4o = c $2.p. +c2$25+cz $2p, UL = C, P2, +2, +2, +Cs P2P, We = c +2p. + c3 2s +,022, Since the s orbital is spherically symmetric, it is assumed it contributes equally to all three hybrid orbitals. Likewise, since altogether only one 2s orbital is available for hybridization, the condition C +Cg? + Ce=1 must be met. Finally, to match the orbitals as shown in the figure, the spherical symmetry of the s orbital means that all these coefficients must be negative. The textbook also reasons that: 1) Because the orbital w, is aligned along the z axis, cs0. 2) Because the hybrid orbital V. points along the positive z axis, C > 0. 3) Because of relative location and symmetry, co > 0 and Co --C. 4) Because of relative location and symmetry, CA = Cy and both are negative.

Answer 1

we know » Emn = 1; if m=n = 0.34 min. and Emn = 4 m/ VnY; Im and Un are "To normalized wave fn. i fore Vasil, and he T40 1 4 ay = 1 LY 614 b7 = 1 24 e 14 c)=1

PO and h4 a 16) = 0 L4 a 14 c) = 0 also <^P0/%) =0 From TV a 14a) = 1 X 4 Popz 11 23 02's a Pepe - 1/2 2 2 2 2 7 = 1 2 <Q 2pz | Pop) - el pap=1&22) - el goal tape) +( <22x1920) = 1 → 4+1+3.1 .0 - 4.0= 1. [resing Emin = 0, man 1 = 1, man >> 92+ 1 9 2 = 1 => 4

using Taal &by=0. → K7 Popx - 320 C4 topx - 1 / 3 / 25 to Popn) = 0 - D 4 tapz|9app) - Jasa Popz Ifa) - 1 ts pz 18 px - 1 2 3 4 X 4 25 | top) + (-1/2) 2 29 28/42e) + chd 20 | Popa) = 0 → vyd-2.0-4. 0 0 +1 0 V3 . I using mn = om tnt - 1 man J. => 0 + - 0. (4 = to From L 46 | 4b) = 1, we would get C42 + 2 +42-1 (by similar poweeswa ХЛ - 2 = 1 - 16 - 175 7062-21

ata = / P2pz - 13 925. 4 - topz - 42-vz popu Me = -1/2 Papx - 1 / 2 1 2 4 + 1/ 2 Capa. bu Fore Va to be normalized; tra&ay = 1 12 loman = (13) ? 220px | Tepe) - 12 14 2px|94) - a 2824 182p) + ( 3 ) 2 14 26 / 428) 2.1-2.0-12.0+ 2 pomno, mex" + - 3 = 1 = RHS. Kota/Pa) = 4(proved).

Fooi 4 and to to be outhogonal; T461407 = 0. LHS. ektep 2 - - eg & - xdog en fly er og r - (ve|eby * * * <*dep|*tby (94-) = fotky zdoby (p/e87 (4) + ( stress Joopy &. & Retezlogy of a (repletozy. u. 4 + { stoppedepy r. 4 + - <Papa 14opad = 0 = – 1 toro-0 + 1 - 0 toto-1 = 0. [ a smn = 0, mtn. 6 - 1 , man ==+= 1 + 2 -3 N. NU 3-3 = 0 = RHS. Karl Key=0Cproved) briore