Question

Table 9C.1 Overlap integrals between hydrogenic orbitals Orbitals Overlap integral 1s,1s S = {1 + (ZR/a0) + 1/3 (ZR/a0)^2} e^{-ZR/a0} 2s,2s S = {1 + (ZR/2a0) + 1/12 (ZR/a0)^2 + 1/240 (ZR/a0)^4} e^{-ZR/2a0} 2p_x,2p_x (?) S = {1 + (ZR/2a0) + 1/10 (ZR/a0)^2 + 1/120 (ZR/a0)^3} e^{-ZR/2a0} 2p_z,2p_z (?) S = -{1 + (ZR/2a0) + 1/20 (ZR/a0)^2 + 1/60 (ZR/a0)^3 + 1/240 (ZR/a0)^4} e^{-ZR/2a0}

          Table 9C.1 Overlap integrals between hydrogenic orbitals

Orbitals	Overlap integral

1s,1s	S = {1 + (ZR/a0) + 1/3 (ZR/a0)^2} e^{-ZR/a0}

2s,2s	S = {1 + (ZR/2a0) + 1/12 (ZR/a0)^2 + 1/240 (ZR/a0)^4} e^{-ZR/2a0}

2p_x,2p_x (?)	S = {1 + (ZR/2a0) + 1/10 (ZR/a0)^2 + 1/120 (ZR/a0)^3} e^{-ZR/2a0}

2p_z,2p_z (?)	S = -{1 + (ZR/2a0) + 1/20 (ZR/a0)^2 + 1/60 (ZR/a0)^3 + 1/240 (ZR/a0)^4} e^{-ZR/2a0}

Table 9C.1 Overlap integrals between hydrogenic orbitals

Orbitals Overlap integral

1s,1s S = 1 + (ZR/a0) + 1/3 (ZR/a0)^2 e^-ZR/a0

2s,2s S = 1 + (ZR/2a0) + 1/12 (ZR/a0)^2 + 1/240 (ZR/a0)^4 e^-ZR/2a0

2px,2px (?) S = 1 + (ZR/2a0) + 1/10 (ZR/a0)^2 + 1/120 (ZR/a0)^3 e^-ZR/2a0

2pz,2pz (?) S = -1 + (ZR/2a0) + 1/20 (ZR/a0)^2 + 1/60 (ZR/a0)^3 + 1/240 (ZR/a0)^4 e^-ZR/2a0

Added by Trinidad B.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Sri K

12/15/2022

Step 1

For example, if you are looking at the overlap integral for \(1s, 1s\) orbitals, the expression is: \[ S = \left\{ 1 + \frac{ZR}{a_0} + \frac{1}{3} \left( \frac{ZR}{a_0} \right)^2 \right\} e^{-ZR/a_0} \] Show more…

Show all steps