Question

Use the Schrodinger equation solution of the H atom corresponding to its wave function for the 3dxy orbital to explain why this orbital has no radial node.

Questions to consider:
(j) What is the value of the wave function and thus the radial part of the function at a node?

(ii) What factor of the radial part of the wave function, containing r, can equal your value in (i) and thus allow you to obtain a value for r?

Answer 1

Answer-

As we know complete wave function is product of radial wave function and angular wave function which is is given below-

?3dxy= R3d × Y3dxy (where R3d is radial wave function for 3d orbital and  Y3dxy is angular wave function which is different for each of d-orbiatals like dxy,dyz,dzx,dx2-dy2,dz2 ) R3d= (1/9?30) × ?2 × Z3/2 × e-?/2   Y3dxy = ?(60/4)xy/r2 × (1/4?)1/2 Y3dxz = ?(60/4)xz/r2 × (1/4?)1/2 Y3dyz = ?(60/4)yz/r2 × (1/4?)1/2 Y3dx2-y2 = ?(15/4)(x2 - y2)/r2 × (1/4?)1/2 and  Y3dz2= ?(5/4){2z2-(x2 + y2)}/r2 × (1/4?)1/2 Z = effective nuclear charge ,? = 2Zr/n where n is the principal quantum number,r = radius ?3dxy=((1/9?30) × ?2 × Z3/2 × e-?/2) (?(60/4)xy/r2 × (1/4?)1/2) There are two types of nodes, angular and radial nodes. Angular nodes are typically flat plane ,Azimuthal quantum no ( ? ) determines the number of angular nodes in an orbital. Radial nodes are spheres (at fixed radius) and equal to n- ? -1. The total nodes of an orbital is the sum of angular and radial nodes (n?1) In a plot of radial probability versus radius (distance from nucleus), number of radial nodes = n-1 for ns-level, n-2 for np-level and n-3 for 3d level so radial node=3-3=0 ( reason may be due to effective nuclear charge tends to zero in radial part expression so it becomes zero.