(i) How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
(ii) How many unpaired electrons are present in the ground state of
(a) `Cr^(3+)+(Z = 24)` (b) Ne (Z = 10)

Explanation

(i) Formula for total number of nodes = n-1
For 2s orbital: Number of radial nodes = 1.
For 4p orbital: Number of radial nodes = n-1-1.
= 4-1-1 = 2
Number of angular nodes = l
`therefore` Number of angular nodes - 1. So, 4p orbital has 2 radial nodes and 1 angular node.
For 5d orbital: Total number of nodes = n-1
= 5-1 = 4 nodes
Number of radial nodes = n-l-1
= 5-2-1 = 2 radial nodes.
Number of angular nodes = l = 2
`therefore` 5d orbital have 2 radial nodes and 2 angular nodes.
For 4f orbital: Total number of nodes = n-1
= 4-1 = 3 nodes
Number of radial nodes = n-l-1
= 4-3-1 = 0 node.
Number of angular nodes =l
= 3 nodes
`therefore` 4f orbital have 0 radial node and 3 angular nodes.
(ii) (a) `Cr (Z=24)1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)`
  `Cr^(3+)-1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(4)`.
It contains 4 unpaired electrons.
(b) `Ne(Z=10)1s^(2)2s^(2)2p^(6)`. No unpaired electrons in it.