(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)

(i) How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How

1.	(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)
Answer» Solution :(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.