Which of the following statements about nodes are true?

1 Answer

• Q1)

  A: TRUE

  The total number of nodes for a given principal quantum number n is (n-1).

  B: TRUE

  The number of angular nodes is given by the orbital quantum number ℓ.

  Note that:

  ℓ=0 is s-orbital

  ℓ=1 is p-orbital

  ℓ=2 is d-orbital

  ℓ=3 is f-orbital

  Examples:

  For s-orbitals, ℓ=0, thus they have ZERO angular nodes (Testifies B)

  1s orbital has ZERO nodes (Testifies A, since n-1 = 1 – 1 = 0 total)

  2s orbital has ONE radial node (Testifies A, since n-1 = 2 – 1 = 1 total)

  3s orbital has TWO radial nodes (Testifies A, since n-1 = 3 – 1 = 2 total)

  4s orbital has THREE radial nodes (Testifies A, since n-1 = 4 – 1 = 3 total)

  For p-orbitals, ℓ=1, thus they have ONE angular node (Testifies B)

  2p orbital has ONE angular node + ZERO radial node (Testifies A, since n-1 = 2 – 1 = 1 total)

  3p orbital has ONE angular node + ONE radial node (Testifies A, since n-1 = 3 – 1 = 2 total)

  4p orbital has ONE angular node + TWO radial node (Testifies A, since n-1 = 4 – 1 = 3 total)

  For d-orbitals, ℓ=2, thus they have TWO angular node (Testifies B)

  3d orbital has TWO angular node + ZERO radial node (Testifies A, since n-1 = 3 – 1 = 2 total)

  4d orbital has TWO angular node + ONE radial node (Testifies A, since n-1 = 4 – 1 = 3 total)

  For f-orbitals, ℓ=3, thus they have THREE angular node (Testifies B)

  4f orbital has THREE angular node + ZERO radial node (Testifies A, since n-1 = 4 – 1 = 3 total)

  ———————-

  Q2) From the examples above, we can see that:

  3d orbital has TWO angular node + ZERO radial node

  Source(s): I teach chemistry