Question

Gas phase transition metals, when ionized, lose their s-electrons first and the remaining electrons redistribute themselves in the d-orbitals (with some exceptions as discussed in class). To understand this phenomenon, calculate the Zeff for the 4s and 3d-electrons separately for the following atoms: Sc, Ti, V, Mn, Fe and Ni. On the basis of the Zeff, which of the electrons (4s or 3d) would be removed first? Ionize each of these atoms to their monopositive ions (i.e. Ti to Ti+ ). Provide their new electron configuration (use an energy axis to show lowering of orbitals / reorganization of electrons)

Answer 1

The effective nuclear charge, symbolized as Z_eff or Z* is the net positive charge experienced by an electron in a multi-electron atom.

Effective nuclear charge, Z_eff = Z - σ Where, Z= Atomic number, σ = Shielding or screening constant. To calculate the effective nuclear charge (Z*) we need the value of screening constant (σ) which can be calculated by using following rules.

First we have to write the electronic configuration of the element as shown below.

(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) (5d)…

·         Fill the electrons according to Aufbau principle.

·         Any electrons to the right of the electron of interest do not contribute to shielding constant.

·         The shielding constant for each group is formed as the sum of the following contributions:

All other electrons in the same group as the electron of interest shield to an extent of 0.35 nuclear charge units except 1s group, where the other electron contributes only 0.30.

If the group is of the [s, p] type, an amount of 0.85 from each electron (n-1) shell and an amount of 1.00 for each electron from (n-2) and lower shell.

If the group is of the [d] or [f] type then an amount of 1.00 for each electron from all lying left to that orbital.

Calculating the effective nuclear charge in Sc for 4s electron & for 3d electron.

Electronic configuration- (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d¹)(4s²).

For 4s electron:

σ = (0.35 × 1) + (0.85 × 18) + (1 × 1) = 16.65

Z_eff = Z – σ = 21 – 16.65 = 4.35

For 3d electron:

σ = (0.35 × 0) + (1 × 18) = 18

Z_eff = Z – σ = 21 – 18 = 3

Calculating the effective nuclear charge in Ti for 4s electron & for 3d electron.

Electronic configuration- (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d²)(4s²).

For 4s electron:

σ = (0.35 × 1) + (0.85 × 18) + (1 × 2) = 17.65

Z_eff = Z – σ = 22 – 17.65 = 4.35

For 3d electron:

σ = (0.35 × 1) + (1 × 18) = 18.35

Z_eff = Z – σ = 22 – 18 = 3.65

Calculating the effective nuclear charge in V for 4s electron & for 3d electron.

Electronic configuration- (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d³)(4s²).

For 4s electron:

σ = (0.35 × 1) + (0.85 × 18) + (1 × 3) = 18.65

Z_eff = Z – σ = 23 – 18.65 = 4.35

For 3d electron:

σ = (0.35 × 2) + (1 × 18) = 18.7

Z_eff = Z – σ = 23 – 18.7 = 4.3

Calculating the effective nuclear charge in Mn for 4s electron & for 3d electron.

Electronic configuration- (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d⁶)(4s²).

For 4s electron:

σ = (0.35 × 1) + (0.85 × 18) + (1 × 6) = 21.65

Z_eff = Z – σ = 26 – 21.65 = 4.35

For 3d electron:

σ = (0.35 × 5) + (1 × 18) = 19.75

Z_eff = Z – σ = 26 – 19.75 = 6.25

Calculating the effective nuclear charge in Fe for 4s electron & for 3d electron.

Electronic configuration- (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d⁵)(4s²).

For 4s electron:

σ = (0.35 × 1) + (0.85 × 18) + (1 × 5) = 20.65

Z_eff = Z – σ = 25 – 20.65 = 4.35

For 3d electron:

σ = (0.35 × 4) + (1 × 18) = 19.4

Z_eff = Z – σ = 25 – 19.4 = 5.6

Calculating the effective nuclear charge in Ni for 4s electron & for 3d electron.

Electronic configuration- (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d⁸)(4s²).

For 4s electron:

σ = (0.35 × 1) + (0.85 × 18) + (1 × 8) = 23.65

Z_eff = Z – σ = 28 – 23.65 = 4.35

For 3d electron:

σ = (0.35 × 7) + (1 × 18) = 20.45

Z_eff = Z – σ = 28 – 20.45 = 7.55

4s electrons would be removed first, because the inner core electrons shield the valence electron from the nucleus so the outer most electrons only experience an effective nuclear charge.

Sc → Sc⁺ → Electronic configuration: (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d¹)(4s¹).

Ti → Ti⁺ → Electronic configuration: (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d²)(4s¹).

V → V⁺ → Electronic configuration: (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d³)(4s¹).

Mn → Mn⁺ → Electronic configuration: (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d⁶)(4s¹).

Fe → Fe⁺ → Electronic configuration: (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d⁵)(4s¹).

Ni → Ni⁺ → Electronic configuration: (1s²) (2s², 2p⁶)(3s², 3p⁶)(3d⁸)(4s¹).