Ionization energy — first, second, …; trends and anomalies (e.g. B - Be): Exercises

The one machine we use again and again:

Level 1 — Recognition

Exercise 1.1

State, without calculation, which member of each pair has the higher first ionization energy, and give the one-line reason. (a) Na or Cl (b) Li or Cs (c) Ne or Na

Recall Solution 1.1

(a) Cl. Same period (period 3), Cl is further right, more protons entering the same shell ⇒ $Z_{eff} ↑$ , radius $↓$ ⇒ higher IE. See Atomic Radius Trends. (b) Li. Same group (Group 1), Li is higher up ⇒ smaller $n$ , less shielding ⇒ the valence electron is closer and more tightly held ⇒ higher IE. (c) Ne. Ne is a noble gas with a filled $2 p^{6}$ shell; Na starts a new shell ( $3 s^{1}$ ) that is far out and well shielded. Ne's electron is gripped hard; Na's is loose ⇒ Ne has the far higher IE.

Exercise 1.2

For the element magnesium, order these correctly with the $<$ sign: $I E_{1}, I E_{2}, I E_{3}$ . Explain in one sentence why the order can never reverse.

Recall Solution 1.2

$I E_{1} < I E_{2} < I E_{3}$ Each removal leaves a species with the same nuclear charge $Z$ but fewer electrons, so the remaining electrons feel a stronger effective pull and repel each other less. By Coulomb's Law a more positive core grips the next electron harder — so successive IEs always rise for the same element.

Level 2 — Application

Exercise 2.1

Use the Coulomb model to estimate $I E_{1}$ of sodium. Na $= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$ , so $Z = 11$ and the electron removed is the $3 s^{1}$ . Use Slater's rules: electrons in shells with $n$ two or more below count $1.00$ each; electrons in the shell just below ( $n - 1$ ) count $0.85$ each; same-shell electrons count $0.35$ each.

Recall Solution 2.1

Step 1 — count the shielders of the $3 s$ electron.

Inner-inner ( $1 s^{2}$ , that's $n = 1$ , two below $n = 3$ ): $2 \times 1.00 = 2.00$ .
Just-below shell ( $2 s^{2} 2 p^{6}$ , that's $n = 2$ ): $8 \times 0.85 = 6.80$ .
Same shell as the $3 s^{1}$ : no other electrons in $n = 3$ , so $0 \times 0.35 = 0$ .

Step 2 — add them. $S = 2.00 + 6.80 + 0 = 8.80$ . Why: $S$ is the total blocking of the nucleus.

Step 3 — effective charge. $Z_{eff} = 11 - 8.80 = 2.20$ .

Step 4 — plug in ( $n = 3$ ). $I E \approx 13.6 \times \frac{2.2 0 ^{2}}{3 ^{2}} = 13.6 \times \frac{4.84}{9} \approx 7.31 eV .$ Check: experimental $I E_{1} (Na) = 5.14 eV$ — same order of magnitude. The model over-shoots because Slater treats the $3 s$ orbital as a clean hydrogen shell; the real electron spends time nearer the nucleus. Physics captured, precision approximate.

Exercise 2.2

For a $2 p$ electron in boron ( $Z = 5$ , $1 s^{2} 2 s^{2} 2 p^{1}$ ), compute $Z_{eff}$ by Slater's rules. Same-shell $(2 s, 2 p)$ electrons shield $0.35$ ; $1 s$ electrons shield $0.85$ (they are the shell just below).

Recall Solution 2.2

The removed electron is the lone $2 p$ .

Same shell ( $n = 2$ ): the two $2 s$ electrons count $0.35$ each ⇒ $2 \times 0.35 = 0.70$ . (No other $2 p$ electron exists.)
Shell below ( $1 s^{2}$ ): $2 \times 0.85 = 1.70$ . $S = 0.70 + 1.70 = 2.40, Z_{eff} = 5 - 2.40 = 2.60.$ This number ( $2.60$ ) is what we compare against Be in the next level.

Level 3 — Analysis

Exercise 3.1

Beryllium and boron sit next to each other, yet $I E_{1} (B) < I E_{1} (Be)$ . Compare their removed electrons and explain the dip. (Refer to the figure.)

Figure — Ionization energy — first, second, …; trends and anomalies (e.g. B - Be)

Recall Solution 3.1

Be $= 1 s^{2} 2 s^{2}$ : the removed electron leaves a full $2 s$ subshell — a compact, low-energy, symmetric pair.
B $= 1 s^{2} 2 s^{2} 2 p^{1}$ : the removed electron sits in a $2 p$ orbital, which is higher in energy than $2 s$ and is shielded by the $2 s^{2}$ pair beneath it.

Look at the red level in the figure: the $2 p$ electron starts higher and feels a weaker effective pull, so less energy lifts it free. Even though B has one more proton, the orbital-energy + shielding penalty outweighs the extra $Z$ . Hence the trend dips: $I E_{1} (B) < I E_{1} (Be)$ . This is Half-filled and Fully-filled Stability in disguise: the before element (Be) is unusually stable, not that B is broken.

Exercise 3.2

Nitrogen and oxygen are adjacent, but $I E_{1} (O) < I E_{1} (N)$ . Give the two-part reason. (Refer to the figure.)

Recall Solution 3.2

N $= \dots 2 p^{3}$ : each of the three $2 p$ orbitals holds one electron (up-up-up). This half-filled set is symmetric and maximises exchange stabilisation — an extra-stable arrangement, so N clings to its electrons.
O $= \dots 2 p^{4}$ : one $2 p$ orbital is now doubly occupied (up-down). Those two electrons share one small region of space and repel each other strongly (see the orange "repulsion" spring in the figure).

Removing an electron from O's paired orbital relieves that repulsion, so it comes out with less effort. Two effects both point the same way — N is stabilised, O is destabilised — so $I E_{1} (O) < I E_{1} (N)$ .

Level 4 — Synthesis

Exercise 4.1

Arrange these five period-2 elements by increasing $I E_{1}$ : B, Be, C, N, O. Then mark where the two anomalies bend the line.

Recall Solution 4.1

Smooth-trend expectation (left→right rises): Be < B < C < N < O. Now apply the two dips:

B < Be (Anomaly 1): B falls below Be.
O < N (Anomaly 2): O falls below N.

Putting the real experimental order together: $B < B e < C < O < N$ Reading it: B is the lowest (shielded $2 p$ ), then Be (full $2 s$ ), then C climbs normally, then O dips below N (pairing), and N is the highest of these five (half-filled $2 p^{3}$ ).

Exercise 4.2

Silicon has successive IEs (kJ/mol): $786, 1577, 3232, 4356, 16091, 19805, \dots$ Which group does Si belong to, and how do the numbers tell you? Convert $I E_{1}$ to eV to sanity-check ( $1 eV = 96.49 kJ mol^{- 1}$ ).

Recall Solution 4.2

Find the big jump. Ratios of consecutive IEs: $\frac{1577}{786} \approx 2.0, \frac{3232}{1577} \approx 2.0, \frac{4356}{3232} \approx 1.3, \frac{16091}{4356} \approx 3.7 (!) .$ The jump from $I E_{4}$ to $I E_{5}$ is by far the largest (×3.7). So after removing 4 electrons we hit a stable noble-gas core $[Ne]$ . Four valence electrons ⇒ Group 14. (Si is indeed $[Ne] 3 s^{2} 3 p^{2}$ .) Sanity check: $I E_{1} = 786/96.49 \approx 8.15 eV$ — a reasonable metalloid value, well above metals like Na (5.1 eV) and below noble gases.

Level 5 — Mastery

Exercise 5.1

An unknown period-3 element has these successive IEs (kJ/mol): $577, 1817, 2745, 11577, 14842, \dots$ Identify the group, deduce the outer configuration, and name the element.

Recall Solution 5.1

Step 1 — locate the core-break by ratios. $\frac{1817}{577} \approx 3.1, \frac{2745}{1817} \approx 1.5, \frac{11577}{2745} \approx 4.2 (biggest) .$ The largest ratio is at $I E_{3} \to I E_{4}$ . Removing 3 electrons reaches the noble-gas core ⇒ 3 valence electrons ⇒ Group 13. Step 2 — period 3, Group 13 ⇒ outer configuration $[Ne] 3 s^{2} 3 p^{1}$ , three removable valence electrons. Step 3 — name it: period 3, Group 13 = aluminium (Al). Cross-check: the first ratio $I E_{2} / I E_{1} \approx 3.1$ is also fairly large — that reflects Al's Anomaly-1 style low $I E_{1}$ (the lone shielded $3 p^{1}$ is easy to pull, just like B), so the very first step is unusually cheap. Consistent with Al being the Group-13 analogue of boron.

Exercise 5.2

Using the Coulomb model, explain quantitatively why $I E_{1} (Be)$ exceeds $I E_{1} (B)$ is not predicted by $Z_{eff}$ alone — i.e. show that $Z_{eff}$ actually rises Be→B, and state what extra physics reverses the trend.

Recall Solution 5.2

Be $= 1 s^{2} 2 s^{2}$ , remove a $2 s$ electron ( $Z = 4$ ):

Same shell ( $n = 2$ ): one other $2 s$ electron ⇒ $1 \times 0.35 = 0.35$ .
Below ( $1 s^{2}$ ): $2 \times 0.85 = 1.70$ . $S = 2.05, Z_{eff} (Be) = 4 - 2.05 = 1.95.$ B $= 1 s^{2} 2 s^{2} 2 p^{1}$ , remove the $2 p$ electron ( $Z = 5$ ): from Exercise 2.2, $Z_{eff} (B) = 2.60$ . The paradox. $Z_{eff}$ rose from $1.95$ to $2.60$ . Plugging naively into $I E \propto Z_{eff}^{2} / n^{2}$ (both $n = 2$ ) predicts B's IE should be higher: $\frac{I E ( B )}{I E ( Be )} \approx \frac{2.6 0 ^{2}}{1.9 5 ^{2}} \approx 1.78 > 1.$ So the simple Coulomb/Slater model gets the anomaly wrong. The missing physics: Slater's rules crudely lump $2 s$ and $2 p$ together and ignore that $2 p$ is a genuinely higher-energy orbital with different penetration — the $2 p$ electron spends more time outside the $2 s^{2}$ shell, feeling a much smaller true pull than $Z_{eff} = 2.60$ suggests. This orbital-energy difference (a quantum-mechanical effect Slater does not capture) is what makes B's electron easy to remove and reverses the trend to $I E (B) < I E (Be)$ .

Wrap-up recall

Recall One-line takeaways (hide and test)

Biggest ratio in successive IEs marks the core-break ::: ⇒ number of valence electrons ⇒ group number. Why Be > B despite lower $Z$ ::: B's electron is a higher-energy, $2 s^{2}$ -shielded $2 p$ ; Be has a stable full $2 s^{2}$ . Why N > O despite lower $Z$ ::: N's half-filled $2 p^{3}$ is exchange-stabilised; O's $2 p^{4}$ pairing repulsion makes one electron easy to remove. Real order of B, Be, C, N, O by increasing $I E_{1}$ ::: $B < B e < C < O < N$ . Slater $Z_{eff}$ for Na's $3 s$ electron ::: $Z_{eff} = 11 - 8.80 = 2.20$ .