Level 4 — ApplicationPeriodic Trends

Periodic Trends

50 marksprintable — key stays hidden on paper

Time: 60 minutes Total Marks: 50 Instructions: Attempt all questions. Show all reasoning. Use Slater's rules and periodic trend logic. Constants where needed are given in the question.

Q1. Effective nuclear charge and its consequences (12 marks)

(a) Using Slater's rules, calculate $Z_{\text{eff}}$ experienced by a $3p$ electron in a sulfur atom ( $Z=16$ ). Show the grouping and each screening contribution. (4)

(b) Calculate $Z_{\text{eff}}$ for a $4s$ electron and a $3d$ electron in a chromium atom ( $Z=24$ , configuration $[\text{Ar}]3d^5 4s^1$ ). (5)

(c) Using only your two results in (b), explain why the $4s$ electron (not a $3d$ electron) is the first to be ionised from chromium, despite $4s$ being filled before $3d$ . (3)

Q2. Isoelectronic and radius reasoning (10 marks)

Consider the isoelectronic series: $\text{S}^{2-},\ \text{Cl}^-,\ \text{K}^+,\ \text{Ca}^{2+}$ .

(a) Arrange them in order of increasing ionic radius and justify quantitatively using the trend in nuclear charge. (4)

(b) A student claims that because all four ions have the same electron count, they must have identical $Z_{\text{eff}}$ for their outermost electrons. Use Slater's rules to compute $Z_{\text{eff}}$ for the outermost ( $3p$ ) electron in $\text{S}^{2-}$ and in $\text{Ca}^{2+}$ , and use the numbers to refute the claim. (6)

Q3. Ionization energy anomalies applied (10 marks)

The successive ionization energies (kJ/mol) of an unknown third-period element X are: $IE_1 = 578,\quad IE_2 = 1817,\quad IE_3 = 2745,\quad IE_4 = 11577,\quad IE_5 = 14842$

(a) Identify element X and justify from the pattern of jumps. (4)

(b) Predict, with reasoning, which of $IE_1$ for X is expected to be lower than $IE_1$ of its immediate left neighbour in the period, and name the electronic-structure principle responsible for this anomaly. (4)

Q4. Diagonal relationship prediction (10 marks)

Beryllium and aluminium show a diagonal relationship.

(a) Explain, in terms of a specific periodic property whose value is similar for the two elements, why the diagonal relationship arises. Support with the direction of the two competing trends. (4)

(b) Predict two distinct chemical properties/behaviours in which Be resembles Al rather than its own group member Mg. Give the chemical reasoning for each. (4)

(c) The charge-to-radius ratio is often invoked. Given ionic radii $r(\text{Be}^{2+})=27\ \text{pm}$ , $r(\text{Al}^{3+})=53\ \text{pm}$ , $r(\text{Mg}^{2+})=72\ \text{pm}$ , compute the charge/radius ratio (in $e/\text{pm}$ ) for each and use it to support your answer to (a). (2)

Q5. Integrated trend prediction (8 marks)

Fluorine has a lower electron gain enthalpy magnitude than chlorine, contrary to the general down-a-group trend.

(a) Explain the anomaly quantitatively in terms of the physical size/electron density argument. (3)

(b) Using the Mulliken definition $\chi_M = \dfrac{IE + EA}{2}$ , and the data $IE(\text{F})=1681$ , $EA(\text{F})=328$ , $IE(\text{Cl})=1251$ , $EA(\text{Cl})=349$ (all kJ/mol), compute the Mulliken electronegativity (in kJ/mol) of F and Cl and comment on which is larger and why the electron affinity anomaly does not reverse the electronegativity order. (5)

Answer keyMark scheme & solutions

Q1 (12)

(a) Sulfur $Z=16$ : $1s^2\,|\,2s^2 2p^6\,|\,3s^2 3p^4$ . For a $3p$ (n=3) electron, Slater groups $(3s,3p)$ together:

Same group $(3s3p)$ : 5 other electrons × 0.35 = 1.75
n−1 shell (n=2): 8 × 0.85 = 6.80
n−2 and inner (n=1): 2 × 1.00 = 2.00

Total screening $S = 1.75 + 6.80 + 2.00 = 10.55$ . $Z_{\text{eff}} = 16 - 10.55 = \mathbf{5.45}$ . (4: grouping 1, three terms 2, final 1)

(b) Cr $[\text{Ar}]3d^5 4s^1$ , $Z=24$ .

4s electron: group $(4s)$ , others in same group = 0; the $3d^5$ counts as inner (for an $s/p$ electron, electrons in lower groups screen). Grouping: $1s^2|2s2p^8|3s3p^8|3d^5|4s^1$ .

Same group ( $4s$ ): 0 × 0.35 = 0
n−1 (n=3: $3s,3p,3d$ = 8+5 =13): 13 × 0.85 = 11.05
n−2 & lower (n≤2: 2+8=10): 10 × 1.00 = 10.00

$S = 21.05$ ; $Z_{\text{eff}}(4s) = 24 - 21.05 = \mathbf{2.95}$ . (2.5)

3d electron: for a d electron, all electrons in same group $(3d)$ contribute 0.35; everything to the left (lower groups) contributes 1.00.

Same group $(3d^5)$ : 4 others × 0.35 = 1.40
All electrons in inner groups (n=1,2,3s3p): $2+8+8 = 18$ × 1.00 = 18.00

$S = 19.40$ ; $Z_{\text{eff}}(3d) = 24 - 19.40 = \mathbf{4.60}$ . (2.5)

(c) $Z_{\text{eff}}(3d)=4.60 > Z_{\text{eff}}(4s)=2.95$ : the $3d$ electrons are held more tightly (higher effective attraction), while the $4s$ electron is more weakly bound and higher in energy once the $3d$ subshell is populated. Hence the $4s$ electron is removed first. (3)

Q2 (10)

(a) All have 18 electrons; nuclear charges: $S(16) < Cl(17) < K(19) < Ca(20)$ . More protons pull the same electron cloud inward, so radius decreases with increasing Z. Increasing radius: $\text{Ca}^{2+} < \text{K}^+ < \text{Cl}^- < \text{S}^{2-}$ . (4: order 2, nuclear-charge justification 2)

(b) Outermost $3p$ electron, screening constant identical because electron configuration is identical ( $[\text{Ar}]$ -like $3s^23p^6$ ): $S = (7 \times 0.35) + (8 \times 0.85) + (2 \times 1.00) = 2.45 + 6.80 + 2.00 = 11.25$ .

$\text{S}^{2-}$ : $Z_{\text{eff}} = 16 - 11.25 = \mathbf{4.75}$
$\text{Ca}^{2+}$ : $Z_{\text{eff}} = 20 - 11.25 = \mathbf{8.75}$

Since $S$ is the same but $Z$ differs, $Z_{\text{eff}}$ differs by $\Delta Z = 4$ . The claim is false: identical electron count does not give identical $Z_{\text{eff}}$ ; the nucleus differs. Higher $Z_{\text{eff}}$ for $\text{Ca}^{2+}$ explains its smaller radius. (6: S value 2, two Z_eff 2, refutation 2)

Q3 (10)

(a) Look for the large jump: $IE_3 = 2745 \to IE_4 = 11577$ (>4× increase). The big jump after removing 3 electrons means the 4th electron comes from a noble-gas core → element has 3 valence electrons → Group 13. Third period ⇒ X = Aluminium (Al). (4)

(b) $IE_1$ of Al is lower than that of its left neighbour Mg. Reason: Al removes a $3p^1$ electron (higher energy, partially shielded by the $3s^2$ pair), whereas Mg removes from a filled, stable $3s^2$ subshell. The extra stability of the filled subshell + easier removal of the singly-occupied p electron makes $IE_1(\text{Al}) < IE_1(\text{Mg})$ — the Be < ... anomaly / filled-subshell (screening) principle. (4)

(c) Common oxidation state +3 (loss of the three easily removed electrons before the core). (2)

Q4 (10)

(a) On moving down a group, atoms get larger / less electronegative; on moving right across a period they get smaller / more electronegative. Going diagonally (Be → Al) the two opposing changes roughly cancel, so properties tied to charge/radius ratio (polarising power) and electronegativity are similar. Be and Al have comparable polarising power and comparable electronegativity (~1.5–1.6). (4)

(b) Any two, with reasoning:

Both Be and Al form covalent/polymeric halides (e.g. $\text{BeCl}_2$ , $\text{AlCl}_3$ ) due to high charge/radius ratio and strong polarisation, unlike ionic $\text{MgCl}_2$ .
Both are amphoteric — oxides/hydroxides dissolve in both acid and alkali (e.g. $\text{Be(OH)}_2$ , $\text{Al(OH)}_3$ ), whereas $\text{Mg(OH)}_2$ is basic.
Both form covalent carbides/complex fluoride ions ( $\text{BeF}_4^{2-}$ , $\text{AlF}_6^{3-}$ ); both passivated by conc. nitric acid — accept valid alternatives. (4: 2 each)

(c) Charge/radius:

Be²⁺: $2/27 = 0.0741\ e/\text{pm}$
Al³⁺: $3/53 = 0.0566\ e/\text{pm}$
Mg²⁺: $2/72 = 0.0278\ e/\text{pm}$

Be²⁺ (0.074) is far closer to Al³⁺ (0.057) than to Mg²⁺ (0.028), confirming similar polarising power → diagonal relationship. (2)

Q5 (8)

(a) F is very small; the incoming electron enters a compact $2p$ subshell where electron–electron repulsion is severe. This repulsion partly offsets the nuclear attraction, so the energy released ( $|E_{eg}|$ ) is smaller than expected. In Cl the larger $3p$ orbital spreads charge out, reducing repulsion, so Cl releases more energy. (3)

(b) Mulliken $\chi_M = (IE+EA)/2$ :

F: $(1681 + 328)/2 = 2009/2 = \mathbf{1004.5}$ kJ/mol
Cl: $(1251 + 349)/2 = 1600/2 = \mathbf{800}$ kJ/mol

F > Cl. Although $EA(\text{Cl}) > EA(\text{F})$ , the difference (21 kJ/mol) is tiny compared with the difference in ionization energy ( $IE(\text{F})-IE(\text{Cl}) = 430$ kJ/mol). The IE term dominates the Mulliken average, so F remains the more electronegative element. (5: two values 3, comment 2)

[
  {"claim":"Z_eff of 3p electron in S is 5.45","code":"S=5*0.35+8*0.85+2*1.00; Zeff=16-S; result=abs(Zeff-5.45)<1e-9"},
  {"claim":"Cr 4s Zeff=2.95 and 3d Zeff=4.60","code":"S4s=13*0.85+10*1.00; Z4s=24-S4s; S3d=4*0.35+18*1.00; Z3d=24-S3d; result=(abs(Z4s-2.95)<1e-9) and (abs(Z3d-4.60)<1e-9)"},
  {"claim":"Isoelectronic 3p Zeff: S2- =4.75, Ca2+ =8.75","code":"S=7*0.35+8*0.85+2*1.00; result=(abs((16-S)-4.75)<1e-9) and (abs((20-S)-8.75)<1e-9)"},
  {"claim":"Mulliken chi F=1004.5, Cl=800, F larger","code":"chiF=(1681+328)/2; chiCl=(1251+349)/2; result=(chiF==1004.5) and (chiCl==800) and (chiF>chiCl)"}
]