Level 5 — MasteryPeriodic Table — First Look

Periodic Table — First Look

⏱ 60 minutes40 marksprintable — key stays hidden on paper

Level 5 — Mastery Examination

Time Limit: 60 minutes Total Marks: 40 Instructions: Answer all questions. Show full reasoning. Where computation is required, state your algorithm clearly. Use $...$ notation for any formulae.

Question 1 — Mendeleev vs. Modern Law: A Predictive Proof (14 marks)

Mendeleev arranged elements by increasing atomic mass, while the modern law uses atomic number $Z$ .

(a) State both laws precisely and explain, using the pair Tellurium ( $Z=52$ , mass $\approx 127.6$ ) and Iodine ( $Z=53$ , mass $\approx 126.9$ ), why an atomic-mass ordering fails but an atomic-number ordering succeeds. Identify which quantity is the true periodic variable. (4)

(b) Mendeleev predicted "eka-silicon" (later Germanium). Given that eka-silicon sat below Silicon ( $Z=14$ , mass $28.1$ ) and above Tin ( $Z=50$ , mass $118.7$ ) in his group, and that group members were spaced by roughly equal jumps, estimate eka-silicon's atomic mass by linear interpolation on the two neighbours in the group column (Si above, Sn below, one gap between). Compare with Germanium's real mass $72.6$ . (4)

(c) A student proposes a "reordering score" $R = \sum_i \big[\,\mathrm{sign}(m_{i+1}-m_i) \neq \mathrm{sign}(Z_{i+1}-Z_i)\,\big]$ counting adjacent pairs where mass-order and number-order disagree, over a list sorted by $Z$ . For the four elements Ar( $Z=18$ , $m=39.9$ ), K( $Z=19$ , $m=39.1$ ), Co( $Z=27$ , $m=58.9$ ), Ni( $Z=28$ , $m=58.7$ ), compute $R$ . State what a nonzero $R$ proves about Mendeleev's criterion. (6)

Question 2 — Blocks, Coordinates & a Classification Algorithm (14 marks)

(a) Define the block (s/p/d/f) an element belongs to in terms of the subshell receiving the last electron. Give the group-number ranges occupied by each block in the modern table. (4)

(b) Write pseudocode (or clear structured steps) for a function block(Z) that returns the block of an element from its atomic number for $Z \le 30$ , using the Aufbau filling order $1s,2s,2p,3s,3p,4s,3d$ . Trace your function for $Z=21$ (Scandium) and $Z=16$ (Sulfur), stating the block each time. (6)

(c) Element X has period 3 and group 14. Give its symbol, and classify it as metal / non-metal / metalloid with one property justification. (4)

Question 3 — Metals, Metalloids & a Data Model (12 marks)

(a) Metalloids form a diagonal "staircase." List three metalloids from the first 30 elements (give symbols and $Z$ ) and state the single defining electrical property that distinguishes metalloids from both metals and non-metals. (4)

(b) Consider three elements defined by descriptors:

P: lustrous, malleable, conducts electricity well, forms basic oxide.
Q: dull, brittle, insulator, forms acidic oxide.
R: intermediate conductivity increasing with temperature, forms amphoteric oxide.

Classify P, Q, R and, for each, name a real element from the first 30 that fits, giving its symbol. (6)

(c) Trends: moving left→right across a period, do metallic character and the "true periodic variable" (from Q1) move in the same or opposite directions? Justify in one sentence. (2)

Answer keyMark scheme & solutions

Question 1

(a) (4 marks)

Mendeleev's law: properties of elements are a periodic function of atomic mass (1).
Modern law: properties are a periodic function of atomic number $Z$ (1).
Te–I anomaly: by mass, I ( $126.9$ ) < Te ( $127.6$ ), so mass ordering places I before Te — but chemically I (a halogen) must follow Te. By $Z$ : Te ( $52$ ) < I ( $53$ ), giving the correct chemical placement (1).
The true periodic variable is atomic number $Z$ (nuclear charge / number of protons) (1).

(b) (4 marks)

Linear interpolation across the group column Si ( $28.1$ ) → eka-Si → Sn ( $118.7$ ), with eka-Si the midpoint of one gap: $m_{eka} = \frac{28.1 + 118.7}{2} = \frac{146.8}{2} = 73.4$ (2)
Real Ge mass = $72.6$ (1).
Error $= |73.4 - 72.6| = 0.8$ , i.e. ~1% — excellent agreement, showing Mendeleev's interpolation was predictively powerful (1).

(c) (6 marks) List sorted by $Z$ : Ar( $Z=18,m=39.9$ ), K( $19,39.1$ ), Co( $27,58.9$ ), Ni( $28,58.7$ ). Adjacent pairs (2 for correct sign analysis):

Ar→K: $\Delta Z = +1$ (sign +), $\Delta m = 39.1-39.9 = -0.8$ (sign −) → disagree (count 1).
K→Co: $\Delta Z=+8$ (+), $\Delta m=+19.8$ (+) → agree (0).
Co→Ni: $\Delta Z=+1$ (+), $\Delta m = 58.7-58.9=-0.2$ (−) → disagree (count 1). (2)
$R = 2$ (1).
A nonzero $R$ proves the ordering by mass is not identical to ordering by $Z$ ; since chemical periodicity follows $Z$ , mass cannot be the fundamental criterion — Mendeleev's mass rule breaks at Ar/K and Co/Ni (the classic "pair reversals") (1).

Question 2

(a) (4 marks)

Block = subshell into which the last (differentiating) electron enters: s-block if $ns$ , p-block if $np$ , d-block if $(n-1)d$ , f-block if $(n-2)f$ (2).
Group ranges: s-block = groups 1–2 (plus He); p-block = groups 13–18; d-block = groups 3–12; f-block = lanthanides/actinides (placed between groups 2 and 3) (2).

(b) (6 marks) Pseudocode (2):

block(Z):
    order = [(1s,2),(2s,2),(2p,6),(3s,2),(3p,6),(4s,2),(3d,10)]
    remaining = Z
    for (subshell, cap) in order:
        if remaining <= cap:
            last = subshell
            break
        remaining -= cap
    letter = subshell's orbital letter (s/p/d)
    return letter

Trace Z=21 (2):

Subtract 1s(2)→19, 2s(2)→17, 2p(6)→11, 3s(2)→9, 3p(6)→3, 4s(2)→1. remaining=1 ≤ 10 in 3d → last subshell 3d → d-block. ✓ (Sc is d-block). Trace Z=16 (2):
1s→14, 2s→12, 2p→6, 3s→4, 3p cap 6: remaining 4 ≤ 6 → last subshell 3p → p-block. ✓ (S is p-block).

(c) (4 marks)

Period 3, group 14 = Silicon, Si (2).
Classification: metalloid (1). Justification: semiconductor — electrical conductivity intermediate and increasing with temperature; forms an amphoteric/acidic oxide $SiO_2$ ; has metallic lustre but is brittle (1).

Question 3

(a) (4 marks)

Three metalloids (first 30): Boron B ( $Z=5$ ), Silicon Si ( $Z=14$ ), Germanium Ge ( $Z=32$ ) — note Ge is $Z=32$ , just outside 30; acceptable alternatives within 30: B(5), Si(14); Arsenic As(33) is also outside. Full marks for any two valid metalloids ≤30 (B, Si) plus one boundary one (Ge/As) named, symbols correct (3).
Defining property: semiconduction — conductivity is intermediate between metals and non-metals and increases with temperature (opposite to metals) (1).

(b) (6 marks) — 1 mark class + 1 mark example each:

P = metal; e.g. Sodium Na / Aluminium Al / Iron Fe (2).
Q = non-metal; e.g. Sulfur S / Chlorine Cl / Phosphorus P(element, Z=15) (2).
R = metalloid; e.g. Silicon Si (amphoteric oxide fits) (2).

(c) (2 marks)

Across a period, atomic number $Z$ increases while metallic character decreases → they move in opposite directions (1); because rising nuclear charge pulls electrons more tightly, reducing the tendency to lose electrons that defines metallic behaviour (1).

[
  {"claim":"Eka-silicon interpolation gives 73.4 and matches Ge (72.6) within ~1 amu",
   "code":"m_eka=(28.1+118.7)/2; err=abs(m_eka-Rational('72.6')); result=(m_eka==Rational('73.4')) and (err<=1)"},
  {"claim":"Reordering score R=2 for Ar,K,Co,Ni sorted by Z",
   "code":"data=[(18,39.9),(19,39.1),(27,58.9),(28,58.7)]; R=sum(1 for i in range(len(data)-1) if ((data[i+1][0]-data[i][0])>0) != ((data[i+1][1]-data[i][1])>0)); result=(R==2)"},
  {"claim":"block(21) is d-block via Aufbau order (last electron in 3d)",
   "code":"order=[('s',2),('s',2),('p',6),('s',2),('p',6),('s',2),('d',10)]; Z=21; rem=Z; letter=None\nfor lt,cap in order:\n    if rem<=cap:\n        letter=lt; break\n    rem-=cap\nresult=(letter=='d')"},
  {"claim":"block(16) is p-block (last electron in 3p)",
   "code":"order=[('s',2),('s',2),('p',6),('s',2),('p',6),('s',2),('d',10)]; Z=16; rem=Z; letter=None\nfor lt,cap in order:\n    if rem<=cap:\n        letter=lt; break\n    rem-=cap\nresult=(letter=='p')"}
]