Periodic Table — First Look

Q1 (10)

(a) Mendeleev's Law: "The properties of elements are a periodic function of their atomic masses." (2)

(b) Ordering strictly by mass sometimes placed an element in a group whose properties didn't match. Chosen pair — Ar and K:

• Atomic mass Ar ≈ 39.95, K ≈ 39.10. By mass, K (39.10) < Ar (39.95), so mass-ordering would put K before Ar. (2)
• But Ar is a noble gas (Group 18) and K is a reactive alkali metal (Group 1); putting K before Ar wrecks the group pattern. So Mendeleev had to swap them against mass order. (2) (Accept Co(58.93)–Ni(58.69) or Te(127.6)–I(126.9) with correct numbers.)

(c) Modern Periodic Law: "Properties of elements are a periodic function of their atomic numbers." (2)

• By atomic number: Ar $Z=18$, K $Z=19$. So Ar naturally comes before K — the swap Mendeleev forced by hand is automatic and correct with $Z$-ordering; anomaly vanishes. (2)

Q2 (8)

Correct list (6 marks, deduct pro-rata; ~0.2/element): H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca, Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn.

Non-obvious symbols (any THREE, 2 marks):

• Na (sodium) — from Latin natrium.
• K (potassium) — from Latin kalium.
• Fe (iron) — from Latin ferrum.
• Cu (copper) — from Latin cuprum. (He/Ne etc. are fine as first letters; award for genuinely Latin-derived ones.)

Q3 (10)

(a) $Z=17$: $1s^2 2s^2 2p^6 3s^2 3p^5$. Last-filled = $3p$ → p-block. (2)

(b) $Z=20$: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$. Last-filled = $4s$ → s-block. (2)

(c) $Z=26$: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$. Last (differentiating) electron = $3d$ → d-block. (3) (1 for config total = 26, 1 for identifying 3d filling, 1 for block.)

(d) Rule: the block is named after the subshell that receives the last (differentiating) electron: s-filling → s-block, p-filling → p-block, d-filling → d-block, f-filling → f-block. (3)

Q4 (8)

(a) $Z=15$ (P): $1s^2 2s^2 2p^6 3s^2 3p^3$. (2)

(b) (4)

• Period = highest principal quantum number $n$ present = 3 (the $3s,3p$). (2)
• Group: p-block group number = $10 + (\text{valence } s + p \text{ electrons})$. Valence electrons $= 3s^2 3p^3 = 5$ ⇒ group $= 10+5 =$ Group 15. (2)

(c) Non-metal. (2) Justification (any two): high electronegativity/ionisation energy in upper-right region; tends to gain/share electrons (forms $P^{3-}$/covalent oxides); poor electrical conductor; located on non-metal side of the metalloid staircase.

Q5 (8)

(a) Table (6 marks, ~0.67 per correct property cell, min 3 each):

(b) Two metalloids in first 30: B (Z=5) and Si (Z=14). (1) Rebuttal: metalloids are not weak metals — they are a distinct category with semiconductor behaviour and amphoteric oxides, properties intermediate between metals and non-metals, not merely a lesser degree of metallic character. (1)

Q6 (6)

(a) X = Na ($Z=11$), Y = Mg ($Z=12$); both Period 3. Na has larger radius. (3) Reasoning: same shell (n=3), but Mg has one more proton (+12 vs +11) with no extra shell, so greater effective nuclear charge pulls the electron cloud inward ⇒ radius decreases left→right; Na (fewer protons) is larger.

(b) Na is more reactive (as a metal). (3) Reasoning: metallic reactivity ∝ ease of losing electron(s); Na loses one $3s$ electron with lower ionisation energy (larger radius, less nuclear pull), whereas Mg holds its two $3s$ electrons more tightly ⇒ Na ionises more readily and is more reactive.

[
  {"claim":"Ar(Z=18) precedes K(Z=19) by atomic number, fixing the mass anomaly","code":"Ar_Z=18; K_Z=19; result = Ar_Z < K_Z"},
  {"claim":"Z=15 valence electrons = 5, giving group 15 (10+5)","code":"valence=2+3; group=10+valence; result = group==15"},
  {"claim":"Z=26 electron count sums to 26 for 1s2 2s2 2p6 3s2 3p6 4s2 3d6","code":"total=2+2+6+2+6+2+6; result = total==26"},
  {"claim":"Na has fewer protons than Mg so larger radius trend holds (11<12)","code":"Na=11; Mg=12; result = Na < Mg"}
]