Level 1 — RecognitionPeriodic Trends

Periodic Trends

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level 1 Test Paper: Recognition

Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice (1 mark each) [10 marks]

Choose the ONE correct option.

Q1. Using Slater's rules, the effective nuclear charge $Z_{eff}$ experienced by a valence electron in sodium (Na, Z = 11) is approximately: (a) 1.0 (b) 2.2 (c) 6.85 (d) 11.0

Q2. Across a period from left to right, the atomic radius generally: (a) increases (b) decreases (c) stays constant (d) first increases then decreases

Q3. In the isoelectronic series $O^{2-}$ , $F^-$ , $Na^+$ , $Mg^{2+}$ , the species with the largest ionic radius is: (a) $O^{2-}$ (b) $F^-$ (c) $Na^+$ (d) $Mg^{2+}$

Q4. The first ionization energy of boron is lower than that of beryllium because: (a) B has a higher nuclear charge (b) the 2p electron of B is easier to remove than a paired 2s electron (c) Be has a half-filled shell (d) B is a metal

Q5. The element with the most negative (most exothermic) electron gain enthalpy among the halogens is: (a) F (b) Cl (c) Br (d) I

Q6. Which pair shows a diagonal relationship? (a) Na/Mg (b) Li/Mg (c) Be/B (d) C/N

Q7. On the Pauling scale, the most electronegative element is: (a) O (b) Cl (c) F (d) N

Q8. Metallic character generally increases: (a) down a group and across a period left→right (b) up a group and down a period (c) down a group and across a period right→left (d) it does not change

Q9. A cation is always ________ its parent atom: (a) larger than (b) smaller than (c) the same size as (d) unrelated to

Q10. The Mulliken electronegativity of an atom is defined as the average of its: (a) ionization energy and atomic radius (b) ionization energy and electron affinity (c) electron affinity and electronegativity (d) atomic number and mass number

Section B — Matching (1 mark each) [6 marks]

Q11. Match each element/species in Column I with its property in Column II.

Column I	Column II
(i) Cl	(P) largest atomic radius in period 3
(ii) Na	(Q) highest first ionization energy in period 2
(iii) Ne	(R) most negative electron gain enthalpy in group 17
(iv) Al	(S) common oxidation state +3

Write the matched pairs (e.g. i–R). (4 marks; ½ each and 2 for full correctness)

Q12. Match the electronegativity scale with its basis. [2 marks]

Column I (Scale)	Column II (Basis)
(i) Pauling	(P) $\frac{IE + EA}{2}$
(ii) Mulliken	(Q) bond dissociation energies

Section C — True/False WITH Justification (2 marks each) [14 marks]

State True or False and give a one-line justification. (1 mark answer + 1 mark justification)

Q13. An anion is larger than its parent neutral atom.

Q14. The second ionization energy of an element is always greater than its first ionization energy.

Q15. Van der Waals radius is smaller than the covalent radius for the same atom.

Q16. Beryllium resembles aluminium due to a diagonal relationship.

Q17. Across period 2, electronegativity decreases from Li to F.

Q18. In the isoelectronic series, increasing nuclear charge decreases ionic radius.

Q19. Non-metallic character increases from left to right across a period.

Answer keyMark scheme & solutions

Section A

Q1. (b) 2.2 — Na config $1s^2\,2s^2 2p^6\,3s^1$ . For the 3s electron: same group (none other in 3s) contributes 0; the eight n−1 electrons (2s,2p) give $8\times0.85=6.80$ ; the two 1s electrons give $2\times1.00=2.00$ . Screening $S=8.80$ . $Z_{eff}=11-8.80=2.2$ . (1 mark)

Q2. (b) decreases — nuclear charge rises while electrons enter the same shell, pulling them closer. (1)

Q3. (a) $O^{2-}$ — all have 10 electrons; lowest nuclear charge (Z=8) holds electrons least tightly → largest radius. (1)

Q4. (b) — B loses a 2p electron (higher energy, better shielded) than Be's paired 2s electron; the fully filled 2s² of Be is extra stable. (1)

Q5. (b) Cl — F's small size causes electron–electron repulsion in the compact 2p, making Cl's electron gain enthalpy more negative. (1)

Q6. (b) Li/Mg — classic diagonal pair. (1)

Q7. (c) F — F = 3.98, highest on Pauling scale. (1)

Q8. (c) — metallic character rises down a group and toward the left of a period. (1)

Q9. (b) smaller than — losing electrons reduces repulsion and often removes a shell; $Z_{eff}$ /electron rises. (1)

Q10. (b) ionization energy and electron affinity — $\chi_M=\frac{IE+EA}{2}$ . (1)

Section B

Q11. i–R, ii–P, iii–Q, iv–S.

Cl → most negative electron gain enthalpy in group 17 (R)
Na → largest atomic radius in period 3 (P)
Ne → highest first IE in period 2 (Q, noble gas)
Al → common oxidation state +3 (S) (½ each = 2; +2 if all four correct = 4 marks)

Q12. i–Q (Pauling: bond dissociation energies), ii–P (Mulliken: (IE+EA)/2). (1 each = 2)

Section C

Q13. True. Adding electron(s) increases electron–electron repulsion and lowers $Z_{eff}$ per electron, so the electron cloud expands. (1+1)

Q14. True. After removing the first electron, the cation has higher effective nuclear charge per remaining electron, so more energy is needed. (1+1)

Q15. False. Van der Waals radius (non-bonded contact) is larger than covalent radius (bonded, overlapping). (1+1)

Q16. True. Be (period 2, group 2) and Al (period 3, group 13) are diagonally placed with similar charge/size ratio and properties. (1+1)

Q17. False. Electronegativity increases Li→F across period 2 (rising $Z_{eff}$ , smaller size). (1+1)

Q18. True. Same electron count; higher Z pulls electrons in more strongly, shrinking the ion. (1+1)

Q19. True. Increasing $Z_{eff}$ and electronegativity across a period enhance non-metallic (electron-gaining) character. (1+1)

[
  {"claim":"Z_eff for Na 3s electron by Slater = 2.2","code":"Z=11; S=8*0.85+2*1.00; Zeff=Z-S; result = abs(Zeff-2.2)<1e-9"},
  {"claim":"Slater screening total for Na = 8.80","code":"S=8*0.85+2*1.00; result = abs(S-8.80)<1e-9"},
  {"claim":"Isoelectronic species all have 10 electrons","code":"Oe=8+2; Fe=9+1; Nae=11-1; Mge=12-2; result = (Oe==10 and Fe==10 and Nae==10 and Mge==10)"},
  {"claim":"Mulliken chi example: IE=1251, EA=349 kJ/mol gives mean 800","code":"IE=1251; EA=349; chi=(IE+EA)/2; result = abs(chi-800)<1e-9"}
]