Chemistry • Year 11 • Module 1 • Lesson 18
Periodic Trends: Electronegativity and Reactivity
Apply electronegativity values to classify bonds, interpret reactivity data, and reason about periodic trends in real chemical contexts.
1. Classify bonds using electronegativity difference
The table below lists eight bonds. Using the Pauling electronegativity values provided, calculate |Δχ|, classify each bond, and indicate which atom (if any) carries δ−. 16 marks (2 each)
Pauling values: H = 2.2 · C = 2.6 · N = 3.0 · O = 3.4 · F = 4.0 · Na = 0.9 · Mg = 1.3 · Al = 1.6 · Si = 1.9 · P = 2.2 · S = 2.6 · Cl = 3.2 · K = 0.8 · Ca = 1.0
| Bond | χ values | |Δχ| | Bond type | Atom with δ− (if any) |
|---|---|---|---|---|
| H–F | H = 2.2 ; F = 4.0 | |||
| Na–Cl | Na = 0.9 ; Cl = 3.2 | |||
| O–H | O = 3.4 ; H = 2.2 | |||
| C–H | C = 2.6 ; H = 2.2 | |||
| S–O | S = 2.6 ; O = 3.4 | |||
| Mg–O | Mg = 1.3 ; O = 3.4 | |||
| N–H | N = 3.0 ; H = 2.2 | |||
| K–F | K = 0.8 ; F = 4.0 |
1.1 One of the bonds above (C–H) sits right at the borderline between non-polar and polar covalent with |Δχ| = 0.4. Using the lesson’s bond classification thresholds, state the classification of the C–H bond. Explain why this bond is described as “borderline” and why the lesson notes that methane is generally treated as non-polar overall. (2 marks)
2. Interpret graph — electronegativity across Period 3
The bar graph below shows Pauling electronegativity values for elements Na through Cl (Period 3). 7 marks
Figure 2. Pauling electronegativity values for Period 3 elements (Na to Cl). Cl bar highlighted in teal. Illustrative data from Pauling (1932).
2.1 Describe the overall trend in electronegativity across Period 3 and explain the underlying atomic structural reason. (3 marks)
2.2 Using the data in the graph, classify the bond between Na and Cl as ionic, polar covalent, or non-polar covalent. Show your |Δχ| calculation. (2 marks)
2.3 A student predicts: “Because Cl has the highest electronegativity in Period 3, Cl2 must be a polar molecule.” Identify the flaw in this reasoning. (2 marks)
3. Compare reactivity trends — metals and non-metals
Complete the table by describing the reactivity trend and the structural explanation for each group and direction. 8 marks (1 per cell)
| Element type & direction | Reactivity trend | Structural explanation | Example pair (more reactive first) |
|---|---|---|---|
| Metal — across period (L to R) | |||
| Metal — down group | |||
| Non-metal — across period (L to R) | |||
| Non-metal — down group |
4. Predict and justify — halogen displacement in a school laboratory
A student adds three different halogen solutions (chlorine water, bromine water, and iodine solution) to separate test tubes each containing potassium bromide (KBr) solution. Results are recorded below.
| Halogen added | Observation |
|---|---|
| Chlorine water (Cl2(aq)) | Orange-brown colour develops (Br2 produced) |
| Bromine water (Br2(aq)) | No colour change |
| Iodine solution (I2(aq)) | No colour change |
4.1 Using these observations, rank Cl2, Br2, and I2 in order of decreasing reactivity as oxidising agents. Explain what the observations tell us about relative reactivity. (3 marks)
4.2 Predict what would happen if iodine solution were added to potassium chloride (KCl) solution. Justify your prediction using the reactivity trend. (2 marks)
4.3 Explain why the halogen reactivity trend (Cl2 > Br2 > I2) is the opposite of the metal reactivity trend for Group 1 (Li < Na < K < Rb). Reference atomic structure in your answer. (3 marks)
Q1 — Bond classification table
H–F: |Δχ| = 4.0 − 2.2 = 1.8 ≥ 1.7 → ionic (or highly polar covalent at the boundary); δ− on F. Na–Cl: |Δχ| = 3.2 − 0.9 = 2.3 → ionic; δ− on Cl (full charge transfer: Cl−). O–H: |Δχ| = 3.4 − 2.2 = 1.2 → polar covalent; δ− on O. C–H: |Δχ| = 2.6 − 2.2 = 0.4 → borderline non-polar covalent; very small δ− on C (or accept “none”). S–O: |Δχ| = 3.4 − 2.6 = 0.8 → polar covalent; δ− on O. Mg–O: |Δχ| = 3.4 − 1.3 = 2.1 → ionic; δ− on O (full charge transfer: O2−). N–H: |Δχ| = 3.0 − 2.2 = 0.8 → polar covalent; δ− on N. K–F: |Δχ| = 4.0 − 0.8 = 3.2 → ionic; δ− on F (full charge: F−).
1.1 C–H bond at the borderline: |Δχ| = 0.4, which is exactly at the threshold between non-polar (< 0.4) and polar covalent (0.4–1.7) [1]. It is described as “borderline” because the threshold is an approximate guide — a difference of 0.4 can be argued either way depending on the source. The lesson notes that methane is generally treated as non-polar overall because the |Δχ| sits right at (or just below, depending on rounding) the 0.4 boundary and is so small that the bond is considered essentially non-polar in practice [1].
Q2.1 — Period 3 EN trend (3 marks)
Electronegativity increases steadily from left to right across Period 3 (Na = 0.9 to Cl = 3.2) [1]. Across a period, each element has one additional proton added to the nucleus but electrons are added to the same (third) principal shell, so inner-shell shielding remains roughly constant [1]. The increasing Zeff (net nuclear charge) means the nucleus pulls bonding electrons more strongly with each successive element → electronegativity increases [1].
Q2.2 — NaCl bond classification (2 marks)
|Δχ| = 3.2 (Cl) − 0.9 (Na) = 2.3 [1]. Since 2.3 ≥ 1.7, the bond is classified as ionic [1].
Q2.3 — Flaw in Cl2 polarity prediction (2 marks)
The flaw is that the student confuses atomic electronegativity with bond polarity [1]. Both atoms in Cl2 are identical (same element), so |Δχ| = 0. The bonding electrons are shared equally → non-polar covalent bond → Cl2 is non-polar. High electronegativity of an element does not automatically make its diatomic molecule polar [1].
Q3 — Compare reactivity trends table
Metal across period (L to R): Reactivity decreases. Ionisation energy increases across a period, making it harder to lose the valence electron; metals become less reactive. Example: Na (more reactive) > Mg > Al.
Metal down group: Reactivity increases. Going down a group, atomic radius increases and Zeff experienced by the valence electron decreases → ionisation energy decreases → valence electron more easily lost → more reactive. Example: K (more reactive) > Na (Group 1).
Non-metal across period (L to R): Reactivity increases. Electronegativity increases across a period → atoms attract incoming electrons more strongly → more reactive non-metal. Example: Cl (more reactive) > S > P.
Non-metal down group: Reactivity decreases. Going down a group, atomic radius increases → incoming electron would enter a more distant, more shielded valence shell → weaker nuclear attraction → harder to gain an electron → less reactive. Example: Cl (more reactive) > Br > I (Group 17).
Q4.1 — Halogen reactivity ranking (3 marks)
Decreasing reactivity (as oxidising agents): Cl2 > Br2 > I2 [1]. Only Cl2 displaces Br− from KBr solution, forming orange-brown Br2 [1]. Br2 and I2 cannot displace Br−, confirming they are less reactive than Cl2. A more reactive halogen can oxidise the halide ion of a less reactive halogen [1].
Q4.2 — Iodine + KCl prediction (2 marks)
No reaction would occur — no colour change [1]. I2 is less reactive than Cl2, so iodine cannot displace Cl− from KCl solution. A less reactive halogen cannot displace the halide ion of a more reactive halogen [1].
Q4.3 — Opposite trends for halogens vs Group 1 metals (3 marks)
Both trends are caused by the same structural change going down a group: atomic radius increases and inner-shell shielding increases [1]. For Group 1 metals, reactivity depends on losing an electron; the larger atom holds its valence electron more loosely → easier to lose → greater metallic reactivity going down [1]. For Group 17 non-metals, reactivity depends on gaining an electron; the larger atom’s nucleus attracts an incoming electron more weakly → harder to gain → lesser non-metallic reactivity going down [1]. Same structural change, opposite outcomes because one type reacts by losing and the other by gaining electrons.