2017 H2 Chem P1 Q7 — Periodic table Solution

The answer is B.
Across Period 3 from Mg to P, the nuclear charge increases while the shielding by inner electrons stays about the same, so the effective nuclear charge increases. The valence electrons are pulled closer, so atomic radius decreases (Data Booklet: Mg 0.160, Al 0.143, Si 0.117, P 0.110 nm), and the atom attracts a shared pair more strongly, so electronegativity increases (Pauling values: Mg 1.31, Al 1.61, Si 1.90, P 2.19).
So on the graph electronegativity must fall steadily as atomic radius increases: P at the top left (smallest radius, highest electronegativity), then Si, Al and Mg at the bottom right. P and Si are close together in radius but differ by about 0.3 in electronegativity, so the P to Si section is the steepest. Graph B shows exactly this.
• A is wrong: it shows P with a much lower electronegativity than Si and Al, level with Mg, and electronegativity rising from P to Si. P is the most electronegative of the four, so it must be the highest point.
• C is wrong: it shows electronegativity increasing with atomic radius, with P lowest and Mg highest. This is the reverse of the real trend.
• D is wrong: although the overall trend is downward, it shows P and Si with almost the same electronegativity and a large drop from Si to Al, and it places Si far from P in radius. In fact the electronegativity falls by a similar amount (about 0.3) at each step, and Si (0.117 nm) is much closer in radius to P (0.110 nm) than to Al (0.143 nm).