Questions: Circle the atom in each group that has the smallest atomic radius.
i) Al, N, P, or B
iii) Mg, K, Na, Be
ii) O, Cl, F, or He
iv) C, Cs, Mg, or Si
Transcript text: Circle the atom in each group that has the smallest atomic radius.
i) $\mathrm{Al}, \mathrm{N}, \mathrm{P}$, or B
iii) $\mathrm{Mg}, \mathrm{K}, \mathrm{Na}, \mathrm{Be}$
ii) $\mathrm{O}, \mathrm{Cl}, \mathrm{F}$, or He
iv) $\mathrm{C}, \mathrm{Cs}, \mathrm{Mg}$, or Si
Solution
Solution Steps
Step 1: Understanding Atomic Radius Trends
The atomic radius generally decreases across a period (left to right) and increases down a group (top to bottom) in the periodic table. This is due to the increasing nuclear charge across a period, which pulls electrons closer to the nucleus, and the addition of electron shells down a group, which increases the distance of the outer electrons from the nucleus.
Step 2: Analyzing Each Group
Group i: $\mathrm{Al}, \mathrm{N}, \mathrm{P}$, or B
Aluminum (Al) and Phosphorus (P) are in the third period.
Nitrogen (N) and Boron (B) are in the second period.
Within the same period, atomic radius decreases from left to right.
Group ii: $\mathrm{O}, \mathrm{Cl}, \mathrm{F}$, or He
Oxygen (O) and Fluorine (F) are in the second period.
Chlorine (Cl) is in the third period.
Helium (He) is in the first period.
Helium, being in the first period, has the smallest atomic radius.
Group iii: $\mathrm{Mg}, \mathrm{K}, \mathrm{Na}, \mathrm{Be}$
Magnesium (Mg) and Sodium (Na) are in the third period.
Potassium (K) is in the fourth period.
Beryllium (Be) is in the second period.
Beryllium, being in the second period, has the smallest atomic radius.
Group iv: $\mathrm{C}, \mathrm{Cs}, \mathrm{Mg}$, or Si
Carbon (C) and Silicon (Si) are in the second and third periods, respectively.
Magnesium (Mg) is in the third period.
Cesium (Cs) is in the sixth period.
Carbon, being in the second period, has the smallest atomic radius.