Question

Identify the correct orders regarding atomic radii
i. $\text{Cl}>\text{F}>\text{Li}$
ii. $\text{P}>\text{C}>\text{N}$
iii. $\text{Tm}>\text{Sm}>\text{Eu}$
iv. $\text{Sr}>\text{Ca}>\text{Mg}$

• A. i, ii, iii only
• B. ii, iv only
• C. ii, iii, iv only
• D. iii, iv only

Hint:

Atomic radii generally decrease across a period and increase down a group. For vertical groups, the increasing number of shells (statement iv) is the dominant factor. For horizontal trends, the increasing effective nuclear charge is dominant (statement ii is a mix, but the given order is correct).

Answer 1

The Correct Option is B

Step 1: State the general trends for atomic radii.
1. Across a period (left to right): Atomic radius generally decreases (due to increasing effective nuclear charge).
2. Down a group (top to bottom): Atomic radius generally increases (due to increasing number of electron shells).
3. Exceptions exist due to screening effect, noble gas configuration, etc.

Step 2: Analyze statement i: $\text{Cl>\text{F}>\text{Li}$.}
The elements are Li (Group 1, Period 2), F (Group 17, Period 2), and Cl (Group 17, Period 3).
Li and F are in the same period. Since F is further right, $r_{\text{Li}}>r_{\text{F}}$.
F and Cl are in the same group. Since Cl is further down, $r_{\text{Cl}}>r_{\text{F}}$.
The actual order of atomic radii (in pm) is: Li (152)>Cl (99)>F (72).
The statement $\text{Cl}>\text{F}>\text{Li}$ is incorrect.

Step 3: Analyze statement ii: $\text{P>\text{C}>\text{N}$.}
The elements are P (Group 15, Period 3), C (Group 14, Period 2), and N (Group 15, Period 2).
P and N are in the same group (15), so $r_{\text{P}}>r_{\text{N}}$.
C and N are in the same period (2). $r_{\text{C}}>r_{\text{N}}$ due to C being further left.
The general trend for atomic radii in this group of elements (in pm): P (110)>C (77)>N (75).
The order $\text{P}>\text{C}>\text{N}$ is correct.

Step 4: Analyze statement iii: $\text{Tm>\text{Sm}>\text{Eu}$.}
These are Lanthanides (Period 6). Lanthanide contraction causes radii to decrease gradually across the series.
The order of atomic numbers is $\text{Sm} (62), \text{Eu} (63), \text{Tm} (69)$.
The radius is expected to decrease with atomic number, so the expected order is $r_{\text{Sm}}>r_{\text{Eu}}>r_{\text{Tm}}$.
However, Eu has a particularly large radius (204 pm) due to its half-filled $f$ subshell ($4f^7$) configuration in its common $+2$ oxidation state, and Sm (180 pm) is larger than Tm (175 pm) as expected from the contraction.
The actual order is $r_{\text{Eu}}>r_{\text{Sm}}>r_{\text{Tm}}$.
The statement $\text{Tm}>\text{Sm}>\text{Eu}$ is incorrect.

Step 5: Analyze statement iv: $\text{Sr>\text{Ca}>\text{Mg}$.}
These are Alkaline Earth Metals (Group 2): Mg (Period 3), Ca (Period 4), Sr (Period 5).
Atomic radius increases down a group.
The expected and actual order is $r_{\text{Sr}}>r_{\text{Ca}}>r_{\text{Mg}}$.
The statement $\text{Sr}>\text{Ca}>\text{Mg}$ is correct.

Step 6: Conclude the final correct answer.
Only statements ii and iv are correct.