Question

1. which element will have the smallest atomic radius? *

○ na
○ ca
○ ti
○ k

1. which has the least electronegativity? *

○ f
○ n
○ o
○ c

Explanation:

Question 3

To determine the element with the smallest atomic radius, we use the periodic trend: atomic radius generally decreases across a period (left to right) and increases down a group (top to bottom). Let's analyze the positions of Na (sodium), Ca (calcium), Ti (titanium), and K (potassium):

• Na (Sodium): Period 3, Group 1.
• K (Potassium): Period 4, Group 1. Since K is below Na in Group 1, K has a larger atomic radius than Na.
• Ca (Calcium): Period 4, Group 2.
• Ti (Titanium): Period 4, Group 4.

Now, compare the elements in Period 4 (Ca, Ti, K) and Period 3 (Na). Across Period 4, from left to right (K, Ca, Ti), the atomic radius decreases because the effective nuclear charge increases, pulling electrons closer. So among K, Ca, Ti, Ti has the smallest atomic radius in Period 4. Now compare Ti (Period 4) with Na (Period 3). Na is in Period 3, which is one period above Period 4. However, we also need to consider the group. Na is in Group 1, Ti in Group 4. Wait, actually, the trend across a period is more significant. Wait, no—let's correct:

Wait, the atomic radius trend: as we move from left to right across a period, the atomic radius decreases because the number of protons increases (higher effective nuclear charge) while electrons are added to the same valence shell. As we move down a group, the atomic radius increases because electrons are added to a new shell (larger principal quantum number, \( n \)).

Let's list the atomic numbers and electron configurations:

• Na: Atomic number 11, electron configuration \( [Ne] 3s^1 \) (Period 3, Group 1)
• K: Atomic number 19, electron configuration \( [Ar] 4s^1 \) (Period 4, Group 1)
• Ca: Atomic number 20, electron configuration \( [Ar] 4s^2 \) (Period 4, Group 2)
• Ti: Atomic number 22, electron configuration \( [Ar] 3d^2 4s^2 \) (Period 4, Group 4)

Now, compare the principal quantum number (\( n \)) of the valence electrons:

• Na: \( n = 3 \)
• K, Ca, Ti: \( n = 4 \)

Wait, but \( n = 3 \) is smaller than \( n = 4 \), so Na's valence electrons are in a smaller shell. But wait, no—because Na is in Group 1, and Ti is in Group 4. Wait, maybe I made a mistake. Let's check the actual atomic radii (approximate values in picometers):

• Na: ~186 pm
• K: ~227 pm
• Ca: ~197 pm
• Ti: ~147 pm

Wait, Ti has a smaller atomic radius than Na? That's because even though Na is in Period 3, Ti is in Period 4 but has a much higher effective nuclear charge (more protons) pulling the electrons closer, overriding the period difference? Wait, no—wait, the atomic radius of Ti is actually smaller than Na. Let's confirm:

Sodium (Na): atomic radius ~186 pm

Titanium (Ti): atomic radius ~147 pm

Yes, because Ti has more protons (22 vs. 11) and the electrons are in the 3d and 4s orbitals, but the effective nuclear charge is higher, so the radius is smaller. So among Na, K, Ca, Ti, Ti has the smallest atomic radius. Wait, but let's re-express the trend:

Across a period (left to right), atomic radius decreases. So in Period 4, from K (Group 1) to Ca (Group 2) to Ti (Group 4), the radius decreases (K > Ca > Ti). Down a group (top to bottom), radius increases (Na < K, Na < Ca? Wait, no—Na is in Period 3, Ca in Period 4. Wait, Na's atomic radius is ~186 pm, Ca's is ~197 pm. So Na is smaller than Ca. But Ti is ~147 pm, which is smaller than Na. So the order of atomic radii from largest to smallest is: K > Ca > Na > Ti. Therefore, Ti has the smallest atomic radius.

To determine the element with the least electronegativity, we use the periodic trend: electronegativity generally increases across a period (left to right) and decreases down a group (top to bottom). The elements given are F (fluorine), N (nitrogen), O (oxygen), and C (carbon). All are in Period 2. Let's list their positions:

• C (Carbon): Group 14
• N (Nitrogen): Group 15
• O (Oxygen): Group 16
• F (Fluorine): Group 17

Across Period 2 (left to right: C, N, O, F), electronegativity increases because the effective nuclear charge increases, making it easier to attract electrons. So the order of electronegativity from least to most is: C < N < O < F. Therefore, C has the least electronegativity.

Answer:

Ti (Titanium)

Question 4