Question

within a period:
electronegativity increases from (left to right, right to left) because atoms on the
right side need (less, more) electrons to complete their octet (8 electrons).

within a group:
electronegativity increases from (top to bottom, bottom to top) because the
negatively charged outermost electrons of smaller atoms are drawn (nearer, farther) to the positively charged
nucleus.

encircle the more electronegative element.

1. be or n because it is found on the ______
2. f or o because it is found on the ______
3. n or c because it is found on the ______
4. as or ca because it is found on the ______
5. na or s because it is found on the ______

Explanation:

Part 1: Electronegativity Trends

Within a Period:

Electronegativity increases from LEFT to RIGHT because atoms on the right need LESS electrons to complete their octet (they are closer to having 8 valence electrons).

Within a Group:

Electronegativity increases from BOTTOM to TOP because the outermost electrons of smaller (top) atoms are drawn nearer to the nucleus (smaller atomic radius, stronger nuclear pull).

Part 2: More Electronegative Element (Correcting Errors)

1. N (Nitrogen) is more electronegative than Be (Beryllium) because N is on the right of Be in the same period.
2. F (Fluorine) is more electronegative than O (Oxygen) because F is on the right of O in the same period.
3. N (Nitrogen) is more electronegative than C (Carbon) because N is on the right of C in the same period.
4. As (Arsenic) is more electronegative than Ca (Calcium) because As is on the right of Ca in the same period (or top of Ca in some groups, but period trend is stronger here).
5. S (Sulfur) is more electronegative than Na (Sodium) because S is on the right of Na in the same period.

Final Answers (Corrected)

• Within a Period: LEFT to RIGHT; LESS
• Within a Group: BOTTOM to TOP; nearer
• Encircle:

1. $\boldsymbol{\text{N}}$ (right of Be)
2. $\boldsymbol{\text{F}}$ (right of O)
3. $\boldsymbol{\text{N}}$ (right of C)
4. $\boldsymbol{\text{As}}$ (right of Ca)
5. $\boldsymbol{\text{S}}$ (right of Na)

Answer:

Part 1: Electronegativity Trends

Within a Period:

Electronegativity increases from LEFT to RIGHT because atoms on the right need LESS electrons to complete their octet (they are closer to having 8 valence electrons).

Within a Group:

Electronegativity increases from BOTTOM to TOP because the outermost electrons of smaller (top) atoms are drawn nearer to the nucleus (smaller atomic radius, stronger nuclear pull).

Part 2: More Electronegative Element (Correcting Errors)

1. N (Nitrogen) is more electronegative than Be (Beryllium) because N is on the right of Be in the same period.
2. F (Fluorine) is more electronegative than O (Oxygen) because F is on the right of O in the same period.
3. N (Nitrogen) is more electronegative than C (Carbon) because N is on the right of C in the same period.
4. As (Arsenic) is more electronegative than Ca (Calcium) because As is on the right of Ca in the same period (or top of Ca in some groups, but period trend is stronger here).
5. S (Sulfur) is more electronegative than Na (Sodium) because S is on the right of Na in the same period.

Final Answers (Corrected)

• Within a Period: LEFT to RIGHT; LESS
• Within a Group: BOTTOM to TOP; nearer
• Encircle:

1. $\boldsymbol{\text{N}}$ (right of Be)
2. $\boldsymbol{\text{F}}$ (right of O)
3. $\boldsymbol{\text{N}}$ (right of C)
4. $\boldsymbol{\text{As}}$ (right of Ca)
5. $\boldsymbol{\text{S}}$ (right of Na)