Question:
In which of the following elements are correctly arranged in the increasing order of their electronegativity values?
In which of the following elements are correctly arranged in the increasing order of their electronegativity values?
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F, O, N, and Cl are the most electronegativity elements in the periodic table.
Second-period elements are always more electronegative than third-period elements of the same or neighboring groups due to their very small atomic size.
Second-period elements are always more electronegative than third-period elements of the same or neighboring groups due to their very small atomic size.
Updated On: Jul 22, 2026
- $\text{Li} \lt \text{Be} \lt \text{Na} \lt \text{Mg}$
- $\text{P} \lt \text{Si} \lt \text{C} \lt \text{N}$
- $\text{Cl} \lt \text{S} \lt \text{N} \lt \text{O}$
- $\text{P} \lt \text{S} \lt \text{N} \lt \text{O}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The question asks us to identify the correct increasing order of electronegativity for the given elements.
We need to apply the periodic trends of electronegativity across periods and down groups.
Step 2: Key Formula or Approach:
Electronegativity is the tendency of an atom to attract a shared pair of electrons.
It generally increases from left to right across a period due to increasing nuclear charge and decreasing atomic radius.
It decreases down a group because of increasing atomic radius and shielding effect.
Step 3: Detailed Explanation:
• Let us analyze the electronegativity values on the Pauling scale:
• Phosphorus (P): $2.19$
• Sulfur (S): $2.58$
• Nitrogen (N): $3.04$
• Oxygen (O): $3.44$
• Comparing these values:
P is in Group 15, Period 3.
S is in Group 16, Period 3. Since electronegativity increases across Period 3, we have $\text{P} \lt \text{S}$.
N is in Group 15, Period 2. Because of its smaller size compared to Sulfur, Nitrogen has a higher electronegativity value than Sulfur ($3.04 \gt 2.58$).
O is in Group 16, Period 2. Since electronegativity increases across Period 2, Oxygen is more electronegative than Nitrogen ($3.44 \gt 3.04$).
• Putting it all together, we get:
\[ \text{P} (2.19) \lt \text{S} (2.58) \lt \text{N} (3.04) \lt \text{O} (3.44) \]
Step 4: Final Answer:
The correct increasing order of electronegativity values is $\text{P} \lt \text{S} \lt \text{N} \lt \text{O}$.
The question asks us to identify the correct increasing order of electronegativity for the given elements.
We need to apply the periodic trends of electronegativity across periods and down groups.
Step 2: Key Formula or Approach:
Electronegativity is the tendency of an atom to attract a shared pair of electrons.
It generally increases from left to right across a period due to increasing nuclear charge and decreasing atomic radius.
It decreases down a group because of increasing atomic radius and shielding effect.
Step 3: Detailed Explanation:
• Let us analyze the electronegativity values on the Pauling scale:
• Phosphorus (P): $2.19$
• Sulfur (S): $2.58$
• Nitrogen (N): $3.04$
• Oxygen (O): $3.44$
• Comparing these values:
P is in Group 15, Period 3.
S is in Group 16, Period 3. Since electronegativity increases across Period 3, we have $\text{P} \lt \text{S}$.
N is in Group 15, Period 2. Because of its smaller size compared to Sulfur, Nitrogen has a higher electronegativity value than Sulfur ($3.04 \gt 2.58$).
O is in Group 16, Period 2. Since electronegativity increases across Period 2, Oxygen is more electronegative than Nitrogen ($3.44 \gt 3.04$).
• Putting it all together, we get:
\[ \text{P} (2.19) \lt \text{S} (2.58) \lt \text{N} (3.04) \lt \text{O} (3.44) \]
Step 4: Final Answer:
The correct increasing order of electronegativity values is $\text{P} \lt \text{S} \lt \text{N} \lt \text{O}$.
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