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Chemistry Question 123 – AP-EAMCET 2026

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Which of the following sets are correctly matched? OrderPropertyI. K>Li>C>FAtomic radiusII. F>C>Li>KFirst ionization enthalpyIII. F>C>K>LiElectronegativity The correct answer is

Recall the general trends for atomic radius, ionization enthalpy, and electronegativity across a period and down a group in the periodic table.

Step 1: Analyze the Elements' Positions✦ Active

First, identify the positions of the given elements in the periodic table. This will help in applying periodic trends correctly.

K (Potassium):Group 1, Period 4Li (Lithium):Group 1, Period 2C (Carbon):Group 14, Period 2F (Fluorine):Group 17, Period 2
💡 Teacher's Secret Hint

Remember that elements in the same period are arranged horizontally, and those in the same group are arranged vertically.

Step 2: Evaluate Statement I: Atomic Radius○ Expand

Atomic radius generally decreases across a period (from left to right) and increases down a group. Let's check the order K > Li > C > F.

1. Comparing K and Li (Group 1): K is below Li, so K>Li. This is correct. 2. Comparing Li, C, F (Period 2): As we move from left (Li) to right (F), the atomic radius decreases, so Li>C>F. This is correct. 3. Combining these, K (Period 4) is significantly larger than any element in Period 2. Therefore, the overall order K>Li>C>F is correct.

💡 Teacher's Secret Hint

Electrons in higher principal energy shells (larger period number) are farther from the nucleus, leading to a larger atomic radius.

Step 3: Evaluate Statement II: First Ionization Enthalpy○ Expand

First ionization enthalpy generally increases across a period (from left to right) and decreases down a group. Let's check the order F > C > Li > K.

1. Comparing F, C, Li (Period 2): As we move from left (Li) to right (F), ionization enthalpy increases, so F>C>Li. This is correct (Li: 520 kJ/mol, C: 1086 kJ/mol, F: 1681 kJ/mol). 2. Comparing Li and K (Group 1): K is below Li, so Li>K. This is correct (Li: 520 kJ/mol, K: 419 kJ/mol). 3. Combining these, F and C (Period 2) have higher ionization enthalpies than Li (also Period 2), and Li has a higher ionization enthalpy than K (Period 4). Therefore, the overall order F>C>Li>K is correct.

💡 Teacher's Secret Hint

Ionization enthalpy is the energy required to remove an electron. Elements with a stable electron configuration or smaller atomic size tend to have higher ionization enthalpies.

Step 4: Evaluate Statement III: Electronegativity○ Expand

Electronegativity generally increases across a period (from left to right) and decreases down a group. Fluorine is the most electronegative element. Let's check the order F > C > K > Li.

1. Comparing F, C, Li (Period 2): As we move from left (Li) to right (F), electronegativity increases, so F>C>Li. This is correct (F: 3.98, C: 2.55, Li: 0.98). 2. Comparing Li and K (Group 1): K is below Li, so Li>K. This is correct (Li: 0.98, K: 0.82). 3. The proposed order is F>C>K>Li. This order incorrectly places K before Li. Based on periodic trends, Li (Period 2) should be more electronegative than K (Period 4). The correct order should be F>C>Li>K. Therefore, statement III is incorrect.

💡 Teacher's Secret Hint

Electronegativity measures an atom's ability to attract electrons in a chemical bond. Fluorine is the benchmark for high electronegativity.

Step 5: Conclusion○ Expand

Based on the analysis, statements I and II are correct, while statement III is incorrect. Therefore, the sets I and II are correctly matched.

💡 Teacher's Secret Hint

Always double-check your trend applications, especially when comparing elements from different periods and groups.

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