MCQEasyJEE 2026Group 14 Elements

JEE Chemistry 2026 Question with Solution

Choose the INCORRECT statement

  • A

    Carbon exhibits negative oxidation states along with +4+4 and +2+2.

  • B

    CO2\text{CO}_2 is the most acidic oxide among the dioxides of group 1414 elements.

  • C

    Among the isotopes of carbon, 13C^{13}\text{C} is a radioactive isotope.

  • D

    Carbon cannot exceed its covalency more than four.

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Four statements about carbon are given, and we must identify the incorrect one.

Find: Which option contains the incorrect statement.

Check each statement one by one:

  1. Option A: Carbon can show negative oxidation state 4-4 in CH4\text{CH}_4 and positive oxidation states such as +2+2 in CO\text{CO} and +4+4 in CO2\text{CO}_2. So this statement is correct.

  2. Option B: Among the dioxides of group 1414 elements, CO2\text{CO}_2 is the most acidic oxide. So this statement is correct.

  3. Option C: 13C^{13}\text{C} is a stable isotope, not a radioactive one. The radioactive isotope of carbon is 14C^{14}\text{C}. Therefore, this statement is incorrect.

  4. Option D: Carbon has no vacant dd-orbitals and cannot expand its octet, so its maximum covalency is four. Hence this statement is correct.

Therefore, the incorrect statement is option C.

Common mistakes

  • Confusing 13C^{13}\text{C} with 14C^{14}\text{C}. This is incorrect because 13C^{13}\text{C} is stable, whereas 14C^{14}\text{C} is radioactive. Always remember the stability of common carbon isotopes before choosing the option.

  • Assuming carbon can expand its octet like heavier elements. This is wrong because carbon belongs to the second period and has no vacant dd-orbitals. Therefore, its covalency does not exceed four.

  • Ignoring oxidation states of carbon in common compounds. This leads to rejecting option A incorrectly. Check examples such as CH4\text{CH}_4, CO\text{CO}, and CO2\text{CO}_2 to verify the statement.

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