Question:
Choose the INCORRECT statement
Choose the INCORRECT statement
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Always remember: $^{12}$C and $^{13}$C are stable isotopes, while $^{14}$C is radioactive.
Updated On: Feb 24, 2026
- Carbon exhibits negative oxidation states along with +4 and +2.
- CO$_2$ is the most acidic oxide among the dioxides of group 14 elements.
- Among the isotopes of carbon, $^{13}$C is a radioactive isotope.
- Carbon cannot exceed its covalency more than four.
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The Correct Option is C
Solution and Explanation
Step 1: Checking statement (A).
Carbon shows negative oxidation states (–4 in CH$_4$) as well as positive oxidation states like +2 (CO) and +4 (CO$_2$). Hence, this statement is correct.
Step 2: Checking statement (B).
Among group 14 dioxides, CO$_2$ is the most acidic due to its small size and high electronegativity of carbon. Therefore, this statement is correct.
Step 3: Checking statement (C).
$^{13}$C is a stable isotope of carbon. The radioactive isotope of carbon is $^{14}$C. Hence, this statement is incorrect.
Step 4: Checking statement (D).
Carbon does not have vacant d-orbitals and cannot expand its octet, so its maximum covalency is four. This statement is correct.
Step 5: Conclusion.
The incorrect statement is option (C).
Carbon shows negative oxidation states (–4 in CH$_4$) as well as positive oxidation states like +2 (CO) and +4 (CO$_2$). Hence, this statement is correct.
Step 2: Checking statement (B).
Among group 14 dioxides, CO$_2$ is the most acidic due to its small size and high electronegativity of carbon. Therefore, this statement is correct.
Step 3: Checking statement (C).
$^{13}$C is a stable isotope of carbon. The radioactive isotope of carbon is $^{14}$C. Hence, this statement is incorrect.
Step 4: Checking statement (D).
Carbon does not have vacant d-orbitals and cannot expand its octet, so its maximum covalency is four. This statement is correct.
Step 5: Conclusion.
The incorrect statement is option (C).
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