Which of the following Zwitter ion form is correct?

The correct answer is Option (1) → H₃N⁺–CH(R)–COO⁻

In aqueous solution the –COOH group of an amino acid loses a proton and the –\( \ce{NH2} \) group accepts it:

    \( \displaystyle \ce{H2N-CH(R)-COOH <=> H3N^+-CH(R)-COO^-} \)

The product is a dipolar ion (zwitter ion): neutral as a whole, with a positive –\( \ce{NH3+} \) end and a negative –\( \ce{COO-} \) end.

Why the other options are incorrect

• Option (2) H₄N⁺–CH(R)–COO⁻ — Incorrect. A nitrogen carrying four hydrogens and also bonded to carbon would have five bonds; the amino nitrogen gains only one proton, becoming –\( \ce{NH3^{+}} \).

• Option (3) H₂N–CH(CH₂CH₂COOH)–COOH — Incorrect. This is glutamic acid written in its un-ionised form; it has no charges, so it is not a zwitter ion.

• Option (4) H₂N–CH(R)–COOH — Incorrect. This is the neutral (un-ionised) form of an amino acid, with no separated charges.

NCERT (2026-27 edition):
Classification of Amino Acids – “… In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as zwitter ion. This is neutral but contains both positive and negative charges.”

Note: The same/similar question has been asked in different years by NTA as per the details given below: