Given below are two statements: Statement I: The identification of Ni²⁺ is carried out by dimethyl glyoxime in the…

🧠 Two Facts From the DMG Test

The dimethylglyoxime (DMG) test for $\mathrm{Ni^{2+}}$ is one of the most-asked qualitative tests in JEE.

Statement I — Identification of $\mathrm{Ni^{2+}}$ uses DMG in the presence of $\mathrm{NH_4OH}$: TRUE. The alkaline medium deprotonates one $-\mathrm{OH}$ on each DMG, allowing the chelate to form. Without $\mathrm{NH_4OH}$ the rosy-red precipitate $\mathrm{Ni(DMG)_2}$ does not form.

Statement II — DMG is a bidentate neutral ligand: TRUE in JEE convention. The neutral DMG molecule (before deprotonation) coordinates through two N atoms — bidentate. (After deprotonation in the actual nickel complex it's anionic, but as a free ligand it's classified as a neutral bidentate per NCERT.)

Both correct → option (3).

🗺️ The Ni–DMG Picture

Two DMG molecules wrap around $\mathrm{Ni^{2+}}$ in a square-planar geometry. Each DMG donates through 2 N atoms → CN = 4. The deprotonated $-\mathrm{O^-}$ of one DMG hydrogen-bonds to $-\mathrm{OH}$ of the other, locking the planar structure. Net result: insoluble rosy-red $\mathrm{Ni(DMG)_2}$.

⚡ The "Rosy-Red Precipitate" Anchor

DMG + $\mathrm{Ni^{2+}}$ + $\mathrm{NH_4OH}$ → rosy-red precipitate. This colour-cue is itself an answer choice in many MCQs.

⚠️ Bidentate, Not Monodentate

Each DMG has two nitrogens (the two oxime nitrogens). Counting only one — or counting through oxygens — is the typical slip. Stick to N-N as the donor pair.

$\boxed{\text{Answer: (3) Both statements true}}$