In structure (4), the methylene group (CH2) between two carbonyl groups (C=O) is highly acidic due to the resonance stabilization of the conjugate base. When the hydrogen is removed, the negative charge on the carbon is delocalized between the two carbonyl groups, making the conjugate base more stable. This increased resonance makes the hydrogen highly acidic.
Match List-I with List-II: List-I
The correct increasing order of stability of the complexes based on \( \Delta \) value is:
List I (Molecule) | List II (Number and types of bond/s between two carbon atoms) | ||
A. | ethane | I. | one σ-bond and two π-bonds |
B. | ethene | II. | two π-bonds |
C. | carbon molecule, C2 | III. | one σ-bonds |
D. | ethyne | IV. | one σ-bond and one π-bond |
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)