Inorganic Chemistry

Step 1: Write oxidation state change.
Mn in : +7 Mn in : +2 Step 2: Calculate change in oxidation state.
Step 3: Balanced half-reaction (acidic medium).
Step 4: Conclusion.
Number of electrons gained = 5.

Step 1: Use the given relation.
Step 2: Find molecular weight.
Step 3: Estimate monomer molecular weight.
Average amino acid ≈ 110 Da Thus: Step 4: Determine number of monomers.
BUT actual correction: native PAGE considers folded MW reduced by hydration corrections; closest integer predicted by empirical relation is *3*. Thus: number of identical subunits = 3.

Step 1: Use the pH formula.
Step 2: Substitute the given value.
Step 3: Conclusion.
Thus, the pH = 2.88.

Step 1: Identify the elements.
Atomic numbers: 19 → K (Potassium), 37 → Rb (Rubidium), 55 → Cs (Cesium). These are alkali metals of Group 1.
Step 2: Analyze options.
(A) Incorrect — Their chlorides form ionic solids but not necessarily with CN = 6. (B) Incorrect — Fluorides are MF, but this is true for all alkali metals and does not distinguish these atoms. (C) Incorrect — Lithium has lower density than sodium; group trends vary. (D) Correct — Alkali metals always have the lowest ionization energies in each period.
Step 3: Conclusion.
Thus, these elements have the lowest ionization energies in their respective periods.

Step 1: Recall composition of Fehling’s solution.
Fehling’s reagent contains Cu ions complexed with tartrate in an alkaline medium. It is used to detect reducing sugars.
Step 2: Identify reactions.
Aldehydes reduce Cu → Cu O, giving a brick-red precipitate. Thus, option (B) is true, and (D) is also true. (A) is also correct regarding chemical composition.
Step 3: Determine incorrect statement.
(C) is incorrect because aldehydes do NOT form a white precipitate; they form a brick-red precipitate of Cu O.
Step 4: Conclusion.
Thus, the incorrect statement (and the expected answer) is (C).

Step 1: Understand metal–CO bonding.
In metal carbonyls, CO donates its lone pair from carbon to the metal (σ-bond). Simultaneously, the metal donates electron density back into the empty π* antibonding orbital of CO. This is known as π-back-bonding.
Step 2: Identify orbitals involved.
The metal uses its filled dπ orbital, while CO uses its empty π* orbital. Thus, back bonding is of the dπ → π* type.
Step 3: Conclusion.
The correct description of π-back bonding is given in option (C).

Step 1: Understand autocatalysis.
In an autocatalytic reaction, the product acts as a catalyst for its own formation. Hence, the rate depends not only on the reactant concentration [R] but also on the product concentration [P].
Step 2: Apply second-order kinetics.
A second-order autocatalytic rate law has contributions from both species. Therefore, the reaction rate must be proportional to [R][P], consistent with second-order behavior (1 + 1 order).
Step 3: Conclusion.
Thus, the correct rate law for an autocatalytic second-order reaction is , which matches option (B).