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The energy of an electron in the ground state ($n = 1$) for $\text{He}^+$ ion is $-x$ J, then that for an electron in $n = 2$ state for $\text{Be}^{3+}$ ion in J is:
$E \propto Z^2/n^2$. He⁺: $4/1 = 4$. Be³⁺ at $n=2$: $16/4 = 4$. Ratio is 1, so energy is $-x$ J.
Match List I with List II.
| List I (Reactions) | List II (Reagents/Condition) |
|---|---|
| A. Bicyclohexyl → 2 cyclohexanone | I. PhCOCl / Anhyd. AlCl₃ |
| B. Benzene → diphenyl ketone | II. CrO₃ |
| C. Cyclohexanol → cyclohexanone | III. KMnO₄ / KOH, Δ |
| D. Ar–CH₂CH₃ (with –COOK) → ozonolysis product | IV. (i) O₃ (ii) Zn–H₂O |
Choose the correct answer from the options given below:
A-III (KMnO₄/KOH), B-I (Friedel–Crafts acylation), C-II (CrO₃), D-IV (ozonolysis).
Activation energy of any chemical reaction can be calculated if one knows the value of:
Arrhenius: $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$ — need $k$ at two $T$.
The most stable carbocation among the following is:
(A) is a 3° carbocation with maximum hyperconjugation and inductive stabilisation.
Given below are two statements:
Statement I: The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane
Statement II: When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.
In the light of the above statements, choose the most appropriate answer from the options given below:
Both standard NCERT facts — II is the correct explanation of I.
Which one of the following alcohols reacts instantaneously with Lucas reagent?
Lucas test: 3° alcohols react instantaneously (immediate turbidity).
The Henry's law constant ($K_H$) values of three gases (A, B, C) in water are $145$, $2\times 10^{-5}$ and $35$ kbar, respectively. The solubility of these gases in water follow the order:
Solubility $\propto 1/K_H$; B has the smallest $K_H$, A the largest.
Which plot of $\ln k$ vs $\dfrac{1}{T}$ is consistent with Arrhenius equation? (Options are sketches of straight lines through the first quadrant: A — decreasing line with negative slope; B–D — increasing lines.)
$\ln k = \ln A - E_a/(RT)$ → slope $= -E_a/R < 0$.
On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as:
Direct solid → vapour transition is sublimation.
Intramolecular hydrogen bonding is present in
Only in the ortho isomer are –OH and –NO₂ close enough to form an intramolecular H-bond.
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