ether (iii) H₂O→ Product.","options":["phenol","benzene","C₆H₅MgBr","4-bromophenol"],"answer":2,"why":"Sandmeyer gives bromobenzene, Mg/ether gives the Grignard reagent, and water protonates it to benzene.","figureNote":"Product structures drawn.","simPath":"Class-12/Chemistry/Chapter-9/problem-9.11-complete-reactions-aniline-diazonium","simTitle":"Diazonium Reactions"},{"qNum":60,"topic":"p-Block Elements","question":"Assertion A: Helium is used to dilute oxygen in diving apparatus. Reason R: Helium has high solubility in O₂. Choose the correct answer.","options":["Both A and R are true and R is the correct explanation of A","Both A and R are true but R is NOT the correct explanation of A","A is true but R is false","A is false but R is true"],"answer":3,"why":"Helium is used because of its very low solubility in blood (preventing bends); R about high solubility in O₂ is false.","simPath":"Class-11/Chemistry/Chapter-3/prompt-42-oxidizing-property-order","simTitle":"Noble Gases"},{"qNum":61,"topic":"Solid State","question":"A compound is formed by two elements A and B. Element B forms a cubic close packed structure and atoms of A occupy 1/3 of the tetrahedral voids. If the formula of the compound is AₓBᵧ, then the value of x + y is:","options":["5","4","3","2"],"answer":1,"why":"B (ccp) = 1, tetrahedral voids = 2; A = (1/3)(2) = 2/3, so A:B = 2:3, formula A₂B₃, x+y = 5.","simPath":"Class-11/Chemistry/Chapter-1/prompt-31-limiting-reagent-ab2","simTitle":"Voids & Formula"},{"qNum":62,"topic":"Biomolecules","question":"Statement I: A unit formed by the attachment of a base to the 1' position of sugar is known as a nucleoside. Statement II: When a nucleoside is linked to phosphoric acid at the 5'-position of the sugar moiety, we get a nucleotide. Choose the correct answer.","options":["Both Statement I and Statement II are true","Both Statement I and Statement II are false","Statement I is true but Statement II is false","Statement I is false but Statement II is true"],"answer":1,"why":"A nucleoside = base + sugar; a nucleotide = nucleoside + phosphate — both statements are correct.","simPath":"Class-12/Chemistry/Chapter-10/problem-10.22-nucleoside-vs-nucleotide","simTitle":"Nucleoside vs Nucleotide"},{"qNum":63,"topic":"Atomic Structure","question":"The relation between nₘ (the number of permissible values of magnetic quantum number m) for a given value of the azimuthal quantum number l, is:","options":["l = (nₘ − 1)/2","l = 2nₘ + 1","nₘ = 2l² + 1","nₘ = l + 2"],"answer":1,"why":"For a given l, m has (2l+1) values, so nₘ = 2l+1 ⇒ l = (nₘ − 1)/2.","simPath":"Class-11/Chemistry/Chapter-2/prompt-29-quantum-numbers-n3","simTitle":"Magnetic Quantum Number"},{"qNum":64,"topic":"Chemical Bonding","question":"Amongst the following, the total number of species NOT having eight electrons around the central atom in its outermost shell is: NH₃, AlCl₃, BeCl₂, CCl₄, PCl₅.","options":["3","2","4","1"],"answer":1,"why":"AlCl₃ (6e), BeCl₂ (4e) and PCl₅ (10e) do not have an octet, so 3 species.","simPath":"Class-11/Chemistry/Chapter-4/prompt-09-octet-rule","simTitle":"Octet Exceptions"},{"qNum":65,"topic":"Chemical Bonding","question":"The correct order of energies of molecular orbitals of the N₂ molecule is:","options":["σ1s < σ*1s < σ2s < σ*2s < (π2pₓ = π2p_y) < σ2p_z < (π*2pₓ = π*2p_y) < σ*2p_z","σ1s < σ*1s < σ2s < σ*2s < σ2p_z < (π2pₓ = π2p_y) < (π*2pₓ = π*2p_y) < σ*2p_z","σ1s < σ*1s < σ2s < σ*2s < σ2p_z < σ*2p_z < (π2pₓ = π2p_y) < (π*2pₓ = π*2p_y)","σ1s < σ*1s < σ2s < σ*2s < (π2pₓ = π2p_y) < (π*2pₓ = π*2p_y) < σ2p_z < σ*2p_z"],"answer":1,"why":"For N₂ (and lighter diatomics) the π2p orbitals lie below σ2p_z due to s-p mixing.","simPath":"Class-11/Chemistry/Chapter-4/prompt-38-lcao-molecular-orbital","simTitle":"MO Energy Order of N₂"},{"qNum":66,"topic":"Chemical Bonding","question":"The number of σ bonds, π bonds and lone pairs of electrons in pyridine, respectively, are:","options":["11, 2, 0","12, 3, 0","11, 3

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