Chemistry — NEET Complete Yearwise Papers MCQs with Solutions

Solution: As the atomic number increases across a period the effective nuclear charge rises, pulling the electrons in, so the atomic radius decreases from left to right. Li, Be, B and C sit in that order in period 2, so the radii fall in the same order: Li > Be > B > C.

Solution: Let Rate $= k[A]^x[B]^y$. Comparing rows 1 and 2 ($[B]$ fixed): doubling $[A]$ doubles the rate, so $2^x = 2$ and $x = 1$. Comparing rows 2 and 3 ($[A]$ fixed): doubling $[B]$ raises the rate from $4 \times 10^{-3}$ to $1.6 \times 10^{-2}$, a factor of 4, so $2^y = 4$ and $y = 2$. So the order is 1 with respect to $A$ and 2 with respect to $B$.

Solution: Hydroboration is ANTI-Markovnikov: boron adds to the less substituted carbon. So propene plus diborane gives the n-propyl borane $\ce{(CH3CH2CH2)3B}$, not the isopropyl compound — Statement I is incorrect. Statement II is correct on its own terms: if you did start from $\ce{((CH3)2CH)3B}$, oxidation with $\ce{H2O2}$/NaOH would replace boron by $\ce{-OH}$ at the same carbon and give propan-2-ol.

Solution: Baeyer's reagent is a cold, dilute, aqueous (alkaline) solution of potassium permanganate. It is used to test for unsaturation — an alkene decolourises its purple colour while being converted into a vicinal diol.

Solution: Dipole moment depends on both shape and bond dipole. CCl₄ (tetrahedral), CO₂ (linear) and BF₃ (trigonal planar) are all symmetrical, so their bond dipoles cancel exactly and $\mu = 0$. HI is a simple diatomic with two different atoms, so its bond dipole cannot cancel — $\mu = 0.38$ D.

Solution: Step (i) $\ce{Cl2}$/$\ce{FeCl3}$ chlorinates the ring ortho to the activating ethyl group. Step (ii) Sn/HCl reduces the $\ce{-NO2}$ group to $\ce{-NH2}$. Step (iii) $\ce{NaNO2}$/HCl at 273–278 K converts the amine into the diazonium salt. Step (iv) KI replaces the diazonium group by iodine (Sandmeyer-type). The iodine therefore ends up exactly where the nitro group was — para to the ethyl group — with the chlorine ortho to the ethyl group, which is option (c).

Solution: Oxalic acid is a weak acid and sodium hydroxide a strong base, and phenolphthalein is the indicator for a weak acid–strong base titration. Phenolphthalein is colourless in acidic medium and pink in alkaline medium, so at the end point the colour changes from colourless to pink.

Solution: Nitrogen — the element with the highest ionisation enthalpy in group 15 (III). Fluorine molecule — the strongest oxidising agent (IV). Oxygen molecule — paramagnetic, because of its two unpaired electrons in the antibonding $\pi^*$ orbitals (I). Xenon — the most reactive element of group 18 (II).

Solution: Corrosion slowly coats the surface of a metallic object with oxides or other salts of the metal — rusting of iron, tarnishing of silver and the green patina on copper and bronze are all examples. Electrolysis of water to produce hydrogen is not corrosion at all; it is a decomposition of water driven by an external supply.

Solution: $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. Zn²⁺ (d¹⁰): $n = 0$; Ti²⁺ (d²): $n = 2$ — different. Cr²⁺ (d⁴): $n = 4$; Fe²⁺ (d⁶): $n = 4$ — SAME ($\sqrt{24}$). Ti³⁺ (d¹): $n = 1$; Cu²⁺ (d⁹): $n = 1$ — SAME ($\sqrt{3}$). V²⁺ (d³): $n = 3$; Cu⁺ (d¹⁰): $n = 0$ — different. So B and C only.

Solution: Reversing the first equilibrium inverts its constant: $2A_3B \rightleftharpoons 3A_2 + B_2$, $K' = \dfrac{1}{K_1}$ Halving all the coefficients takes the square root: $A_3B \rightleftharpoons \dfrac{3}{2}A_2 + \dfrac{1}{2}B_2$, $K_2 = \sqrt{K'} = \dfrac{1}{\sqrt{K_1}}$

Solution: Chloroform is an organic, essentially non-polar solvent, so "like dissolves like" — the more polar the solute, the LESS soluble it is. In order of decreasing polarity: NaCl (ionic) > CH₃OH > CH₃CN > cyclohexane (non-polar). So the increasing order of solubility is NaCl < CH₃CN < CH₃OH < cyclohexane.

Solution: PCl₅ is trigonal bipyramidal and sp³d hybridised, with axial-to-equatorial angles of $90^\circ$ and equatorial angles of $120^\circ$ — so (a), (c) and (d) are all correct. Statement (b) is the incorrect one: the two AXIAL bonds are longer than the three equatorial bonds, because the axial bonds suffer more repulsion. That is exactly why PCl₅ is so reactive.

Solution: The expansion is into a VACUUM, so the external pressure is zero: $W = -P_{ext}\Delta V = 0$ Since it is isothermal and the gas is ideal, $\Delta T = 0$ and $\Delta U = nC_v\Delta T = 0$. From the first law, $\Delta U = q + W \Rightarrow 0 = q + 0 \Rightarrow q = 0$. So both the work done and the heat are zero.

Solution: Numbering the ring to give the substituents the lowest possible locant set, fluorine takes position 1, the nitro group position 2 and the ethyl group position 4. Listing the substituents alphabetically gives 4-ethyl-1-fluoro-2-nitrobenzene.

Solution: The most stable oxidation state of the lanthanoids is +3. Ions in the +4 state are therefore easily reduced back to +3, which makes them good OXIDISING agents — Ce⁴⁺ and Tb⁴⁺. (By the same argument, the +2 ions Eu²⁺ and Yb²⁺ are reducing agents.)

Solution: Reaction (b) is written incorrectly. LiAlH₄ is a strong REDUCING agent: benzamide treated with (i) LiAlH₄ and (ii) H₂O is reduced to benzylamine, $\ce{C6H5CH2NH2}$ — not oxidised to benzoic acid as shown. The other three are all correct: acid chloride hydrolysis to the acid, and the oxidations of ethanol and propan-1-ol to their carboxylic acids.

Solution: In the lanthanoids the absorption bands are NARROW, because the transitions are f–f transitions within the deeply buried 4f level, which is well shielded from the ligand environment.

Solution: EDTA⁴⁻ binds a central metal ion through SIX donor atoms — the two nitrogen atoms of the ethylenediamine backbone and the four carboxylate oxygen atoms. Six donor sites makes it a hexadentate ligand.

Solution: Molarity $M = \dfrac{w_2 \times 1000}{M_2 \times V(\text{mL})}$ $\dfrac{1}{20} = \dfrac{w_2 \times 1000}{180 \times 250}$ $w_2 = \dfrac{180 \times 250}{20 \times 1000} = 2.25\ \text{g}$

Solution: Statement (a) is the incorrect one — it has the order backwards. The acidic strength of the hydrogen halides is HF < HCl < HBr < HI, because the H−X bond enthalpy DECREASES down the group, so the proton is released more easily. The other three statements are all correct.

Solution: By Le Chatelier's principle: The reaction is exothermic, so a LOW temperature favours the forward direction. The forward reaction reduces the number of gas moles (4 → 2), so a HIGH pressure favours it. Increasing the concentration of a reactant such as H₂ also drives it forward. A catalyst only speeds up the approach to equilibrium; it does not shift the position.

Solution: SOCl₂ converts methanol to methyl chloride (A). KCN in aqueous ethanol substitutes the chloride to give acetonitrile, $\ce{CH3CN}$ (B). Na(Hg)/EtOH reduces the nitrile to the primary amine, ethylamine $\ce{CH3CH2NH2}$ (C). (i) NaNO₂ + HCl then (ii) H₂O is the reaction of a primary ALIPHATIC amine with nitrous acid: the unstable diazonium salt loses nitrogen and gives the alcohol — ethanol, $\ce{CH3CH2OH}$ (D).

Solution: Ce ($Z = 58$) — a lanthanoid (I). Mn ($Z = 25$) — a d-block element (IV). As ($Z = 33$) — a p-block element (II). Cs ($Z = 55$) — an s-block element (III).

Solution: An ambidentate ligand has two DIFFERENT donor atoms, either of which can bind the metal. SCN⁻ can bind through S or N, NO₂⁻ through N or O, and CN⁻ through C or N — all ambidentate. The oxalate ion $\ce{C2O4^2-}$ binds only through oxygen. It has two donor sites but both are the same kind of atom, so it is bidentate, not ambidentate.

Solution: Energy is ordered by $(n + l)$; when two are equal, the one with the larger $n$ has the higher energy. I: $4d$, $n + l = 6$ II: $3d$, $n + l = 5$ III: $4p$, $n + l = 5$ IV: $3p$, $n + l = 4$ So I is highest and IV lowest. Between II and III (both 5), III has the larger $n$, so III > II. Order: I > III > II > IV.

Solution: Alcoholic KOH with heat eliminates HBr to give propene, $\ce{CH3CH=CH2}$ (A). HBr adds across the double bond following Markovnikov's rule, putting the bromine on the middle carbon: 2-bromopropane (B). Aqueous KOH substitutes that bromine by $\ce{-OH}$, giving propan-2-ol (C).

Solution: The carbylamine test is positive only for PRIMARY amines, which rules out N-methylaniline (secondary), triethylamine (tertiary) and pyridine. Aniline is a primary amine, so it gives the test. And its nitrogen lone pair forms a salt with the Lewis acid AlCl₃, deactivating the ring so that Friedel-Crafts alkylation fails.

Solution: An endothermic reaction ABSORBS heat from the surroundings. Heat absorbed at constant pressure is positive, so $q_p$ is positive (D), and since $\Delta_r H = q_p$ at constant pressure, $\Delta_r H$ is positive too (B).

Solution: Moles of H₂ $= \dfrac{1}{2} = 0.5$, so the number of molecules is $0.5\,N_A$. 14 g of N₂ $= \dfrac{14}{28} = 0.5$ mol $= 0.5\,N_A$ molecules — a match. (18 g of H₂O is 1 mol, 16 g of CO is $\frac{4}{7}$ mol and 28 g of N₂ is 1 mol.)