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Chemistry — NEET Complete Yearwise Papers MCQs with Solutions

Free NEET Complete Yearwise Papers Chemistry MCQs with step-by-step solutions (50 questions). Part of NEET 2023 Paper. Practise online on Prepizo — no login needed.

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Questions with solutions

Q1 — Chemistry · medium · theory
Amongst the following, the total number of species NOT having eight electrons around central atom in its outer most shell, is NH₃, AlCl₃, BeCl₂, CCl₄, PCl₅
A. 4
B. 1
C. 3  ✓ Correct
D. 2
Solution: Counting the electrons around the central atom: NH₃ — 8 (octet complete) AlCl₃ — 6, electron deficient BeCl₂ — 4, electron deficient CCl₄ — 8 (octet complete) PCl₅ — 10, an expanded octet So three species (AlCl₃, BeCl₂ and PCl₅) do not have eight electrons.
Q2 — Chemistry · medium · theory
Some tranquilizers are listed below. Which one from the following belongs to barbiturates?
A. valium
B. Veronal  ✓ Correct
C. Chlordiazepoxide
D. Meprobamate
Solution: Veronal, amytal, nembutal, luminal and seconal are all derivatives of barbituric acid — the barbiturates, an important class of tranquilizers. Valium and chlordiazepoxide are benzodiazepines, and meprobamate is a carbamate.
Q3 — Chemistry · medium · theory
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: A reaction can have zero activation energy. Reason R: The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy. In the light of the above statements, choose the correct answer from the options given below:
A. A is true but R is false
B. A is false but R is true
C. Both A and R are true and R is the correct explanation of A.
D. Both A and R are true and R is NOT the correct explanation of A.  ✓ Correct
Solution: Both statements are true. A reaction CAN have zero activation energy, when the reactant molecules already possess the threshold energy. And the Reason correctly defines activation energy as the minimum extra energy that must be absorbed to reach the threshold. But the definition does not by itself explain why a zero-activation-energy reaction is possible, so R is not the correct explanation of A.
Q4 — Chemistry · medium · theory
The given compound is an example of:
A. allylic halide  ✓ Correct
B. vinylic halide
C. benzylic halide
D. aryl halide
Solution: In an allylic halide the halogen is attached to an $sp^3$ carbon which is next to a carbon-carbon double bond. Here X sits on the $sp^3$ carbon adjacent to the CH=CH double bond, so the compound is an allylic halide.
Q5 — Chemistry · medium · theory
The number of $\sigma$ bonds, $\pi$ bonds and lone pair of electrons in pyridine, respectively, are:
A. 11, 3, 1  ✓ Correct
B. 12, 2, 1
C. 11, 2, 0
D. 12, 3, 0
Solution: Pyridine, $\ce{C5H5N}$, is a six-membered aromatic ring with five C−H bonds. Ring bonds: 6 sigma; C−H bonds: 5 sigma — giving 11 $\sigma$ bonds in all. The aromatic ring has 3 $\pi$ bonds. The nitrogen carries 1 lone pair, which lies in the ring plane and is not part of the aromatic sextet.
Q6 — Chemistry · medium · numerical
The right option for the mass of CO₂ produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40) $\ce{CaCO3 ->[1200 K] CaO + CO2}$
A. 2.64 g
B. 1.32 g
C. 1.12 g
D. 1.76 g  ✓ Correct
Solution: 100 g of CaCO₃ gives 44 g of CO₂. For 20 g of pure CaCO₃: $\dfrac{20}{100} \times 44 = 8.8\ \text{g}$ of CO₂. But the limestone is only 20% pure, so: $\dfrac{20}{100} \times 8.8 = 1.76\ \text{g}$
Q7 — Chemistry · medium · theory
Which one is an example of heterogenous catalysis?
A. Decomposition of ozone in presence of nitrogen monoxide.
B. Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron.  ✓ Correct
C. Oxidation of sulphur dioxide into sulphur trioxide in the presence of oxides of nitrogen.
D. Hydrolysis of sugar catalyzed by H⁺ ions.
Solution: In heterogeneous catalysis the catalyst is in a DIFFERENT phase from the reactants. $\ce{N2(g) + 3H2(g) ->[Fe(s)] 2NH3(g)}$ — the gases react over a solid iron catalyst, so the phases differ. This is the Haber process. The other three all have catalyst and reactants in the same phase.
Q8 — Chemistry · medium · theory
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic. Reason R: The deep blue solution is due to the formation of amide. In the light of the above statements, choose the correct answer from the options given below.
A. A is true but R is false  ✓ Correct
B. A is false but R is true
C. Both A and R are true and R is the correct explanation of A.
D. Both A and R are true and R is NOT the correct explanation of A.
Solution: The Assertion is true — alkali metals dissolve in liquid ammonia to give deep blue, paramagnetic solutions. The Reason is false: the blue colour is due to AMMONIATED ELECTRONS, not to the formation of an amide. The ammoniated electron is also what makes the solution paramagnetic.
Q9 — Chemistry · medium · theory
Which one of the following statements is correct?
A. The bone in human body is an inert and unchanging substance.
B. Mg plays roles in neuromuscular function and interneuronal transmission.
C. The daily requirement of Mg and Ca in the human body is estimated to be 0.2 - 0.3 g.  ✓ Correct
D. All enzymes that utilise ATP in phosphate transfer require Ca as the cofactor.
Solution: The daily requirement of magnesium and calcium in the human body is about 200–300 mg, i.e. 0.2–0.3 g — statement (c) is correct. The others are wrong: bone is a dynamic, constantly remodelled tissue; it is CALCIUM that plays the role in neuromuscular function and interneuronal transmission; and enzymes using ATP in phosphate transfer require MAGNESIUM as the cofactor.
Q10 — Chemistry · medium · theory
Identify the product in the following reactions:
A.
B.
C.
D.  ✓ Correct
Solution: Step (i) $\ce{Cu2Br2}$/HBr is a Sandmeyer-type reaction: the diazonium group is replaced by bromine, giving bromobenzene. Step (ii) Mg in dry ether forms the Grignard reagent, phenylmagnesium bromide. Step (iii) water protonates the Grignard reagent, replacing $\ce{-MgBr}$ by $\ce{-H}$. The product is therefore plain benzene.
Q11 — Chemistry · medium · numerical
For a certain reaction, the rate $= k[A]^2[B]$. When the initial concentration of A is tripled keeping concentration of B constant, the initial rate would:
A. Increase by a factor of nine  ✓ Correct
B. increase by a factor of three
C. decrease by a factor of nine
D. increase by a factor of six
Solution: The rate is second order in A, so with $[B]$ fixed the rate goes as $[A]^2$. $\dfrac{r_f}{r_i} = \dfrac{(3[A])^2}{[A]^2} = 9$ So the rate increases by a factor of nine.
Q12 — Chemistry · medium · theory
Which of the following statements are NOT correct? A. hydrogen is used to reduce heavy metal oxides to metal. B. heavy water is used to study reaction mechanism. C. hydrogen is used to make saturated fats from oils D. The H−H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any element. E. Hydrogen reduces oxides of metals that are more active than iron. Choose the most appropriate answer from the options given below:
A. D, E only  ✓ Correct
B. A, B, C only
C. B, C, D only
D. B and D only
Solution: D is wrong: because of the small size of the hydrogen atoms the s–s overlap is very effective, so the H−H bond dissociation enthalpy is the HIGHEST for a single bond between two atoms of any element, not the lowest. E is wrong: hydrogen only reduces the oxides of metals that are LESS active than iron. Metals above iron in the reactivity series are not reduced by it. A, B and C are all correct.
Q13 — Chemistry · medium · theory
Homoleptic complex from the following complexes is:
A. Pentaamminecarbonatocobalt (III) chloride
B. Triamminetriaquachromium (III) chloride
C. Potassium trixalatoaluminate (III)  ✓ Correct
D. Diamminechloridonitrito - N - platinum (II)
Solution: A homoleptic complex is bound to only ONE kind of ligand. Potassium trioxalatoaluminate(III) is $\ce{K3[Al(C2O4)3]}$ — the aluminium is surrounded by three oxalate ligands and nothing else, so it is homoleptic. Each of the others carries two or more different ligands.
Q14 — Chemistry · medium · theory
In Lassaigne's extract of an organic compound both nitrogen and sulphur are present, which gives red colour with Fe³⁺ due to formation of
A. $\ce{[Fe(CN)5NOS]^4-}$
B. $\ce{[Fe(SCN)]^2+}$  ✓ Correct
C. $\ce{Fe4[Fe(CN)6]\cdot xH2O}$
D. $\ce{NaSCN}$
Solution: When both nitrogen and sulphur are present, sodium fusion gives sodium thiocyanate: $\ce{Na + C + N + S -> NaSCN}$ The thiocyanate then reacts with ferric ion to give the blood-red complex: $\ce{Fe^3+ + SCN^- -> [Fe(SCN)]^2+}$
Q15 — Chemistry · medium · theory
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: In equation $\Delta_r G = -nFE_{cell}$ value of $\Delta_r G$ depends on n. Reason R: $E_{cell}$ is an intensive property and $\Delta_r G$ is an extensive property. In the light of the above statements, choose the correct answer from the options given below:
A. A is true but R is false
B. A is false but R is true
C. Both A and R are true and R is the correct explanation of A.  ✓ Correct
D. Both A and R are true and R is NOT the correct explanation of A.
Solution: Both are true, and the Reason explains the Assertion. $E_{cell}$ is an intensive property — it does not depend on how much substance reacts. $\Delta_r G$ is extensive, so it scales with the amount, i.e. with $n$. That is exactly why $\Delta_r G = -nFE_{cell}$ carries the factor $n$ while $E_{cell}$ does not.
Q16 — Chemistry · medium · theory
Identify product (A) in the following reaction:
A.
B.
C.  ✓ Correct
D.
Solution: Zn–Hg with concentrated HCl is the Clemmensen reduction: it converts a keto group $\ce{C=O}$ all the way to a methylene group $\ce{CH2}$. Both acetyl groups are reduced, so each $\ce{-CO-CH3}$ becomes $\ce{-CH2-CH3}$ — giving the di-ethyl compound of option (c), with the two water molecules released as shown.
Q17 — Chemistry · medium · numerical
The relation between $n_m$ ($n_m$ = the number of permissible values of magnetic quantum number ($m$)) for a given value of azimuthal quantum number ($l$), is:
A. $n_m = 2l^2 + 1$
B. $n_m = l + 2$
C. $l = \dfrac{n_m - 1}{2}$  ✓ Correct
D. $l = 2n_m + 1$
Solution: For a given $l$, the magnetic quantum number runs from $-l$ to $+l$, giving $n_m = 2l + 1$ Rearranging: $2l = n_m - 1$, so $l = \dfrac{n_m - 1}{2}$.
Q18 — Chemistry · medium · theory
The stability of Cu²⁺ is more than Cu⁺ salts in aqueous solution due to:
A. hydration energy  ✓ Correct
B. second ionisation enthalpy
C. first ionisation enthalpy
D. enthalpy of atomization
Solution: Cu²⁺ is smaller and doubly charged, and hydration enthalpy goes as $\dfrac{\text{charge}}{\text{size}}$. So the much larger hydration energy of Cu²⁺ more than compensates for the second ionisation enthalpy, making Cu²⁺ the more stable ion in aqueous solution.
Q19 — Chemistry · medium · theory
Match LIST-I with LIST-II: LIST-I (A) Coke (B) Diamond (C) Fullerene (D) Graphite LIST-II (I) Carbon atoms are $sp^3$ hybridised. (II) Used as a dry lubricant (III) Used as a reducing Agent (IV) Cage like molecules Choose the correct answer from the options given below:
A. A - III, B - I, C - IV, D - II  ✓ Correct
B. A - III, B - IV, C - I, D - II
C. A - II, B - IV, C - I, D - III
D. A - IV, B - I, C - II, D - III
Solution: Coke — used as a reducing agent in metallurgy (III). Diamond — every carbon is $sp^3$ hybridised (I). Fullerene — cage-like molecules such as $\ce{C60}$ (IV). Graphite — used as a dry lubricant, because its layers slide over one another (II).
Q20 — Chemistry · medium · numerical
Weight (g) of two moles of the organic compound which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is:
A. 30
B. 18
C. 16
D. 32  ✓ Correct
Solution: This is decarboxylation (soda-lime reaction): $\ce{2CH3COONa + 2NaOH ->[CaO,\ heat] 2CH4 + 2Na2CO3}$ The organic product is methane, molar mass 16. Two moles weigh $2 \times 16 = 32\ \text{g}$.
Q21 — Chemistry · medium · theory
Complete the following reaction. [C] is:
A.
B.  ✓ Correct
C.
D.
Solution: HCN adds across the ketone carbonyl to give the cyanohydrin [B], with an $\ce{-OH}$ and a $\ce{-CN}$ on the same carbon. Concentrated $\ce{H2SO4}$ with heat then does two things: it dehydrates (losing $\ce{H2O}$ to form the ring double bond) and hydrolyses the nitrile to a carboxylic acid. The product [C] is therefore the unsaturated carboxylic acid of option (b).
Q22 — Chemistry · medium · theory
Given below are two statements: Statement I: A unit formed by the attachment of base to 1′ position of sugar is known as nucleoside. Statement II: When nucleoside is linked to phosphorous acid at 5′ position of sugar moiety we get nucleotide. In the light of the above statements, choose the correct answer from the options given below:
A. Statement I is true but Statement II is false  ✓ Correct
B. Statement I is false but Statement II is true
C. Both Statement I and Statement II are true
D. Both Statement I and Statement II are false
Solution: Statement I is true — a nitrogenous base joined to the 1′ position of the sugar is a nucleoside. Statement II is false in one word: the nucleoside is linked to PHOSPHORIC acid (not phosphorous acid) at the 5′ position to give a nucleotide.
Q23 — Chemistry · medium · theory
Which amongst the following options is correct graphical representation of Boyle's Law?
A.
B.
C.
D.  ✓ Correct
Solution: Boyle's law says $P \propto \dfrac{1}{V}$ at constant temperature, so $P = K\left(\dfrac{1}{V}\right)$. Comparing with $y = mx$, a plot of $P$ against $\dfrac{1}{V}$ is a STRAIGHT LINE through the origin, with the slope rising with temperature. That is graph (d).
Q24 — Chemistry · medium · numerical
A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy $\dfrac{1}{3}$ of tetrahedral voids. If the formula of the compound is $A_xB_y$, then the value of $x + y$ is:
A. 3
B. 2
C. 5  ✓ Correct
D. 4
Solution: In a ccp lattice there are $Z = 4$ atoms of B per unit cell, and twice as many tetrahedral voids, i.e. 8. Atoms of A fill one third of them: $\dfrac{1}{3} \times 8 = \dfrac{8}{3}$. $A : B = \dfrac{8}{3} : 4 = \dfrac{2}{3} : 1 = 2 : 3$ So the formula is $A_2B_3$ and $x + y = 5$.
Q25 — Chemistry · medium · theory
Intermolecular forces are forces of attraction and repulsion between interacting particles that will include A. dipole - dipole forces B. dipole - induced dipole forces C. hydrogen bonding D. covalent bonding E. dispersion forces. Choose the most appropriate answer from the options given below:
A. A, B, C, E are correct  ✓ Correct
B. A, C, D, E are correct
C. B, C, D, E are correct
D. A, B, C, D are correct
Solution: Covalent bonding is an INTRA-atomic (inter-atomic) force within a molecule, not an intermolecular force — so D is the odd one out. Dipole-dipole, dipole-induced dipole, hydrogen bonding and dispersion (London) forces are all genuine intermolecular forces.
Q26 — Chemistry · medium · theory
The correct order of energies of molecular orbitals of N₂ molecule, is
A. $\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < \sigma 2p_z < \sigma^*2p_z(\pi 2p_x = \pi 2p_y) < (\pi^*2p_x = \pi^*2p_y)$
B. $\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < 2p_z(\pi 2p_x = \pi 2p_y) < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z$
C. $\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z$  ✓ Correct
D. $\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z$
Solution: For the lighter second-period diatomics up to N₂, s–p mixing pushes $\sigma 2p_z$ ABOVE the degenerate $\pi 2p_x$ and $\pi 2p_y$ orbitals. So the order is $\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z$ (For O₂ and F₂ the $\sigma 2p_z$ drops below the $\pi$ orbitals.)
Q27 — Chemistry · medium · theory
Taking stability as the factor, which one of the following represents correct relationship?
A. $AlCl > AlCl_3$
B. $TlI > TlI_3$  ✓ Correct
C. $TlCl_3 > TlCl$
D. $InI_3 > InI$
Solution: Because of the inert pair effect, the lower oxidation state becomes more stable down group 13. Thallium is the heaviest of them, so its +1 state is far more stable than its +3 state — hence $TlI > TlI_3$. For the lighter members aluminium and indium, the +3 state is the stable one, so the other options are the wrong way round.
Q28 — Chemistry · medium · numerical
The conductivity of centimolar solution of KCl at $25^\circ C$ is $0.021\ \text{ohm}^{-1}\text{cm}^{-1}$ and the resistance of the cell containing the solution at $25^\circ C$ is 60 ohm. The value of cell constant is
A. $1.26\ \text{cm}^{-1}$  ✓ Correct
B. $3.34\ \text{cm}^{-1}$
C. $1.34\ \text{cm}^{-1}$
D. $3.28\ \text{cm}^{-1}$
Solution: Cell constant $=$ conductivity $\times$ resistance. $= 0.021 \times 60 = 1.26\ \text{cm}^{-1}$
Q29 — Chemistry · medium · theory
Which of the following reactions will NOT give primary amine as the product? (a) $\ce{CH3CONH2 ->[(i) LiAlH4][(ii) H3O+] Product}$ (b) $\ce{CH3CONH2 ->[Br2/KOH] Product}$ (c) $\ce{CH3CN ->[(i) LiAlH4][(ii) H3O+] Product}$ (d) $\ce{CH3NC ->[(i) LiAlH4][(ii) H3O+] Product}$
A. $\ce{CH3CONH2 ->[(i) LiAlH4][(ii) H3O+]}$
B. $\ce{CH3CONH2 ->[Br2/KOH]}$
C. $\ce{CH3CN ->[(i) LiAlH4][(ii) H3O+]}$
D. $\ce{CH3NC ->[(i) LiAlH4][(ii) H3O+]}$  ✓ Correct
Solution: Reduction of an ISOCYANIDE gives a SECONDARY amine — $\ce{CH3NC}$ reduces to $\ce{CH3NHCH3}$, dimethylamine. The other three all give primary amines: the amide reduces to ethylamine, the Hofmann bromamide degradation gives methylamine, and the nitrile reduces to ethylamine.
Q30 — Chemistry · medium · theory
The element expected to form largest ion to achieve the nearest noble gas configuration is:
A. N  ✓ Correct
B. Na
C. O
D. F
Solution: All four form ions isoelectronic with neon. For isoelectronic species the size falls as the nuclear charge rises: $\ce{N^3-} > \ce{O^2-} > \ce{F^-} > \ce{Na+}$ Nitrogen gains three electrons and has the lowest nuclear charge of the set, so $\ce{N^3-}$ is the largest ion.