NEET Chemistry 2023 Important Questions and Answers - Smart Mock Test | NEET 2023 Chemistry Exam

NEET 2023 is approaching, and it is time to do thorough revisions and give your best for the entrance exam. This comprehensive session has provided the strategy to solve the most critical questions of physical, organic and inorganic chemistry for NEET 2023. The interesting part about this smart mock series is that you will get a chance to solve these types of conceptual questions.

Question 1: 21.75 g of MnO₂ on reaction with HCl forms 2.8L of Cl₂ (g) at STP; the percentage purity of MnO₂ is:

(Given: Atomic mass of Mn = 55 u)

MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

1. 80%
2. 75%
3. 33%
4. 50%

Answer: d) 50%

Solution: MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

Moles of MnO₂ = Moles of Cl₂

Given, Cl₂ = 2.8L

∴ Moles of Cl₂ = 2.8/22.4 moles

We know that 1 mole of MnO₂ yields 1 mole of Cl₂

∴ Moles of MnO₂ that yield 2.8/22.4 moles of Cl₂ = 2.8/22.4

∴ Amount of MnO₂ = (2.8/22.4) x 87g (Molar mass of MnO₂ = 87g)

∴ Percentage of MnO₂ = (2.8/22.4) x (87/21.75) x 100 = 50%

Question 2: The radii of the 2nd Bohr orbit of Be³⁺ ion is:

1. 26.45 pm
2. 52.9 pm
3. 79.35 pm
4. 105.8 pm

Answer: b) 52.9 pm

Solution: r_n = a° x (n₂/Z) = 52.9 pm x (2²/4) = 52.9 pm

Question 3: Antiaromatic species among the following is:

Solution: Antiaromatic is a cyclic planar structure that contains 4nπe⁻. All of the carbons are sp² hybridised, and the structure ‘c’ has 4πe⁻, indicating that it is antiaromatic.

Question 4: van der Waals constants (a) for the gases A, B, C and D are 1.25, 3.29, 4.28 and 0.244, respectively. The gas which is most easily liquefied is:

1. A
2. B
3. C
4. D

Answer: c) C

Solution: Van der Waals forces are weak electric forces attracting neutral molecules in gases, liquefied and solidified gases, and almost all organic liquids and solids. The tendency of such persistent dipoles to align with one another generates a net attractive force.

Because the intermolecular force is related to the van der Waals constant ′a′, the gas with the largest value of ′ a ′ is the easiest to liquefy. As a result, the value of constant ‘a’ in C is the highest.

Question 5: For the reaction, CCl₄ (g) + 2H₂O (g) → CO₂ (g) + 4HCl (g), at constant temperature, ∆H – ∆E is:

1. –RT
2. RT
3. –2RT
4. 2RT

Answer: d) 2RT

Solution: ∆H = ∆U + ∆n_gRT = ∆H = ∆E + ∆n_gRT

∴ ∆H – ∆E = ∆n_gRT

Now, ∆n_g = n_(g)P – n_(g)R = 5 – 3 = 2

∴ ∆H – ∆E = 2RT

Question 6: Four monobasic acids, A, B, C and D, have their respective ∆_neutH° values of –11.5, –7.5, –12.4 and –8.9 kcal/mol. Which of the following acids has the highest pK_a value?

1. A
2. B
3. C
4. D

Answer: b) B

Solution: ∆_neutH° ∝ Acidic strength

We know that, pK_a = –log K_a

B has the lowest magnitude, which means it has the highest pK_a value. Thus, B is the correct option.

Question 7: The correct order of ionic radii is represented in:

1. O > O^– > O^2–
2. Al⁺ > Al²⁺ > Al³⁺
3. S^2– > K⁺ > Cl^–
4. Mg²⁺ > Na⁺ > N^3–

Answer: b) Al⁺ > Al²⁺ > Al³⁺

Solution: The positive charge is increasing as we move across Al⁺ > Al²⁺ > Al³⁺ series, which means the effective nuclear charge is also increasing. We know that size decreases when effective nuclear charge increases. Hence, (b) is the correct option.

Question 8: Which of the following pairs of compounds are isostructural?

1. H₂O and SO₃
2. I₃^– and XeF₂
3. NH₃ and BF₃
4. SF₄ and XeF₄

Answer: b) I₃^– and XeF₂

Solution: The compounds I₃^– and XeF₂ are isostructural. The number of valence electrons in both is the same. Both have three lone pairs and two bond pairs of electrons; hence they have linear geometry. The geometry of electron pairs is trigonal bipyramidal, but molecule geometry is linear.

Question 9: The species which does not exist is:

1. Li₂
2. C₂
3. H₂
4. He₂

Answer: d) He₂

Solution: The Bond Order of He₂ is zero. Hence, it does not exist.

Question 10: Formic acid on reaction with concentrated H₂SO₄ at 373 K gives:

1. CO₂
2. HCHO
3. CH₃OH
4. CO

Answer: d) CO

Solution:

When formic acid is heated in the presence of concentric sulfuric acid, carbon monoxide gas and water vapour are produced. The concentrated sulphuric acid serves as a dehydrating agent in this reaction. Formic acid releases a molecule of water, leaving carbon monoxide gas behind. During the reaction, the oxidation number of carbon does not change. Because a water molecule is lost during the reaction of formic acid with concentric sulphuric acid, we can term it a dehydration reaction.

Question 11: The coordination complex which shows linkage isomerism is:

1. [Co(NH₃)₅NO₂]²⁺
2. [Co(NH₃)₆]³⁺
3. [Co(NH₃)₅Br]²⁺
4. [Cr(H₂O)₅Cl]²⁺

Answer: a) [Co(NH₃)₅NO₂]²⁺

Solution: [Co(NH₃)₅NO₂]²⁺ has NO_2, which is an ambidentate ligand. Linkage isomerism in NO₂ takes place due to the presence of an ambidentate ligand. In a given complex, the central metal atom is connected to the ambidentate ligand NO₂. They can be attached to the central atom in two different ways and exhibit linkage isomerism.

Question 12: The compound which is most reactive towards electrophilic substitution reaction is:

Solution: There is a lone pair present in the methoxy (OCH₃) group, which shows the +R effect. It will donate electrons, and hence it is a ring-activating group. This methoxy (OCH₃) group is more ring-activating compared to the CH₃ group. Thus, it is the most reactive towards electrophilic substitution reaction.

Question 13: Which of the following pairs of compounds are metamers of each other?

Solution: This pair of compounds represents metamer. They have the same molecular formula and the same functional groups, but the alkyl groups attached to the functional groups are different.

Question 14: Ethene on reaction with Baeyer’s reagent gives:

1. Ethane-1,2 diol
2. Ethanoic acid
3. Ethanal
4. Ethanol

Answer: a) Ethane-1,2 diol

Solution: Reaction of Baeyer’s reagent with ethene gives:

Related Links:

• NEET Chemistry MCQs – Important Questions and Answers
• NEET Chemistry 2023 – Important Topics and Preparation Tips
• NEET 2023 Chemistry Weightage
• NEET Chemistry Flashcards
• NEET Chemistry Syllabus 2023

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