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 โ MnCl2 + Cl2 + 2H2O
- 80%
- 75%
- 33%
- 50%
Answer: d) 50%
Solution: MnO2 + 4HCl โ MnCl2 + Cl2 + 2H2O
Moles of MnO2 = Moles of Cl2
Given, Cl2 = 2.8L
โด Moles of Cl2 = 2.8/22.4 moles
We know that 1 mole of MnO2 yields 1 mole of Cl2
โด Moles of MnO2 that yield 2.8/22.4 moles of Cl2 = 2.8/22.4
โด Amount of MnO2 = (2.8/22.4) x 87g (Molar mass of MnO2 = 87g)
โด 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:
- 26.45 pm
- 52.9 pm
- 79.35 pm
- 105.8 pm
Answer: b) 52.9 pm
Solution: rn = aยฐ 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 4nฯeโ. All of the carbons are sp2 hybridised, and the structure โcโ has 4ฯeโ, 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:
- A
- B
- C
- 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 โฒaโฒ, the gas with the largest value of โฒ a โฒ 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) โ CO2 (g) + 4HCl (g), at constant temperature, โH โ โE is:
- โRT
- RT
- โ2RT
- 2RT
Answer: d) 2RT
Solution: โH = โU + โngRT = โH = โE + โngRT
โด โH โ โE = โngRT
Now, โng = n(g)P โ n(g)R = 5 โ 3 = 2
โด โH โ โE = 2RT
Question 6: Four monobasic acids, A, B, C and D, have their respective โneutHยฐ values of โ11.5, โ7.5, โ12.4 and โ8.9 kcal/mol. Which of the following acids has the highest pKa value?
- A
- B
- C
- D
Answer: b) B
Solution: โneutHยฐ โ Acidic strength
We know that, pKa = โlog Ka
B has the lowest magnitude, which means it has the highest pKa value. Thus, B is the correct option.
Question 7: The correct order of ionic radii is represented in:
- O > Oโ > O2โ
- Al+ > Al2+ > Al3+
- S2โ > K+ > Clโ
- Mg2+ > Na+ > N3โ
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?
- H2O and SO3
- I3โ and XeF2
- NH3 and BF3
- SF4 and XeF4
Answer: b) I3โ and XeF2
Solution: The compounds I3โ 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:
- Li2
- C2
- H2
- 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:
- CO2
- HCHO
- CH3OH
- 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:
- [Co(NH3)5NO2]2+
- [Co(NH3)6]3+
- [Co(NH3)5Br]2+
- [Cr(H2O)5Cl]2+
Answer: a) [Co(NH3)5NO2]2+
Solution: [Co(NH3)5NO2]2+ has NO2, 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 Baeyerโs reagent gives:
- Ethane-1,2 diol
- Ethanoic acid
- Ethanal
- 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 โ Important Topics and Preparation Tips
- NEET 2023 Chemistry Weightage
- NEET Chemistry Flashcards
- NEET Chemistry Syllabus 2023
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