Top 500 Inorganic Chemistry Questions for JEE Main PDF

Classification of Elements and Periodicity in Properties Top 500 Question Bank for JEE Main Chemistry...

Classification of Elements and Periodicity in Properties Top 500 Question Bank for JEE Main Chemistry MathonGo Q1. The law of triad is applicable to a group of (4) N 3 > O 2 > F > Na + < Mg 2+ (1) Cl, Br, I Q9. Assertion (2) C, N, O PbCl2 is more stable than PbCl 4 (3) Na, K, Rb Reason (4) H, O, N PbCl4 is a powerful oxidising agent Q2. The atomic number of Unnilunium is _____. (1) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion Q3. The element with atomic number 25 will be found in group (2) Both Assertion and Reason are correct but Reason is not the Q4. From the given compounds, if X number of compounds are correct explanation for Assertion acidic in water, find the value of X. (3) Assertion is correct but Reason is incorrect CaO, SO2 , SO3 , Fe2 O3 , Cl2 O7 , CO2 , Na2 O (4) Assertion is incorrect but Reason is correct Q5. How many of the following have greater Z eff than Silicon Q10. The correct increasing order of electronegativity for atom: Sn, Sn 2+ and Sn 4+ species is: (i) Na (ii) Mg (iii) Al (iv) P (v) Cl (vi) S (vii) N (viii) O (ix) F (1) Sn < Sn 2+ < Sn 4+ Q6. Atomic radii of alkali metals follow the order: (2) Sn 4+ < Sn 2+ < Sn (1) Li > Na > K > Cs (3) Sn = Sn 2+ = Sn 4 (2) K > Cs > Li > Na (4) Sn < Sn 2+ = Sn 4+ (3) Na > K > Cs > Li Q11. Find the correct order of electron gain enthalpy (Δ eg H) of (4) Cs > K > Na > Li the given elements. Q7. The first (Δ H and the second (Δ H ionization enthalpies (1) O > B > C > N i 1) i 2) ( in kJmol −1 ) and the Δe electron gain enthalpy (in kJ mol (2) O > C > N > B gH −1 ) of a few elements are given below : (3) O > C > B > N Elements ΔH1 ΔH2 Δeg H (4) O > N > C > B I 52O 7300 −60 II 419 3051 −48 Q12. A → A − + e, E1 and A + → A 2+ + e, E2. The energy III 1681 3374 −328 required to pull out the two electrons are E and E respectively. 1 2 IV 1008 1846 −295 The correct relationship between two energy would be : V 2372 5251 +48 VI 738 1451 −40 (1) E 1 < E2 Which of the above likely to be the least reactive non-metal? (2) E 1 > E2 (3) E 1 = E2 Q8. In the increasing order of their sizes, arrange the following (4) E 1 ≠ E2 ions N 3− , Na + ,F − , Mg 2+ and O 2− (1) Mg 2+ > Na + > F > O 2− < N 3− Q13. The correct decreasing order of electropositive character (2) N 3− < F > O 2 > Na + > Mg 2+ among the following elements is (3) Mg 2+ < Na + < F < O 2− < N 3− (1) Fe > Sc > Rb > Br > Te > F > Ca If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Classification of Elements and Periodicity in Properties Top 500 Question Bank for JEE Main Chemistry MathonGo (2) Ca > Rb > Sc > Fe > Te > F > Br Q15. Which one of the following relations is correct with respect (3) Rb > Ca > Sc > Fe > Br > Te > F to first (I) and second (II) ionization potentials of sodium and (4) Rb > Ca > Sc > Fe > Te > Br > F Magnesium? (1) I Na > IMg Q14. Which of the following element shows maximum valency? (2) I Mg > IINa (1) Carbon (3) II Mg > IINa (2) Barium (4) II Na > IIMg (3) Nitrogen (4) Sulphur If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Classification of Elements and Periodicity in Properties Top 500 Question Bank for JEE Main Chemistry MathonGo Answer Key Q1 (1) Q2 (101) Q3 (7) Q4 (4) Q5 (6) Q6 (4) Q7 (4) Q8 (3) Q9 (1) Q10 (1) Q11 (3) Q12 (1) Q13 (4) Q14 (4) Q15 (4) If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Classification of Elements and Periodicity in Properties Top 500 Question Bank for JEE Main Chemistry MathonGo Q1. The term "effective" is used because the shielding effect of negatively charged electrons prevents According to Dobereiner, the atomic weight of the higher orbital electrons from experiencing the full middle element is nearly the same as the average nuclear charge of the nucleus due to the repelling atomic weights of the other two elements. effect of inner-layer electrons. Q2. It increases across the period as the orbit remains the According to IUPAC convention for naming of elements with same but the number of protons increases as we atomic number more than 100, different digits are written in order move from left to right in the periodic table. and at the end ium is added. For digits following naming is used. As we move down the group effective nuclear 0 − nil charge decreases or almost remains constant as an 1 − un increase in shell number is balanced by the increase 2 − bi in the number of protons in the group. 3 − tri Therefore, the order of Z eff is: and so on... F > O > N > Cl > S > P > Si > Al > Mg > Na So, 6 elements have higher Z eff than Silicon. Q3. For d block elements: Q6. Group number = number of electrons in (n -1)d subshell + number Atomic radii of alkali metals increases from Li to Cs of electrons in nS subshell. due to presence of an extra shell of electrons. Hence, order of atomic radii: The E.C. of element =[Ar] 3d 5 4s 2 ; Group = 7th. Cs > K > Na > Li. Q4. Generally, oxides and hydroxides of non-metals are Q7. The least reactive non-metal is element IV. The element IV acidic in nature as they can release hydrogen ions in has not so high ΔiH1 but it has high negative electron gain a solution. Also, oxides and hydroxides of metals are enthalpy (ΔegH). The given values for element IV match with I, basic in nature as they can release hydroxide ions in (iodine). a solution. Q8. Mg 2+ < Na + < F < O 2− < N 3− Acidic compounds are as given below: SO2 , SO3 , Cl2 O7 , CO2 Q9. X = 4 Q5. Pb 4+ is less stable than Pb 2+ due to the inert pair effect. So, Pb 4+ compounds are very good oxidizing The effective nuclear charge Z , is the net positive eff agents. charge experienced by an electron in an atom. Pb 4+ converts itself into Pb 2+ to attain stability, the process is: If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Classification of Elements and Periodicity in Properties Top 500 Question Bank for JEE Main Chemistry MathonGo Pb 4+ + 2e − → Pb 2+ Metallic character(electropositive) is inversely proportional to Pb 4+ is a very big cation and it's association with 4 ionization energy. anions like I is not possible. Therefore, PbI does − 4 Alkali and alkaline earth metals are most electropositive. Alkali not exist and all other molecules exist. metals are more electropositive than alkaline earth metals. In d-block elements , the elements near the alkaline earth metals are more electropositive than rest of the members. Metalloids are Q10. less electropositive than metals. Halogens (nonmetals) are least Higher the oxidation state, greater is the electronegativity. electropositive. In halogen group electropositive character So, order will be Sn < Sn 2+ < Sn 4+ increases down the group. Q11. Electron gain enthalpy down the group decreases and across Q14. the period it increases. Nitrogen has the positive value of electron Second period elements can not form more than four bonds due to gain enthalpy. unavailability of vacant d-orbitals. Q12. But third period onwards elements can form more than four bonds The second ionization energy is always higher than the first due to availability of vacant d-orbitals. ionization energy as an electron is to be removed from unipositive N and C belong to the second period so they can not expand octet. cation. There is a strong electrostatic attractive force between the Ba is metal and +2 oxidation state is more stable. unipositive cation and the electron. This makes the removal of Sulphur belongs to the third period of Periodic Table hence, it has electron difficult and requires a large amount of energy. maximum valency and maximum valency of S is 6. Always IP2 > IP1 Q15. After losing one electron, Na ion gets noble gas + So E2 > E1 configuration. Q13. If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Chemical Bonding and Molecular Structure Top 500 Question Bank for JEE Main Chemistry MathonGo Q1. What are the formal charges on the carbon and oxygen (3) SbH 3 > AsH3 > NH3 > PH3 atoms and the formal charge difference, ΔFC, in the Lewis (4) NH 3 > PH3 > AsH3 > SbH3 (electron dot) structure of carbon dioxide shown in the figure? Q7. Which of the following is the correct prediction about 0̈ = c = 0̈ observed B − F bond length in BF molecule? 3 (1) +4, −2 and +6 (1) B − F bond length in BF is found to be less than theoretical 3 (2) +4, +4 and 0 value because the electronegativity values of B(2.04) and 4.0) (3) +4, +2 and +2 suggest the bond length to be ionic and hence, the attraction (4) 0, 0 and 0 between oppositely charged ions must decrease the nd length. (2) BF and [BF have equal B − F bond length. − 3 4] Q2. The sum of the number of lone pairs of electrons on each (3) The decrease in the B − F bond length in BF is due to 3 central atom in the following species is delocalised pπ − pπ bonding between vacant 2p orbital of B and [XeF [Atomic number : 2− + − [T eBr6 ] , [BrF2 ] , SN F3 3] filled 2p orbital of F N = 7, F = 9, S = 16, Br = 35, T e = 52, Xe = 54] (4) The correct B − X bond length order is Q3. How many of the following compounds have (pπ − dπ) B − F > B − Cl > B − Br > B − I multiple bonds? Q8. The diamagnetic species is i. SO 2 (1) NO ii. SO 3 (2) NO 2 iii. HSO ⊖ 4 (3) O 2 iv. SO 4 2− (4) CO 2 v. SO 2− 3 Q9. The difference between bond orders of CO and NO is ⊕ x vi. HSO ⊖ 3 2 where x = Q4. Find the number of lone pairs of electrons present in a OF 2 (Round off to the Nearest Integer) molecule. Q10. Out of the given ten pairs of combination total pair(s), Q5. If a molecule MX has zero dipole moment, the sigma 3 which results in zero overlapping is: (Assuming Z is overlapping bonding orbitals used by M (atomic number < 21 ) are axis). (p y + s) , (dxy + s) , (dxy + p ) y (1) Pure p (pz + dxy ) , (pz + s) , (py + Pz ) (s + s), (s + py ) , (py + py ) (2) sp hybridised (3) sp hybridised 2 Q11. Arrange the following in increasing order of melting point: (4) sp hybridised 3 Li2 O, LiF, Li3 N (1) Li 3 N < Li2 O < LiF Q6. Bond angles of NH 3, PH3 , AsH3 and SbH are in the order 3 (2) Li 2O < Li3 N < LiF (1) PH 3 > AsH3 > SbH3 > NH3 (3) LiF < Li 2 O < Li3 N (2) SbH 3 > AsH3 > PH3 > NH3 (4) Li 3 N < LiF < Li2 O If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Chemical Bonding and Molecular Structure Top 500 Question Bank for JEE Main Chemistry MathonGo Q12. Discuss and compare the dipole moments as well as bond angles in NCl and PCl. 3 3 (1) Bond angle NCl 3 > Bond angle PCl 3 (2) Bond angle NCl 3 < Bond angle PCl 3 (3) Bond angle NCl 3 = Bond angle PCl 3 (4) None of these Q13. Which one among the following does not have the (4) hydrogen bond? (1) Phenol Q15. The hydrogen bond is strongest in (2) Water (1) O − H … … F (3) Liquid NH 3 (2) O − H … … H (4) Liquid HCl (3) F − H … … F (4) O − H … …. O Q14. Overlapping leads to the formation of σ bonding molecular orbital in: Q16. Identify the correct order of boiling points of the following compounds: (a) CH 3 CH2 CH2 CH2 OH (b) CH 3 CH2 CH2 CHO (c) CH 3 CH2 CH2 COOH (1) a > b > c (2) c > a > b (1) (3) a > c > b (4) c > b > a Q17. Boiling point of H 2O is higher than that of H 2 S , because the former (2) (1) is capable of forming H - bonds (2) has higher molecular mass (3) has relatively strong covalent bonds (4) is capable of forming co-ordinate bonds with H ions + Q18. The experimental value of the dipole moment of HCl is 1.03D. The length of the H − Cl bond is 1.275Å. The (3) percentage of ionic character in HCl is: [Given: 1D = 10 −18 esu cm] (1) 43% If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Chemical Bonding and Molecular Structure Top 500 Question Bank for JEE Main Chemistry MathonGo (2) 21% (4) 7% (3) 17% Q19. The total mole of valence e in 2.6 g of N − − 3 If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Chemical Bonding and Molecular Structure Top 500 Question Bank for JEE Main Chemistry MathonGo Answer Key Q1 (4) Q2 (6) Q3 (4) Q4 (8) Q5 (3) Q6 (4) Q7 (3) Q8 (4) Q9 (0) Q10 (4) Q11 (1) Q12 (1) Q13 (4) Q14 (2) Q15 (3) Q16 (2) Q17 (1) Q18 (3) Q19 (1) If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Chemical Bonding and Molecular Structure Top 500 Question Bank for JEE Main Chemistry MathonGo Q1. Q3. dπ-pπ bonding is the formation of a π molecular orbital by the The calculation of formal charge (FC) can be overlap of a orbital on one atom with a p or p* orbital on another considered as a process involving electronic book- atom. keeping, it is a hypothetical charge worked out as i. SO2 ⇒ one (pπ − pπ) and one (pπ − dπ) bond. follows: ii. SO3 ⇒ one (pπ − pπ) and two (pπ − dπ) bond. FC=(number of valence electrons)− (number of 1 and iv. both have 2(pπ − dπ) bond. 2 ⊝ ⊝ iii. HSO4 SO4 bonding electrons)−(number of non-bonding v. SO3 2− and HSO 3 2− both have one (pπ − pπ) bond. electrons) Thus, i to iv have (pπ − dπ) multiple bonds. ∴ FC (C)= (4) − 1 2 (8) − 0 = 0 FC (O)= (6) − 1 (4) − 4 = 0 Q4. 2 Hence, ΔFC= 0 − 0 = 0. Q2. Number of σ- Number O atom has = (2 lone pairs) bonds of lone F atom has = (3 lone pairs each) formed pairs on Total lone pairs = 2 + 3 + 3 = 8. by central central metal Q5. metal atom/ The atomic number of the element M is given as less than 21. atom/ ion The electronic configuration for an atom with an atomic number ion of 21 will be [Ar]3s 2 3p 1. This implies that the element M with (i) less than 21 electrons does not have d-orbital electrons. In 6 1 2− [T eBr6 ] (ii) If the molecule MX is non-polar in nature, and contains no d- 3 In 2 2 orbitals, then it must have three sigma (single) bonds. The + [BrF2 ] hybridisation for three sigma bonds is sp. Also, the fact that the 2 (iii) molecule has zero dipole moment confirms that the molecule has In 4 0 SN F3 a symmetrical geometry of triagonal planar arrangement, which (iv) is possible only for an sp hybridisation around M. 2 In 3 3 [XeF3 ] − Q6. ⇒ Total number of lone pairs of electrons = 1 + 2 + 0 + 3 According to Drago's rule, elements belonging to the third = 6 period or below do not show hybridisation. This is accompanied If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Chemical Bonding and Molecular Structure Top 500 Question Bank for JEE Main Chemistry MathonGo by a decrease in bond angle to nearly 90°. Difference = 0 = x 2 However, in the case of ammonia, the bond angle is x = 0 approximately 107°. Q10. NH3 > PH3 > AsH3 > SbH3 Q7. It has partial double character in B − F bond due to pπ − pπ back bonding. Q8. NO & O2 are paramagnetic, which can be understood by the molecular orbital theory, as they have electronic configuration of ∗ ∗ ∗ ∗ NO = σ1s2 σ 2 σ2s2 σ 2 σ2p2 π2p2 = π2p2 π 1 = π 0 1s 2s 2p 2p ∗ ∗ ∗ ∗ O2 = σ1s2 σ 2 σ2s2 σ 2 σ2p2 π2p2 = π2p2 π 1 = π 1 1s 2s 2p 2p As we can see, both have unpaired electrons. So, it will show paramagnetic in nature. Lewis structure of NO : 2 No overlapping is possible in the opposite symmetry orbitals (along Z axis) i.e., S + Py , Py + Pz , dxy + S, dxy + py. So, we can see one unpaired electron over nitrogen, which makes it paramagnetic in nature. Q11. Structure of CO : 2 As we know, higher the ionic character in compound higher the O = C = O melting point because of higher intermolecular attraction. So, no unpaired electrons, and it is diamagnetic in Fajan's rule tells about the covalent character in ionic bond. nature. Covalent character depends upon charge on cation and anion and size of cation and anion. Here in given molecules, cations are same, so size and charge Q9. Bond order of CO = 3 same on cation of each compound. Bond order of NO + = 3 Size of all anions (N −3 , O −2 , F − ) are same because all are isoelectronic species but charge on all anion are different. If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Chemical Bonding and Molecular Structure Top 500 Question Bank for JEE Main Chemistry MathonGo According to Fajan's rule, higher the charge on anion, higher the Q14. polarization power of cation and higher the covalent character; 1. S + py orbital gives non-bonding molecular orbitals. so, lower the ionic character. 2. S + pz orbital give bonding molecular orbitals. Hence, higher the charge on anion, higher the ionic character and 3. Lateral overlaps give Pi molecular orbitals. higher the melting point. Order of melting point Li 3N < Li2 O < LiF. Q15. The strength of hydrogen bond depends upon the Q12. Dipole moment: The magnitude of P − Cl bond moment is coulombic interaction between the much larger than that of N − Cl ( greater electronegativity electronegativity of the attached atom and difference in the former case) bond, therefore, PCl is more 3 hydrogen. Fluorine is the most electronegative polar than NCl 3 element. Thus, hydrogen bond will be strongest in F − H....... F. Q16. In carboxylic acids, molecules are more strongly associated due to strong H− bonding followed by alcohols due to H−bonding (which is weaker than in case of carboxylic acid) and in aldehydes there are dipole-dipole interactions. Hence, the correct order of boiling points is c > a > b. Q17. In H 2 OH − bond is present Bond angle: Cl − N Cl bond angle is greater than Cl − P Cl bond angle. This is due to smaller bond pair - bond pair Q18. repulsion in PCl. 3 % Ionic character = Experimental value of dipole moment × 100 Theoretical value of dipole moment μexp = × 100 Q13. μ theo Where: A hydrogen bond is formed between H and a more μexp = 1.03 D and μ =|q|×d electronegative atom like F,O or N. It is a form of dipole-dipole theo −8 d = 1.275 Å = 1.275 ×10 cm attraction between molecules. −10 |q|= 4.8 ×10 esu In liquid HCl, a hydrogen bond cannot be formed with Cl atom −10 −8 μ = 4.8 ×10 × 1.275 ×10 esu cm due to its greater size and lesser electronegativity as compared to theo −18 = 4.8 ×1.275 ×10 esu cm F,O or N. = 4.8 × 1.275D If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Chemical Bonding and Molecular Structure Top 500 Question Bank for JEE Main Chemistry MathonGo 1D = 10 −18 esu cm Therefore, the number of valence electrons in three % Ionic character = 1.03 4.8×1.275 × 100 = 17% (approx.) atoms of N = 15 Since the valency on the azide ion is −1, it means Q19. it has one extra electron on it. Given weight of N ion = 2. 6 g − 3 So, the total number of valence electrons in azide Number of moles present in 2. 6 g of azide ion. ion is 15 + 1 = 16. Given Weight 2.6 g 2.6 g = Molar Weight = (3×14 g) = 42 g = 0. 061 moles Mole of valence electrons in given mass of azide ion is= 0. 061 × 16 (valence e of N ) − − Number of valence electrons in one atom of N = 5 3 = 0. 99 ≈ 1 If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Hydrogen Top 500 Question Bank for JEE Main Chemistry MathonGo Q1. Very pure hydrogen gas is prepared by the action of pure (3) 3 dilute H 2 SO4 on (4) 4 (1) water Q5. Which of the following atomic and physical properties of (2) sodium hydride hydrogen is false? (3) magnesium ribbon (1) Hydrogen > Deuterium > Tritium; (melting point /K ) (4) aluminium (2) Hydrogen < Deuterium < Tritium; (boiling point /K ) Q2. In context with the industrial preparation of hydrogen (3) Hydrogen < Deuterium < Tritium; (density/ gL −1 ) from water gas (CO + H ), which of the following is the 2 (4) Hydrogen > Deuterium > Tritium; (\% relative correct statement? abundance) (1) CO is oxidised to CO with steam, in the presence of a 2 Q6. Dalda is prepared from oils by catalyst followed by the dissolution of CO in water under 2 (1) Oxidation ssure. (2) Reduction (2) CO and H are fractionally separated using the differences 2 (3) Hydrolysis in their densities. (4) Distillation (3) CO is removed by absorption in aqueous Cu 2 Cl2 solution. (4) H is removed through occlusion with Pd. 2 Q7. Determine the values of A, B, C and D shown in the figure. Q3. Hydrogen can be prepared by the action of dil. H 2 SO4 on :- (1) Cu (2) Fe (3) Pb (4) Hg Q4. Which of the following statements is not true for hydrogen: (1) SnO , 2 PbO3 , 2 ZnO3 , AlO 1) It exists as a diatomic molecule. 3 2 (2) 2SnO 2 2 PbO2 2ZnO2 AlO2 2) It has one electron in the outermost shell. (3) SnO3 3) It can lose an electron to form a cation which can freely 13 (4) 3SnO 3 2 PbO3 2ZnOAlO2 exist 4) It cannot form ionic compounds. Q8. Number of electron-rich hydrides among the following (1) 1 are: CH 4 , NH3 , PH3 , H2 O, H2 S, BH3 , HF, AIH3 , AsH3. (2) 2 If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Hydrogen Top 500 Question Bank for JEE Main Chemistry MathonGo Q9. When electric current is passed through an ionic hydride Q14. Statements about heavy water are given below. in the molten state, A. Heavy water is used in exchange reactions for the study of (1) Hydrogen is liberated at the cathode reaction mechanisms. (2) Hydrogen is liberated at the anode B. Heavy water is prepared by exhaustive electrolysis of (3) Hydride ion migrates towards cathode water. (4) No reaction takes place C. Heavy water has higher boiling point than ordinary water. D. Viscosity of H 2O is greater than D 2O. Q10. A commercial sample of hydrogen peroxide marked as Choose the most appropriate answer from the options given 100 volume H 2 O2 , it means that- below: (1) 1ml of H 2 O2 will give 100mlO at STP2 (1) A and C only (2) 1 L of H 2 O2 will give 100mlO at STP2 (2) A, B and C only (3) 1 L of H 2 O2 will give 22.4 L at STP (3) A and B only (4) 1ml of H 2 O2 will give 1 mole of O at STP 2 (4) A and D only Q11. A sample of Hydrogen peroxide solution labelled as 28 Q15. Permanent hardness can be removed by adding volume has density of 265 g/L. Mark the correct options(S) (1) Cl 2 representing concentration of same solution in other units. (2) Na 2 CO3 (1) Molarity of H 2 O2 = 2.8 (3) Ca(OCl)Cl (2) Percentage w/v = 8.5 (4) K 2 CO3 (3) Mole fraction of H 2 O2 = 0.5 (4) Molality of H 2 O2 = 12.53 Q16. Which of the following ions will cause hardness of sample water? Q12. Consider the following equations: 2+ (1) Na + 2Fe + H2 O2 → xA + yB (2) K + (in basic medium) − + ′ ′ ′ (3) Ca 2+ 2MnO + 6H + 5H2 O2 → x C + y D + z E 4 (4) Li + (in acidic medium) The sum of the stoichiometric coefficients x, y, x , y and z ′ ′ ′ Q17. Which of the following statements about heavy water for products A, B, C, D and E respectively, is are correct? (a) Heavy water is used as a moderator in nuclear reactors. Q13. The reason of bleaching properties of H 2 O2 is (b) Heavy water is more associated than ordinary water. (1) Unstable nature (c) Heavy water is a more effective solvent than ordinary (2) Acidic nature water. (3) Reducing nature (1) (b) and (c) (4) Oxidising nature If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Hydrogen Top 500 Question Bank for JEE Main Chemistry MathonGo (2) (a) and (b) (4) (a) and (c) (3) (a), (b) and (c) If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Hydrogen Top 500 Question Bank for JEE Main Chemistry MathonGo Answer Key Q1 (3) Q2 (1) Q3 (2) Q4 (3) Q5 (1) Q6 (2) Q7 (2) Q8 (6) Q9 (2) Q10 (1) Q11 (2) Q12 (19) Q13 (4) Q14 (2) Q15 (2) Q16 (3) Q17 (2) If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Hydrogen Top 500 Question Bank for JEE Main Chemistry MathonGo Q1. Q6. Very pure hydrogen gas is prepared by the Dalda is an ester of saturated fatty acids, action of pure dilute H 2 SO4 with magnesium. whereas oils are esters of unsaturated fatty acids. Hence oils (liquid glycerides) react with Q2. hydrogen in the presence of a metal catalyst CO is oxidised to CO with steam in the 2 (like nickel) it gives saturated glycerides (semi- presence of a catalyst, followed by the solid glycerides) i.e., fats. Thus, vegetable ghee dissolution of CO in water under pressure. 2 (dalda) is obtained by the hydrogenation catlayst CO + H2 + H2 O −− −−→ CO2 + 2H2 (reduction) of oils. water gas steam Oils + H → Dalda And, 2 CO & H2 are not fractionally separated by Q7. using the differences in their densities. Carbon All the given metals are amphoteric in nature monoxide is separated from hydrogen gas by and thus, can react with aqueous alkali to give passing it through water at a pressure of 30 hydrogen gas and double salts. The reactions atmospheres due to which carbon monoxide is are: absorbed in water unlike hydrogen gas Sn + 2NaOH → H2 + Na2 SnO2 CO is not removed by absorption in aqueous Pb + 2NaOH → H2 + Na2 PbO2 Cu2 Cl2 solution. Zn + 2NaOH → H2 + Na2 ZnO2 H2 is not removed through occlusion with Pd Al + 2NaOH + 2H2 O → 3H2 + NaAlO2 Q3. Fe + H 2 SO4 ⟶ FeSO4 + H2 Q8. Q4. When an atom of hydrogen loses an electron, it becomes a proton that has an extremely small size (less than 10 −15 m Hydrides of group 15, 16 and 17 ), and therefore has difficulty in surviving. Generally, the ( i. e. , NH3 , PH3 , H2 O, H2 S, HF, AsH3 ) proton (H ) is captured by species having a lone pair of + have more electrons than required to form electrons, like with the case of hydronium ions in water. normal covalent bonds and hence are called electron-rich hydrides. These extra electrons are Q5. Property T D H present as lone pair electron on the central Melting point (K) 20.62 > 18.5 > 13.9 metal. These lone pair electron makes atom Boiling point (K) 24.9 > 23.5 > 20.4 electron rich atom. Density (g/L) 0.27 > 0.18 > 0.09 Relative abundance (%) 10 −15 < 0.016 < 99.984 If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Hydrogen Top 500 Question Bank for JEE Main Chemistry MathonGo Q9. M + H − → M + + H − electrolysis of water. hydride ion B. P. of D 2 O = 374. 4 K H − → 1 2 H2 + e − (at anode) B. P. of H 2O = 373 K Viscosity of H 2O = 0. 89 centipoise Q10. Viscosity of D 2 O = 1. 107 centipoise If a commercial sample of hydrogen peroxide Q15. Permanent hardness is removed by precipitating marked as 100 volume H 2 O2 that means at STP, carbonates of Ca 2+ and Mg 2+. 100 ml of oxygen is liberated by 1 ml of this CaCl2 + Na2 CO3 → CaCO3 ↓ +2NaCl solution on heating. Q16. Q11. Water containing soluble bicarbonate, chloride, Given, sulphate, etc. salts of calcium and magnesium is Volume strength is 28. known as hard water. This type of water does volume strength Then, Molarity is = 28 = 2. 5 11.2 11.2 not give lather with soap. Percentage Water which does not contain such salts is volume strength×68 28×68 w /v is = = 8. 5. 22.4 224 known as soft water. This water forms lather Q12. with soap easily. 2+ 3+ − (i) 2Fe + H2 O2 → 2Fe + 2OH Q17. Heavy water is deuterium oxide (D 2 O) that contains − + 2+ (ii) 2 MnO + 5H2 O2 + 6H ⟶ 2 Mn + 5O2 + 8H2 O deuterium​, (an isotope) heavier than the hydrogen. 4 So sum of (x + y + x 1 + y 1 1 + z )= 2 + 2 + 2 + 5 + 8 = 19 The difference increases the strength of water’s hydrogen- oxygen bonds. Heavy water is also a commonly used Q13. Oxidising nature moderator (medium that reduces the speed of fast neutrons). Hence, (a) and (b) statements are correct. Q14. Heavy water is used in exchange reactions for study of reaction mechanisms Heavy water is prepared by exhaustive If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app s Block Elements Top 500 Question Bank for JEE Main Chemistry MathonGo Q1. Alakali metals (IA) are generally soluble with liquid ammonia Q8. ′ X ′ is an alkaline earth metal. It imparts brick red colour to the (Liq. NH ) and provides blue colour solution and 3 Bunsen flame. Its oxide is basic in nature. The atomic shows paramagnetic behavior, so what could be the reason to number of ′ X ′ is provide blue colour and paramagnetic behavior? − Q9. On heating a mixture of one mole each Li 2 CO3 , MgCO 3 and + (1) [M(NH 3 )x ] and [e(NH 3 )y ] Na2 CO3 , the volume of CO evolved under STP condition 2 − (2) [e(NH 3 )y ] will be − + (3) [e(NH 3) y ] and [M(NH 3) x ] (1) 67.2 L + (4) [M(NH 3) x ] (2) 33.6 L (3) 44.8 L Q2. Which of the following elements will form alkaline oxide? (4) 22.4 L (1) K (2) P Q10. A compound X has the following characteristics: (3) S a. It is used as a disinfectant. (4) Cl b. It is used to prepare bleaching powder. c. Its suspension in water is called milk of lime. Q3. The combustion of sodium in excess air yields a higher oxide. Then compound X is What is the oxidation state of the oxygen in the product? (1) Ca(OH) 2 Q4. A solid is a compound of group 1 element and it gives a bright (2) CaO red colour in the flame test. The solid is (3) CaCO 3 (1) LiBr (4) CaSO 4 2H2 O (2) CsCl ∘ Δ,120 C (3) KCl Q11. CaSO4 ⋅ 2H2 O ⟶ A In the above reaction, A and B, ∘ Δ,200 C (4) NaCl CaSO4 ⋅ 2H2 O ⟶ B respectively, are: Q5. Find the sum of bond order between same bonded atoms in Q (1) CaSO 1 4, CaSO4 ⋅ H2 O 2 and R compounds. (2) CaSO ⋅ 1 H2 O, CaSO4 4 2 (3) CaSO 4 ⋅ 1 2 H2 O , Gypsum. (4) Gypsum, CaSO. 4 Q12. Which of the following statements is true about Ca(OH) ? 2 Q6. 3. 88 g of a mixture of Na 2 CO3 and CaCO is heated to a 3 (1) It is used in the preparation of bleaching powder. constant mass of 3.00 g. What was the mass of Na 2 CO3 (in (2) It is a light blue solid. grams) in the mixture? (3) It does not possess disinfectant property. (4) It is used in the manufacture of cement. Q7. In the following reaction, find the number of products formed which contain carbon atom(s): Q13. The metal showing a resemblance with lithium is: Δ Sodium bicarbonate ⟶ Products (1) Be (2) Mg If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app s Block Elements Top 500 Question Bank for JEE Main Chemistry MathonGo (3) Al Q15. Which of the following set contains pair of elements that do (4) Na not belong to same group but show chemical resemblance? (1) Hf , Zr Q14. The moles of hydrocarbon produced by hydrolysis of 1 mole (2) K, Rb Mg C3 2 (magnesium carbide) is (3) Be, Al (4) B, Al If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app s Block Elements Top 500 Question Bank for JEE Main Chemistry MathonGo Answer Key Q1 (2) Q2 (1) Q3 (-1) Q4 (1) Q5 (3.00) Q6 (1.88) Q7 (2) Q8 (20) Q9 (3) Q10 (1) Q11 (2) Q12 (1) Q13 (2) Q14 (1) Q15 (3) If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app s Block Elements Top 500 Question Bank for JEE Main Chemistry MathonGo Q1. The colour is due to the presence of solvated electrons in ∴ (3. 88 − 3. 00 = 0. 88g) of CO is lost during heating. 2 ammonia. It also imparts paramagnetic character to the solution. 0. 88 g CO2 = 0.88 44 = 0. 02 mol CO2 From the reaction stoichiometry, Q2. 0. 02 mol CO2 ≡ 0. 02 mol CaCO3 Alkaline metal have ns Electronic configuration. 1 0. 02 mol CaCO3 = 0. 02 × 100 = 2. 00g CaCO3 Alkali metal are placed in group 1 Of the periodic ∴ Mass of Na 2 CO3 in the mixture = 3. 88 − 2. 00 g = 1. 88 g table. Δ Q7. 2NaHCO 3 ⟶ Na2 CO3 + H2 O + CO2 Among the given elements, potassium is the only Products containing C-atom: Na 2 CO3 and CO 2 element which is placed in group 1 of the periodic table and has 4s Electronic configuration. So, it is an 1 Q8. Element 'X' is calcium. Calcium imparts brick red colour to the alkali, and hence, it forms an alkaline oxide. Bunsen flame. Atomic number of calcium is 20. Phosphorous is placed in group 15, sulphur is placed in Q9. group 16, and chlorine is placed in group 17 of the periodic table, and they are all non-metals. Hence, they Li2 CO3 ​and MgCO ​are thermally unstable and they 3 do not form alkaline oxides. decompose on heating. On the other hand, sodium carbonate is thermally stable, which does not Q3. decompose on heating. 1 Na + O2 → Na2 O, Na2 O + O2 → Na2 O2 2 Li2 CO3 → Li2 O + CO2 (1 mole) It exists as 2 Na and O + 2− 2 MgCO → MgO + CO2 (1 mole) 3 So oxidation state of O is −1 Na2 CO3 does not decompose. Q4. Lithium and strontium salts impart bright red colour to the So, the total of 2 moles of CO gas is evolved by 2 flame and it says that it belongs to alkali metals which means it is heating the given mixture. salt of lithium. The volume of 1 mole of CO gas at STP = 22. 4 L 2 The volume of 2 moles of CO gas 2 = 2 × 22. 4 L = 44. 8 L Q10. Q5. Bleaching powder is prepared by passing chlorine gas Bond order of [O − O] in H 2 O2 = 1.0 (Cl2 ) through a solution of calcium hydroxide Bond order of [O − O] in O 2 = 2.0 Ca(OH)2. So sum of bond order between same bonded atoms in Q and R The reaction that occurs is, compounds = 1 + 2.0 = 3.0. Ca(OH) +Cl2 (g)→CaOCl +H2 O. 2 2 Q6. When an excess of calcium oxide is added to water, Na2 CO3 does not decompose on heating. Ca(OH) 2 forms, CaCO3 decomposes to form CaO and CO. 2 CaO+H2 O→Ca(OH) 2 Δ CaCO3 − −→ CaO + CO2 ∴ Loss in mass is due to loss of CO gas 2 If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app s Block Elements Top 500 Question Bank for JEE Main Chemistry MathonGo It partially dissolves in water to produce a solution 2. Their hardness is higher than the other elements called lime water, which is a moderate base. in their respective groups. 3. They react with nitrogen to form their respective Ca(OH) 2 is highly basic and antibacterial due to nitrides. which it is used as a disinfectant. 4. On heating their carbonates, they yield their Q11. respective oxides along with the liberation of at 1 1 carbon dioxide. CaSO4 ⋅ 2H2 O o CaSO4 ⋅ H2 O + 1 H2 O 120 C 2 2 The commercial name of the compound Calcium Q14. sulphate hemihydrate, i.e., Plaster of Paris. Δ, 200°C MgC 2 on heating gives Mg 2 C3. This carbide contains CaSO4 ⋅ 2H2 O −−−−→ anhydrous CaSO4 + 2H2 O C 4− 3 units and reacts with water to form Anhydrous calcium sulphate formed this way is propyne (methyl acetylene). commonly called dead burnt plaster. Mg C3 + 4H2 O ⟶ 2 Mg (OH) + CH3 − C ≡ CH 2 2 Propyne ( 1 mole ) Q12. Q15. Calcium hydroxide reacts with chlorine to form 1. calcium hypochlorite, which is a component of Hf and Zr both are d block elements, both belonging to the group bleaching powder. 4 of the periodic table. The reaction is as follows: 2. 2 Ca (OH) 2 + 2 Cl2 → Ca (OCl) 2 + CaCl2 + 2H2 O K and Rb both are S block elements, both belonging to the group 1 of the periodic table. Q13. 3. The diagonally adjacent elements of period 2 and 3 Be belongs to the group 2 and Al belongs to the group 13 of the show similarity in their properties, which is known as periodic table. diagonal relationship. This is accounted to the But they show resemblance in properties due to: similarity in the ionic radii of these elements or their a) similar charge/ size ratio charge to size ratio. Examples of such elements are: Li b) similar electronegativity. and Mg; Be and Al. This is due to the diagonal relationship between In Mg, there is an increase in the number of shells, Be and Al. along with the completely fill