ISC/CBSE - Grade -12 - Inorganic Chemistry - Important Q&A

Reason-Based Questions - Inorganic Chemistry- 

One of the important types of questions framed from Inorganic Chemistry is Reason Based. The answers to these questions should consist of their specific cause. You can score well in these questions by pointing out a keyword which should be related to the cause like their physical or chemical property or it can be any law, etc.

Here are some important reasoning based questions with answers from the p - block and d - block of Inorganic Chemistry.

Give Reasons for the following 

1.     ICI is more reactive than I₂

Ans. The bond dissociation energy of I – Cl is less than of I – I. Therefore, I – Cl bond break easily due to its weak nature and readily brings about the reactions.

2.     Helium is used in diving apparatus.

Ans. Due to the low solubility of He, as compared to N₂ in blood, helium is used in diving apparatus along with a mixture of oxygen.

3.     Fluorine shows anomalous behaviour in its group.

Ans. The anomalous behaviour of fluorine is due to 

1. Small size 
2. Highest electro negativity 
3. Low F – F bond dissociation energy 
4. Non-availability of d-orbitals in its valence shell.

4.     Fluorine exhibits only – 1 oxidation state whereas other halogens exhibit higher positive oxidation states also.

Ans. Fluorine being the most electronegative atom has the tendency to accept one electron to acquire stable electronic configuration of the nearest noble gas. Therefore unlike other halogens, it exhibits only – 1 oxidation state. But other halogens being less electronegative can share their respective electron with more electronegative elements and can show a positive oxidation state. Further, fluorine cannot expand its valence shell due to the absence of vacant d-orbitals but other halogens have vacant d-orbitals and can expand their octets by exciting their electrons to nd orbitals.

5.     The acidity of oxoacids of chlorine is HOCl < HClO₂ < HClO₃ < HClO₄

Ans. It is because of the increase in the oxidation number of the halogen atom. It increases from HClO to HClO₄. Again, the stability of the conjugate bases of oxoacids of chlorine increases from ClO^- to ClO^-­₄ as the dispersal of negative charge increases from ClO^- to ClO^-₄ because of the number of oxygen atom increase from ClO^- to ClO^-₄.

6.     Xenon does not form XeF₃ and XeF₅.

Ans. Xe atom has all the filled orbitals with paired electrons and the promotion of one, two or three electrons from 5p filled orbitals to 5d vacant orbitals will give rise to two, four and six half-filled orbital’s i.e., only even number of fluorine atoms can combine with Xe.

7.     Only xenon is able to form chemical compounds.

Ans. Xe has least ionization energy except Radon which is radioactive, in its group. Therefore it is only able to form chemical compounds.

8.     Neon is used for warning signals.

Ans. Neon lights are visible from long distances even in mist and fog. This is why they are used as warning signs.

9.     Noble gases have low boiling points.

Ans. Noble gases are monoatomic and they are held together by weak Vander Waal’s forces. Therefore, they have low boiling points.

10.  Fluorine forms only one oxoacid, HOF.

Ans. Due to high electronegativity and small size of fluorine atom, it cannot behave as the central atom in higher oxoacids like HOFO, HOFO₂ and HOFO₃ in which the oxidation state of F would be +3, +5 and +7 respectively.

11.  Halogens are coloured.

Ans. The molecules of halogens absorb the light in the visible region and excite their electron to higher energy levels while the remaining light is transmitted and its colour is actually the colour of the transmitted light. The intensity of colour depends upon the size of the atom.

12.  Iodine forms I^-₃, ions but fluorine does not form F^-₃ ions.

Ans. Because of the presence of vacant d-orbital’s in iodine, it readily accepts electrons from I^- to form I^-₃ ions but due to the absence of d-orbital’s in fluorine, it does not accept electrons from F^- ions to form F^-₃ ions.

13.  The electron affinity of chlorine is higher than that of fluorine.

Ans. This is due to the small size of the fluorine atom. Its atomic size is comparatively smaller than chlorine atom. Therefore it experiences strong electron repulsions. On further addition of an electron to it, it has fewer tendencies to accept the electron in comparison to the chlorine atom. Hence it releases less energy during the process

 F(g)  + 1e^-   →  F^- (g).

14.  HF is a weaker acid than HI.

Ans. The bond dissociation energy of HF (574 KJ mol^-1) is higher than that of H – I (295KJ mol^-1). Therefore H- F behaves as the weakest acid because it requires a maximum amount of energy to get dissociated in the aqueous solution. It is because of the greater atomic size of iodine than fluorine.

15.  The tendency to show +6 oxidation states diminish from sulphur to polonium.

Ans. Due to inert pair effect, the stability of the +6 oxidation state decreases down the group. Thus, the +6 oxidation state is most stable in case of sulphur and least stable in case of polonium.

16.  Oxygen molecule has formula O₂ whilst sulphur is S₈.

Ans. Because of high electronegativity and small size, oxygen atoms form pπ – pπ double bond with other oxygen atom to form O = O molecule but sulphur does not form pπ – pπ multiple bonds due to its larger size and prefers to form single bonds having S₈ structure.

17.  SO₂ can act as an oxidizing as well as reducing agent but SO₃ can act only as oxidizing agent.

Ans. In SO₂, sulphur atom exhibits +4 oxidation state which is less stable than the oxidation state of +6. Therefore it acts as both oxidizing and reducing agent whereas in SO₃, sulphur atom exhibits +6 oxidation state. Being more stable, it can only behave as an oxidizing agent.

18.  Noble gases form compound like XeF₆ with fluorine or XeOF₂ with oxygen and fluorine.

Ans. Fluorine and oxygen is the most electronegative elements and are very reactive. Therefore they form compounds with noble gases, especially with xenon.

19.  SF₆ is one of the known compounds but SH₆ is not.

Ans. Fluorine being the most electronegative element can easily cause the promotion of electrons from the filled to the vacant 3d orbital’s in sulphur atom but hydrogen being less electronegative than sulphur cannot cause the promotion of electrons. Therefore SF₆ is known but SH₆ is not known.

20.  Interhalogens are more reactive than halogens.

Ans. It is due to weaker bond and lesser bond energy of X–X’ bond in interhalogens than that of halogen (X– X).

21.  Perchloric acid is a stronger acid than sulphuric acid.

Ans. The oxidation state of Cl in per choric acid is +7 while that of S in sulphuric acid is +6. Due to higher oxidation state and higher electronegativity of Cl, ClO₃ part of HClO₄ pulls the electrons of O-H bond more strongly towards itself making the O-H bond weaker than SO₂ part in H₂SO₄. Therefore, perchloric acid is a stronger acid than sulphuric acid  

22.  The bond energy of F₂ is less than that of Cl₂.

Ans. Due to the smaller size, the lone pairs of electron on the fluorine atom repel the bond pair of the F- F bond but due to the bigger size of electrons, the repulsion between bond pair and lone pair is comparatively less in chlorine atom. Hence the bond energy of F₂ is less than that of Cl₂.

23.  ClF₃ exists but FCl₃ does not.

Ans. Cl has vacant d-orbital but F has no d-orbital. Again because of the bigger size of chlorine, it can accommodate three F atoms but due to the smaller size of fluorine, it cannot accommodate large-sized Cl atoms around it.

24.  Why oxygen does not show positive oxidation state like other group members? Also, report the compound where oxygen shows a positive oxidation state.

Ans. Oxygen does not positive oxidation state because it does not have vacant d – orbital. So, there is no place to accommodate the excited electrons. Therefore, no sharing of the electron is possible.

Oxygen is the most electronegative element next to fluorine. It shows a positive oxidation state only in compounds with fluorine, e.g., it has a positive oxidation state in OF₂.

25.  The tendency to show -2 oxidation state decreases from sulphur to polonium in group 16. Why?

Ans. The elements of group 16 have the ground state valence shell electronic configuration is ns² np⁴ and tends to attain noble gas configuration by gaining or sharing two electrons. Oxygen with its high electronegative value tends to complete its octet by gaining two electrons and exhibits -2 oxidation state. The electronegativity of other element is very less and hence, there is a low probability for them to form divalent negative ions. The decreasing electronegativity value of the elements in this family with increasing atomic number suggests that the tendency to show -2 oxidation state decreases from sulphur onward to polonium.

26.  Why is OF₆ compound not known?

Ans. Electronic configuration of oxygen is 1s² 2s² 2p². Since it does not have any 2d orbital’s in its valence shell, so it cannot expand its octet due to non-availability of d orbitals. Hence, oxygen atom cannot exhibit higher positive oxidation state, and hence, OF₆ compound is not known.

27.  H₂S acts only as a reducing agent while SO₂ can act both as a reducing agent and oxidizing agent. Give reasons.

Ans. S in SO₂ has oxidation state +4. It lies in between the minimum oxidation state (-2) and maximum oxidation state (+6) of S. Thus, S in SO₂ can show an increase in its oxidation number (i.e., can act as reductant) or can show a decrease in its oxidation number (i.e., can act as an oxidant). On the other hand, in H₂S, S is in -2 oxidation state and can only increase its oxidation state to act as reductant.

28.  Sulphur dioxide is a more powerful reducing agent in an alkaline medium than in acidic medium. Why?

Ans. The reducing property may be represented as                   SO₂ + 2OH^-    →        SO₄^2- + 2e^-

Addition of acid favours the reverse reaction whereas addition of OH^- favours the forward reaction.

29.  When chlorine is bubbled through a fluorine solution, no reaction takes place. Explain.

Ans. Due to the small size and high electronegativity, fluorine is a stronger oxidizing agent than chlorine. Hence, chlorine cannot displace or oxidize fluorine ions to fluorine. Chlorine shows different oxidation states in its compounds.

30.  Why a metal ion liberates I₂ from KI but the same metal ion does not liberate Cl₂ from KCl?

Ans. It is because I^- is a stronger reducing agent as compared to Cl^- and thus, KI reduces the metal ion to liberate I₂. For example KI on reaction with Cu²⁺ liberates I₂ and Cu²⁺ get reduced to Cu¹⁺.

                                           4KI + 2Cu²⁺ →    I₂ + Cu₂I₂ + 4K⁺

31.  Compare, giving a reason, the oxidizing powers of fluorine and chlorine.

Ans. Because of low bond dissociation energy of fluorine, it is a stronger oxidizing agent.

32.  Why fluorine does not form oxoacids like other halogens?

Ans. Fluorine shows only -1 oxidation state due to the highest electronegativity whereas, in oxoacids, the element involved should have positive oxidation state. Thus, fluorine does not form oxoacids.

33.  Name the halogen having some metallic nature.

Ans. Iodine, possesses metallic lustre, forms ionic compounds such as IPO₄, I(CH₃COO)₃ having a +3 oxidation state.

34.  Pure HI kept in bottle acquires a brown colour after some time. Why?

Ans. HI is a strong reducing agent and on keeping, it gets slowly oxidized to brown coloured iodine by oxygen.

                                            4HI + O₂     →      2I₂ + 2H₂O

35.  Halogens are strong oxidants. Why?

Ans. The electron affinity of halogen is maximum among all groups. Higher is the electron affinity more is the tendency to gain an electron and thus more is the tendency to get itself reduced and more is oxidizing power.

                                            X + e^-    →        X^-                           ∆ H = -E (higher values)

36.  Why is fluorine more reactive than other halogens?

Ans. Fluorine is more reactive than other halogens due to the following reasons:

         Fluorine has a low bond dissociation energy due to repulsion between its non-bonding electrons.
Fluorine cannot form multiple bonds because of the absence of d orbitals.
Fluorine has a small size and high electronegativity, and due to this it has a greater tendency to accept an electron to form a more stable fluoride ion.
         The extent of hydration in F^- is maximum.

37.  What reasons can you give for fluorine existing only the -1 and zero oxidation states whereas iodine exists in +1, 0 and -1 oxidation states?

Ans. Fluorine being the most electronegative element combines with other elements by ionic or polar covalent bonds in which it is negatively polarized. Thus, its oxidation state is -1. But when fluorine atoms combine with each other to form F₂ oxidation state of fluorine is zero.

38.  Explain how the oxidizing character of the halogen varies as we move down the group.

Ans. Due to their high electron affinities, halogens act as strong oxidizing agents. But their oxidizing power goes on decreasing on moving down the group from the fluorine to iodine due to the increase in the size of the atoms. The decreasing trend in oxidizing power is quite evident from their respective standard reduction potential which is maximum for fluorine and minimum for iodine.

39.  Xenon form compounds whereas krypton other noble gas elements do not form compounds. Explain.

Ans. Because xenon has lower ionization energy as compared to other noble gas elements.

40.  What is the basic cause of solubility of noble gases in the water?

Ans. The increase in solubility of noble gases down the group is due to the increase in polarization with the increase in sizes. The basic cause of solubility of noble gases in water is due to dipole induced dipole attraction between water molecules and noble gases.

41.  Why helium and neon, do not form compounds with highly electronegative fluorine and oxygen?

Ans. Helium and neon do not form compounds with highly electronegative fluorine and oxygen because these noble gases have high ionization energy and they do not have d orbitals in their valence shells.

42.  Why the ease of liquefaction of noble gases increases down the group?

Ans. Liquefaction of noble gases depends upon the atomic size. Vander Waal’s forces increase down the group, and therefore, the tendency of liquefaction also increases down the group.

43.  Noble gases (Ne, Ar) have larger radii in their periods.

Ans. In the noble gases, we can measure only Wander Waal’s radii which are larger than covalent radii.

44.  Why is helium used to inflate aeroplane tyres and to fill the weather balloons through its lifting power is only 90% of hydrogen?

Ans. It is because hydrogen gas is highly inflammable; therefore, it may catch fire with an explosion. On the other hand, helium gas is non-inflammable, so there is no danger of fire or explosion.

45.  Why are the elements of group 18 known as noble gas?

Ans. The elements of group 18 have their valence shell orbitals completely filled. As a result, they are fairly non-reactive and hence are called noble gases. However, Xe reacts with highly electronegative elements such as oxygen and fluorine under certain conditions to form oxides, fluorides and ox fluorides.

46.  The enthalpies of atomization of the transition metals are high.

Ans. Transition metals have strong metallic bonds between the atoms of these elements. Therefore a high amount of heat is required to break the metal lattice to give free atoms.

47.  The transition metals generally form coloured compounds.

Ans. The transition metals have in complete d-subshell or have unpaired electrons in their d-subshell due to which d-d transition takes place and they become coloured.

48.  The transition metals and many of their compounds show paramagnetic behaviour.

Ans. The transition metals have unpaired electrons in (n – 1) d-orbital. Therefore their ions and compounds are paramagnetic.

49.  Cu⁺ ion is not stable in aqueous solution.

Ans. Cu⁺² is much more stable than Cu⁺ (aq). It is because although second ionization enthalpy of copper is large but enthalpy of hydration (Δ_hy H) for Cu⁺ (aq) is much more negative than that for Cu⁺² (aq) and hence it more than compensates for the second ionization enthalpy of copper and they readily undergo disproportionately in aqueous solution.        2Cu⁺     →      Cu⁺² + Cu

50.  Actinoid concentration is greater from element to element than Lanthanoid concentration.

Ans. This is due to poor shielding by 5f electrons in the actinoids than that by 4f electrons in the lanthanoids.

51.  The lowest oxide of a transition metal is basic, the highest is acidic.

Ans. In the low oxidation state of the metal, some of the valence electrons of the metal atom are not involved in bonding. Hence it can donate electrons and behave as a base but in higher oxidation state, valence electrons are involved in bonding and are not available. Therefore it can accept electrons and hence behave as acid.

52.  Copper is considered as a transition metal.

Ans. Because copper in the oxidation state of +2, it has electronic configuration of 4s^o3d⁹, i.e., incompletely filled d-subshell.

53.  Scandium forms no coloured ions yet it is regarded as a transition element.

Ans. Scandium in the ground, the state has one electron in the 3d-subshell. Therefore it is regarded as transition element but in +3 oxidation state, it has no electron in 3d subshell (3d^o). Hence it does not form coloured ions.

54.  Transition elements have high enthalpies of hydration.

Ans. Because of transition, elements have a small atomic size and high nuclear charge. Both the properties change as we move from left to right along the transition series.  

55.  Transition elements show variable oxidation states.

Ans. In the transition elements, the energies of (n – 1) d orbitals and ns orbitals are very close or there is little difference in them. Hence electrons from both can take part in bonding.

56.  Transition elements have many irregularities in their electronic configuration.

Ans. Due to the very small difference in the energy of (n – 1)d-subshell and ns subshell, the incoming electron may enter into ns or (n – 1) d subshell. Therefore, they show irregularities in their electronic configuration.  

57.  The paramagnetic character in 3d transition series elements increase up to Mn and then decreases.

Ans. On moving from Sc to Mn, the numbers of unpaired electrons increases and hence paramagnetic character increases. But after Mn, the pairing of electrons in the d-subshell starts and the number of unpaired electrons decreases and hence paramagnetic character decreases.

58.  Zn⁺² salts are white, white Cu⁺² salts are blue.

Ans. Zn⁺² has completely filled d-orbital’s (3d¹⁰) causing no d-d transition but due to the presence of unpaired electrons or incompletely filled d-orbital’s (3d⁹) the d-d transition takes place and Cu⁺² salts are blue.

59.  The melting and boiling point of Zn, Cd and Hg are low.

Ans. Zn, Cd and Hg have completely filled d-subshell i.e. all the electrons in d-subshell are paired. Hence the metallic bonds present in them are weak. Therefore they have low melting and boiling points.

60.  The separation of lanthanide is difficult.

Ans. Due to lanthanide contraction, the change in the atomic or ionic radii of these elements is very small. Hence their chemical properties are similar which makes their separation difficult.

61.  Electronic configuration of lanthanoids is not known with certainty.

Ans. In the lanthanoids, 4f and 5d subshell are very close in energy. The outer most 6s orbital remains filled with 2 electrons (6s²). The electron can easily jump from 4f to 5d or vice-versa. Therefore their electronic configuration is not known with certainty.

62.  Second and third transition series elements show a similar size.

63.  Ans. Due to lanthanide concentration, the size of an atom of the third transition series is almost the same as that of the element present just above in the second transition series. Hence there is similarity in their properties.

64.  Cr, Mn and Fe have nearly the same atomic radii.

Ans. Cr, Mn and Fe have nearly the same effective nuclear charge. Hence they have almost the same atomic radii.

65.  Cd⁺² salts are white.

Ans. Cd⁺² salts are white because it has completely filled d-orbitals (d¹⁰) and there is no possibility of d-d transition.

66.  In permanganate ion, there is covalency between manganese and oxygen.

Ans. In permanganate ion (MnO^-₄) ion, manganese is in its highest oxidation state i.e. + 7 and in its high oxidation state, transition metals form covalent bonds.

67.  Mn⁺² compounds are more stable than Fe⁺² towards oxidation to their +3 state.

Ans: Electronic configuration of Mn⁺² shows that it has half-filled (3d⁵) sub-shell, exhibiting extra stability. Therefore third ionization enthalpy is very high and it cannot lose the third electron easily. In the case of Fe⁺², an electronic configuration is 3d⁶ and it can lose the third electron easily to give stable configuration 3d⁵.

68.  Sodium cyanide is used in the extraction of silver from its ore.

Ans. Sodium cyanide on reaction with the ore of silver forms sodium Argento cyanide. It is a soluble complex from which the silver can be precipitated easily by adding more electropositive metal like zinc to it.

69.  Out of La (OH)₃ and Lu(OH)₃, the former one is more basic

Ans. As the size of the lanthanoid decreases from La⁺³ to Lu⁺³ the covalent character of the hydroxides increases. Hence the basic strength decreases from La(OH)₃ to Lu(OH)₃.

70.  Cr⁺² is reducing but Mn⁺³ is oxidizing when both have d⁴ configuration.

Ans. Cr⁺² have d⁴ configuration and easily changes to Cr⁺³ because d³ has half-filled t_2g orbital and hence more stable. Therefore it is reducing. On the other hand, Mn⁺² is more stable due to half-filled d⁵ configuration and Mn⁺³ can easily be changed to Mn⁺² and therefore it is oxidizing.

71.  H₂S is a better reducing agent than H₂O.

Ans. The H-S bond is weaker than H-O because of the lesser tendency of S for overlapping due to lower electronegativity and larger size of the anion S^2-. Thus, H₂S having weaker bond acts as a better reducing agent.

72.  Sulphur is a solid whereas oxygen is a gas at room temperature.

Ans. Oxygen forms a stable diatomic molecule. In oxygen molecules, two atoms of oxygen have joined together through a double bond (pπ – pπ bonding) as O=O. The multiple bonding in oxygen is possible due to the small size of oxygen atoms. Thus, oxygen is a gas and no more atoms of oxygen can join.

73.  Argon not forms diatomic molecules like oxygen or chlorine?

Ans. Because no unpaired electron is present in its electronic configuration (1s² 2s² 2p⁶ 3s² 3p⁶) and it has very high ionization energy.

74.  The bond dissociation energy of F₂ is less than the bond dissociation energy of chlorine.

Ans. It is due to strong inter-electronic repulsion between nonbonding electrons in small size 2p orbitals of fluorine as compared to big sizes 3p orbitals of chlorine and this makes the bond dissociation energy of F₂ less than that of Cl₂.

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