CBSE Class 12 Chemistry Elements Worksheet

Question. Give chemical reaction in support of the statement that all the bonds in PCI₅ molecule are not equivalent.
Answer : Due to greater bond pair-bond pair repulsions, the two axial P — Cl bonds are less stable than the three equatorial P — Cl bonds. It is because of this reason that when PC1₅ is heated, the less stable axial bonds are Broken to form Cl₂.
                 Δ
PCl₅ --------------→  PCl₃ + Cl_2.

Question. The tendency for catenation decreases down the group 14. Or Why carbon shows catenation but silicon does not
Answer : As we move down the group 14, the atomic size increases and hence the strength of M—M bond decreases steadily. Consequently the tendency for catenation decreases down the group.

Question. PCI₅ is known but NC1₅ is not known. Explain.
Answer : Electronic configuration of P is 1s² 2s² 2p⁶ 3s² 3px¹ 3py¹ 3pz¹ . Thus phosphorus has empty 3d orbitals in which the elecron of the 3s-orbital can be excited but nitrogen has no d-orbitals as its electronic configuration is 1s² 2s² 2px¹ 2py¹ 2pz¹.

Question. Why molecular nitrogen is not reactive ?
Answer : The bond dissociation energy for the triple bond in N = N is very high because of small size of N-atoms and hence small internuclear distance.

Question. H₃PO₃ is diprotic (or)dibasic. Why?
Answer : Since it has only two ionisable H – atoms which are present as OH groups.

Question. Nitrogen exists as diatomic molecule, N₂, whereas phosphorus exists a tetratomic molecule P4. Explain.
Answer : Nitrogen atom being small in size forms multiple bonds with another nitrogen atom but phosphorus being large in size forms only single bonds with other P-atoms.

Question. Phosphine is a weaker base than NH3. Explain.
Answer : Due to bigger size of P than N, P—H bond is much weaker than N—H bond. As a result, PH₃ has less tendency than N to donate its lone pair of electrons to a proton. Consequently, PH₃ is a weaker base than NH₃.

Question. NO (Nitric oxide) is paramagnetic in the gaseous state but diamagnetic in the liquid and solid states. Why?
Answer : NO has an odd number of electrons (11 valence electrons) and hence is paramagnetic in the gaseous state. But in liquid and solid states, it exists as a symmetrical or asymmetrical dimer and hence is diamagnetic in these states.

Q. 9. Nitrogen (N ≡ N). carbon monoxide (C=O) and cyanide ion ( -C = N) are isoelectronic, yet CO and -C=N are reactive but N₂ is not. Explain why.
Answer : N₂ is non-polar and has very high ionization energy.

Question. NC1₃ gets readily hydrolysed while NF₃ does not. Why ?
Answer : In NC1₃, Cl has vacant d-orbitals to accept lone pair of electrons donated by O-atom of H₂O molecule but in NF₃, F does not have d-orbitals.

Question. Oxides of nitrogen have open chain structures while those of phosphorus have closed chain or cage structures. Why is it so? Illustrate with one structural example for each type of oxides.
Answer : Nitrogen because of its small size and higher electronegativity than phosphorus, has a strong tendency to form pπ - pπ. multiple bonds. As a result, nitrogen forms multiple bonds with oxygen and hence the oxides of nitrogen have open chain structures. Phosphorus, on the other hand, because of its large size does not form pπ - pπ multiple bonds with oxygen atoms but instead prefers to form single bonds. As a result, oxides of phosphorus have closed chain or cage structures.

Question. Write balanced chemical equations for the following reactions : (a) Ca₃P₂ + H₂O —>? (b) P₄O₁₀ + H₂O —>? (c) As₄ + C1₂ (excess) —>? (d) P₄ + KOH + H₂O —>?
Answer : (a) Ca3P₂ + 6H₂O -----→  3Ca(OH)₂ + 2PH₃ (Phosphine)
(b) P₄O₁₀ + 6H₂O ---→  4H₃PO₄
(c) As₄ + 10 C1₂ (excess) ----→  4 AsCl₅
(d) P₄ + 3 KOH + 3 H₂O ----→  PH₃ + 3 KH₂PO₂.

Question. What is the oxidation state of phosphorus in the following:
(a) H3PO₃ (b) PC1₅ (c) Ca₃P₂ (d) Na₃PO₄ (e) POF₃.
Answer : (a) + 3
(b) + 5
(c) -3
(d) + 5 (e) + 5

Question. Suggest a method for the laboratory preparation of DC₁. Write a balanced equation for the reaction.
Answer : Hydrolysis of PCl₅ with heavy water (D₂O) gives DCl.
PCl₅ + D₂O -------→  POCl₃ + 2DCl.

Question. Unlike phosphorus, nitrogen shows little tendency for catenation ?
Answer : Due to smaller size, the lone pair of electrons on the N-atoms repel the bond pair of N—N bond. In contrast, because of comparatively larger size of P, the lone pair of electrons on P atoms do not repel the bond pair of the P—P bond to the same extent. As a result, N—N single bond is weaker than P—P single bond. Consequently nitrogen shows little tendency for catenation while P has a distinct tendency for catenation.

Question. Give the chemical reaction to support that +5 oxidation state of Bi is less stable than +3 state.
Answer : Due to inert pair effect, Bi can show +3 and +5 oxidation states. Since the inert pair effect is maximum in case of Bi, therefore, its +5 oxidation state is less stable than +3 oxidation state. This is evident from the observation that BiCI₃ even on prolonged heating with C1₂ does not form BiCl₅.

Question. Give reasons for the following : (i) Nitric oxide becomes brown when released in air. (ii) PC1₅ is ionic in the solid state (iii) Ammonia is a good complexing agent.
Answer : (i) Nitric oxide readily combines with O₂ of the air to form nitrogen dioxide (NO₂) which has brown colour.
2NO + O₂ —>                  2NO₂
(Colourless)                  (Brown)
(ii) PC1₅ is ionic in the solid state because it exists as [PCl₄]+ [PC1₆]- in which the cation is tetrahedral and the an ion is octahedral.
(iii) Due to presence of a lone pair of electrons on N, NH₃ acts as a complexing agent. As a result, it combines with transition metal cations to form complexes. For example : AgCl + 2NH₃ --------→  [Ag(NH₃)₂]Cl. [Diamminesilver (I) chloride]
CuSO₄ + 4NH₃ -----→  [Cu(NH₃)]SO₄. [Tetraamine copper (II) sulphate]
CrCl₃ + 6NH₃ ------------→  [Cr(NH₃)]Cl₃. [Hexaaminechromium (III) chloride].

Question. Oxygen forms diatomic molecules (O₂) while sulphur forms octaatomic molecules (S8). Explain. Or Why O₂ is a gas but S8is solid. Justify.
Answer : Due to small size of oxygen, the non-bonding electrons on the two oxygen atoms in O—O bond strongly repel each other while such repulsions are much less in S—S bond. As a result, O—O bond dissociation energy is much less than S—S bond dissociation energy. In other words, sulphur has much higher tendency for catenation than oxygen. Further, oxygen because of its small size has a high tendency to form pπ - pπ multiple bonds but sulphur because of its bigger size normally does not form pπ - pπ multiple bonds. Oxygen because of its less tendency for catenation and higher tendency for pπ - pπ multiple bonds, forms stable diatomic molecules (O₂) whereas sulphur because of its higher tendency for catenation and lesser tendency for pπ - Pπ. multiple bonds forms octaatomic (S8) molecules having eight membered puckered ring structure.

Question. 44. H₂S acts only as a reducing agent but SO₂ acts both as a reducing agent as well as an oxidising agent. Why ?
Answer : The minimum oxidation number (O.N.) of S is —2 while its maximum O.N. is +6. In SO₂, the O.N. is +4. therefore, it can not only increase its O.N. by losing electrons but can also reduce its O.N. by gaining electrons. As a result, it acts both as a reducing agent as well as an oxidising agent. On the other hand, in H₂S, S has an O.N. of —2. As a result, it can only increase its O.N. by losing electrons and hence acts only as a reducing agent.

Question. Oxygen generally exhibits an oxidation state of—2 only whereas other members of its family show oxidation states of +2, +4 and +6 as well. Explain why.
Answer : The electronic configuration of oxygen is 1s² 2s² 2px² 2py^l 2pz¹, i.e., it has two half-filled orbitals and also there are no d-orbitals available for excitation of electrons. Therefore, it can show an oxidation state of either - 2 or + 2. Since, oxygen is the most electronegative element of its family, therefore, it shows an oxidation state of—2 only. Other elements of this family like sulphur have d-orbitals available for excitation thereby giving four and six half-filled orbitals for bonding. Furthermore they can combine with more electronegative elements. Hence they show oxidation states of +2, +4 and +6 also.

Question. Name two compounds in which oxygen has oxidation state different from —2. Give oxidation states also.
Answer : (i) OF₂, oxidation state of O = + 2 (ii) H₂O₂ , oxidation state of O = —1.

Question. The tendency to show —2 oxidation state diminishes from sulphur to polonium. Why ?
Answer : As we move down the group from sulphur to polonium, the electronegativity decreases. Q.

Question. Why H₂O is a liquid while H₂S is a gas?
Answer : There is H—bonding in H₂O due to high electronegativity and small size of O-atom but there is no H-bonding in H₂S.

Question. Why H₂S is acidic while H₂O is neutral ?
Answer : The S—H bond is weaker than O—H bond because size of S-atom is bigger than that of O-atom. Hence H₂S can dissociate to give H⁺ ions in aqueous solution.

Question. SF6 is known but SH₆ is not known. Explain.
Answer : In the highest oxidation state, sulphur can combine only with highly electronegative elements like oxygen and fluorine.

Question. Why the compounds of fluorine with oxygen are called fluorides of oxygen and not the oxides of fluorine?
Answer : This is because fluorine is more electronegative than oxygen.

Question. Starting from elemental sulphur, how would you prepare (a) H₂SO₄ (b) SC1₂ and (c) SF₆ ? V₂O₅, 720 K; 2 Bar pressure
Answer : (a) S + O₂ ---→  SO₂ ; SO₂ + O₂ -------------------------------------------→  2 SO₃ SO₃ + H₂SO₄ --------→  H₂S₂O₇ ; H₂S₂O₇ + H₂O -------→  2H₂SO₄.
(b) Heating S and Cl₂ gives S₂Cl₂ which upon saturation with Cl₂ gives SCl₂
                      Heat
S₈ + 4 Cl₂ --------→  4S₂Cl₂
S₂Cl₂ + Cl₂ -----→  2 SCl₂.
(c) S₈ + 24 F₂ -------→  8 SF₆.

Question. Write balanced equations for the following reactions : Catalysyt (a) Cu + con. H₂SO₄ ---→  (b) SF₄ + H₂O -----? (c) H₂S + SO₂ --------------→  ? (d) Te + Cl₂ --→  ?
Answer : (a) Cu + 2 con. H₂SO₄ --→  CuSO₄ + SO₂ + 2 H₂O.
(b) SF₄ + 2 H₂O -----→  SO₂ + 4HF. catalyst
(c) 2 H₂S + SO₂ -------------→  3 S + 2 H₂O
(d) Te + 2 Cl₂ ------→  TeCl4.

Question. Why does sulphur in vapour state exhibit paramagnetic behaviour ?
Answer : In vapour state, sulphur partly exists as S₂ molecules which like O₂ has two unpaired electrons in the antibonding molecular orbitals (2 πx* and 2 πy *) and hence exhibits paramagnetism.

Question. Among the hydrides of group 16, water shows unusual physical properties.
Answer : Due to high electronegativity of O, the O—H in H₂O forms strong intermolecular H -bonds. As a result, water exists as an associated molecule while other hydrides of group 16 do not form H-bonds and hence exist as monomers. As a result, water shows unusual physical properties, i.e. high b.p., high thermal stability and weaker acidic character as compared to other hydrides of group 16.

Question. Give reasons for the following observations: (i) SF6 is not easily hydrolysed though thermodynamically it should be. (ii) Sulphur exhibits greater tendency for calenation than selenium.
Answer : (i) in SF6, S is sterically proctected by six F atoms and hence does not allow H₂O molecules to attack the S atom. Furthermore, F does not have f-orbitals to accept the electrons donated by H₂O molecules. As a result of these two reasons, SF6 does not undergo hydrolysis. In contrast, in SF4, S is not sterically protected since it is surround by only four F - atoms. As a result, attack of H₂O molecules can take place easily and hence hydrolysis occurs. (ii) As we move from S to Se, the atomic size increases and hence the strength of M—M bond decreases. As a result, S—S bond is much stronger than Se-—Se bond and consequently, S shows greater tendency for catenation than selenium.

Question. Sulphur disappears when boiled with an aqueous solution of sodium sulphite.
Answer : S reacts with Na₂SO₃ to form sodium thiosuiphate (Na₂S₂O₃) and hence S diappears. Na₂CO₃ + S ---------→  Na₂S₂O₃

Question. The bond energy of F₂ is less than that of C1₂
Answer : Due to smaller size, the lone pairs of electrons on the F-atoms repel the bond pair of the F—F bond. In contrast, because of comparatively larger size of CI atoms, the lone pairs of Cl do not repel the bond pair of CI—Cl bond. As a result, F—F bond energy is lower than that of Cl—Cl bond energy.

Question. Flourine does not undergo disproportionation reactions but other halogens do. Explain.
Answer : Disproportionation means simultaneous oxidation-reduction. F being the most electronegative element undergoes only reduction but not oxidation. As a result, it shows only —1 oxidation state while other halogens show both negative (—-1) and positive (+1, +3, +5, +7) oxidation states. Thus, F does not show disproportionation reactions while other halogens do.

Question. Write an example of a neutral molecule which is isoelectronic with C1O-.
Answer : CIO- ion has 17 + 8 + 1 = 26 electrons. A neutral molecule with 26 electrons is OF₂ (8+2 × 9 = 26). Thus OF₂ is isoelectronic with CIO-.

Question. Both NO and C₁O₂ are odd electron species whereas NO dimerises but C₁O₂ does not. Why?
Answer : In NO, the odd electron on N is attracted by only one O-atom but in C1O₂, the odd electron on Cl is attracted by two O-atoms. As a result, the odd electron on N in NO is localized while the odd electron on Cl in C1O₂ delocalized. Consequently, NO has a tendency to dimerize but C1O₂ does not.

Question. Bleaching of flowers by chlorine is permanent while that by sulphur dioxide is temporary. Explain.
Answer : Cl₂ bleaches coloured material by oxidation : C12 + H₂O -------→  2 HC1 + [O] Coloured material + [O] -----→  Colourless and hence bleaching is permanent. On the other hand, SO₂ bleaches coloured material by reduction and hence bleaching is temporary since when the bleached colourless material is exposed to air, it gets oxidised and the colour is restored.
SO₂ + 2 H₂O ---------→  2 H₂SO₄ + 2 [H] Aerial oxidation
Coloured material + [H] --------→  Colourless material ----------------------------→  Coloured material.

Question. Why fluroine shows an oxidation state of—1 only while other halogens show oxidation states of +1, +3, + 5 and +7 as well ? Or Why fluorine does not show variable valency while other halogens exhibit variable valency ?
Answer : Electronic configruation of fluorine is Is² 2s²2px² 2py² 2pz¹. It has only one half-filled orbital and there is no d-orbital available for excitation of electrons. Therefore, it can show oxidation state of either +1 or -1. Since, it is the most electronegative element, therefore, it shows oxidation state of—1 only. In all other halogens, d-orbitals are available for excitation of electrons. Moreover they can combine with the more electronegative element, oxygen. Hence they show variable positive oxidation states.

Question. F-atom is more electronegative than I-atom, yet HF has lower acid strength than HI. Explain. Or Hydrogen fluoride is a much weaker acid than HCI in aqueous solution.
Answer : F atom being small in size, the bond dissociation energy of H—F is very high as compared to that of H—I bond because I-atom is very large in size.

Question. Why F₂ is a stronger oxidizing agent than C1₂ while electron affinity of fluorine is less than that of chlorine ?
Answer : F₂ is a stronger oxidizing agent than CI₂ because electrode potential of F₂ (+ 2.87 V) is much higher than that of C1₂ (+ 1.36 V). As a result, F₂ is more readily reduced than C1₂ and hence F₂ is a stronger oxidising agent than Cl₂

Question. OF₂ should be called oxygen difluoride and not fhiorine oxide. Why ?
Answer : Because F is more electronegative than oxygen.

Question. Iodine is more soluble in KI than in water. Why?
Answer : Iodine combines with KI to form the soluble complex, KI₃ (KI + I₂ ---→  KI3).

Question. Arrange HC1O, HBrO and HIO in order of decreasing acidic strength giving reason.
Answer : H—O—Cl > H—O—Br > H—O—I. This is because electro negativities decrease in the order Cl > Br > I.

Question. Arrange HCIO₄, HC1O₃, HC1O₂, HC1O in order of (i) acidic strength (ii) oxidizing power. Same as above.
Answer : Acidic strength: HC1O₄ > HC1O₃ > HC₁O₂ > HC1O
Reason: Stability of C1O₄- > ClO₃- > C1O₂- > C1O_-
Oxidizing power: HC₁O₄ < HC1O₃ < HC1O₂ < HCIO
Reason: Same as above.

Question. Why interhalogens are more reactive than halogens?
Answer : This is because the bond in the interhalogen (A—X) is weaker than X— X bond in the halogens. This is due to less effective overlapping between orbitals of dissimilar atoms than those between similar atoms.

Question. Why HF acid is stored in wax coated glass bottles?
Answer : This is because HF does not attack wax but reacts with glass. It dissolves SiO₂ present in glass forming hydrofluoro silicic acid. SiO₂ + 6 HF-----→  H₂SiF₆ + 2 H₂O

Question. Chlorine vapours are evolved when concentration sulphuric acid is added to a mixture of sodium chloride and manganese dioxide. Write a balanced equation for the reaction.
Answer : 2NaCl (s) + 2H₂SO₄ (aq) + 2MnO₂ (s) --→  Na2SO₄ (aq) + MnSO₄ (aq) + 2H₂O

Question. Deduce the molecular shape of BrF-, on the basis of VSEPR theory.
Answer : On the basis of VSEPR theory, a molecule with three bond pairs and two lone pairs must be T-shaped (i.e. distorted bipyramidal) as shown.
The two lone pairs will occupy equatorial positions to minimize lone pair-lone pair and lone pair-bond pair repulsions, which are greater than, bond pair-bond pair repulsions. Further, the axial F atoms will be slightly bent towards the equatorial F atom to minimize the lone pair-lone pair repulsions. Therefore, the shape of BrF3 would be slightly bent T as shown above. Q.

Question. C₁F₃ exists but FC1₃ does not.
Answer : Reasons are : (i) Cl has vacant d-orbitals and hence can show an oxidation state of +3 but F has no d-orbitals, therefore, it cannot show positive oxidation states. Further, since F can show only -1 oxidation state, therefore, FCl₃ does not exist. (ii) Because of bigger size, Cl can accommodate three small F atoms around it while F being smaller cannot accommodate three large sized Cl atoms around it.

Question. Helium and neon do not form compounds with fluorine. Why?
Answer : He and Ne do not contain d-orbitals in their valence shell and hence their electrons cannot be promoted to higher energy levels like that in Xe to form bonds. Therefore, He and Ne do not form compounds with fluorine.

Question. Neon is generally used for warning signals. Why?
Answer : Neon lights are visible from long distances even in fog and mist and hence neon is generally used for warning Signals.

Question. Why do noble gases form compounds with fluorine and oxygen only?
Answer : The electro negativity of both fluorine and oxygen is very high. Moreover, fluorine is a highly reactive element.

Question. What prompted Bartlett to the discovery of noble gas compounds.
Answer : Since PtF6 oxidises O₂ To O₂⁺, Bartlett thought that PtF6 should also oxidise Xe and Xe⁺ because the ionization energies of O₂ (1175 kJ mol^-1) and Xe (1170 kJ mol^-1} are quite close.

Question. Most of the known noble gas compounds are those of xenon.
Answer : Amongst noble gases, Xe has the lowest ionization enthalpy and hence can be easily oxidised by strong oxidising agents tike O₂ and F₂

Question. Give reasons: (i) Xenon does not form fluorides such as XeF₃ and XeF₅. (ii) Out of noble gases, only xenon is known to form real chemical compounds.
Answer : (i) All the filled orbitals of Xe have paired electrons. The promotion of one, two or three electrons from the 5p- filled orbitals to the 5 d-vacant orbitals will give rise to two, four and six half-filled orbitals. So Xe can combine with even but not' odd number of F atoms. Hence, it cannot form XeF₃ and XeF₅. (ii) Except radon which is radioactive. Xe has least ionization energy among noble gases and hence it readily forms chemical compounds particularly with O₂ and F₂.