to Q. (ii) Transition elements and their compounds are generally found to be good catalysts in chemical reactions. Many actinide metals have properties of both d block and f block elements. Solution Show Solution Actinides exhibit larger oxidation states because of very small energy gap between 5f, 6d and 7s sub-shells . (iii) 2MnO4– + 16 H+ + 5C204– ———-> 2Mn+2 + 8H20 + 10CO2, Question 10: Question 61: However, occasionally in solutions or in solid compounds, +2 and +4 ions are also obtained. Therefore, all these three subshells can … Question 7: Question 36. Explain each of the following observations: (i) It is due to presence of vacant d-orbitals of suitable energy, smaller size of cations and higher charge. Write on the similarity between the chemistry of lanthanoids and actinoids. (i) Why is E (Cu 2 + / Cu) 0 value exceptionally positive? ——-> Na2Cr207 + Na2S04 + H20 They also show +2, +3, and +4 oxidation states. (iii) Cobalt (II) is stable in aqueous solution but in the presence of complexing agents, it is easily oxidised. (i) It is because Cu2+ is more stable because hydration energy overcomes 2nd ionisation energy. (ii) It is because energy required to remove electron is more due to greater effective nuclear charge which is due to lanthanoid contraction. (iii) Actinoids show a wide range of oxidation states. Describe the preparation of KMn04 from pyrolusite ore (MnO4). (a) What is meant by the term lanthanoid contraction? (a) Complete the following chemical reaction equations: Answer: Answer: (a) (i) It is because hydration energy of Cu2+ overcomes 2nd ionisation enthalpy, that is why Cu+ changes to Cu2+ and Cu. * This is due to the very small energy gap between 5f, 6d and 7s sub shells. (a) Complete the following chemical equations for reactions: +2. It is attained by removing outermost 2 electrons of 6s electrons and 1 electron from 4f electrons. 1045 Views. MnO. (ii) Mn04– + SO2– + H+ ———> Another way to prevent getting this page in the future is to use Privacy Pass. Question 60: (b) Account for the following: Explain giving reasons: Question 41: to Q.29 (ii). Answer: In the +7 oxidation state, this atom is electronegative enough to react with water to form a covalent oxide, MnO 4-.. (ii) Refer Ans. Plutonium. (ii) Transition metals form a large number of complexes. (b) Complete the following equation: Assign suitable reasons for the following: Lanthanoids show lanthanoid contraction like actinoids contraction. Answer: (ii) It is due to similarity in atomic and ionic size, there is more horizontal similarity. (i) It is because 5f, 6d and 7s have comparable energy. Answer: (b) (i) Refer Ans. Therefore, its salts are white. Books. Question 42: (a) (i) Refer Ans. Changes of oxidation state therefore change the colour of the light absorbed, and so the colour of the light you see. How would you account for the following? Answer the following: (a) What is the general electronic configuration of lanthanoids? (ii) It is due to their small size, high charge and availability of vacant d-orbitals, Question 13: (b) What are the common oxidation states of Cerium (At.no. (iii) It is so because manganese (Mn) has five unpaired electrons and 2 electrons in s orbital which can take part in bond formation. Question 64: Originally, the term was used when oxygen caused electron loss in a reaction. Chemistry. Now, it is known that all of the lanthanides can form +2 complexes in solution. Answer: (ii) Although Co2+ ion appears to be stable, it is easily oxidised to Co3+ ion in the presence of a strong ligand. Question 9:
(ii) Actinoids exhibit greater range of oxidation states than lanthanoids. Question 24: Question 33: Others are colourless due to the absence of unpaired electrons and cannot undergo d-d transitions. (i) Mn04– + C2072- + H+ ——-> (i) Thf enthalpies of atomisation of the transition metals are high. Actinides form basic oxides and hydroxides. 5 (ii). which one of these in the most stable in aqueous solutions? Answer: (i) E° value for the Mn+3/Mn+2 couple is positive (+1.5 V) whereas that of Cr+3/Cr+2 is negative (-0.4 V). (ii) Why is actinoid contraction greater than lanthanoid contraction? Answer: (ii) Actinoids exhibit greater range of oxidation states than lanthanoids. View Answer. It is due to similar energy of (n – 1 )d and ns orbitals, electrons from both can be lost. Though 3+ is the most stable oxidation state, other oxidation states are possible because of the good shielding of f-electrons. Mention its two uses. Actinides are typical metals. It is because neither Zn nor Zn+2. (ii) Transition metals form complex compounds Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Now, it is known that all of the lanthanides can form +2 complexes in solution. The wide range of oxidation states of actinoids is attributed to the fact that the 5f, 6d and 7s energy levels are of comparable energies. There is a greater range of oxidation states, which is in part attributed to the fact that the 5f, 6d and 7s levels are of comparable energies. Actinides show variable oxidation states because of the smaller energy gap between 5f, 6 d and 7s orbitals. (iii) Refer Ans. They have the ability to form complexes with ligands such as chlorides, sulfates, etc. (iii) Most of the transition metal ions exhibit characteristic colours in aqueous solutions (i) It is because of strong metallic bonds due to large number of unpaired electrons in d-orbitals. (iii) It is because 5f, 6d and 7s have comparable energy. to Q.9 (ii). (ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series. The decrease in size with the increase in atomic number among lanthanoids is called lanthanoid contraction. (ii) Refer Ans. to Q.9 (i). (ii) Actinoids exhibit a much larger number of oxidation states than the lanthanoids. For any content/service related issues please contact on this number . (ii) Transition metals form alloys. to Q.36 (ii). (iii) Zn has lowest enthalpy of atomisation. (i) The gradual decrease in size (actinoid contraction) from element to element is greater among the actinoids than that among the lanthanoids (lanthanoid contraction). (ii) It is because neither Zn nor Zn2+ ions have incompletely filled d-orbitals. Answer: Manganese has a very wide range of oxidation states in its compounds. Explain the following observations: (i) They show variable oxidation state. to Q.35 (i). Why? Answer: Question 55: Answer: (i) Fe2+ ions (i) Transition elements and their compounds are known to act as catalysts. Iron has two common oxidation states (+2 and +3) in, for example, Fe 2+ and Fe 3+. (ii) Mn(III) undergoes disproportionation reaction easily. Write the ionic equations for the reactions involved. When present in most compounds, hydrogen has an oxidation state of +1 and oxygen an oxidation state of −2. (ii) Mn has electronic configuration (Ar) 4s2 3d5 and all the electrons in ‘s’ as well as ‘d’ orbitals can take part in bond formation, therefore, it shows + 7 highest oxidation state. to Q.23 (b). to Q.30 (i). Some just lose one electron to form +3 … (ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e. (ii) It is due to lanthanoid contraction. Create free account. (ii) It is due to presence of unpaired electrons which participate in metal-metal bonding in 5d series (3rd series). (iii) Refer Ans. The reason for greater range of oxidation states in actinoids is attributed to. Lanthanoids show limited oxidation states(+2,+3,+4). Question 45: (i) Refer Ans. • Their separation becomes difficult. Find 1 Answer & Solution for the question Why do actinoids show a wide range of oxidation states ? . Although the lanthanides are sometimes called the rare earths, the elements are not particularly rare. * Actinides show a variety of oxidation states from +3 to +6. (b) Name an important alloy which contains some of the lanthanoid metals. For example, uranium, neptunium and plutonium form salts of the Na 2 UO 4 (uranate) and (NH 4) 2 U 2 O 7 (diuranate) types. (b) What is lanthanoid contraction? (iii) The E° value for the Mn3+/Mn2 + couple is much more positive than that for Cr3+/Cr2+ couple. (iii) Refer Ans. (ii) Which transition metal of 3d series has positive E°(M2+/M) value and why? How would you account for the following: It is useful to have a way of distinguishing between the charge on a transition-metal ion and the oxidation state of … Why? (ii) Cr2072 – + 14H+ + 6Fe2+ ———> 2Cr3+ + 7H20 + 6Fe3+. Write two characteristics of the transition elements. (ii) The actinoids exhibit a large number of oxidation states than the corresponding members in the lanthanoid series. In 3d series (Sc to Zn), which element shows the maximum number of oxidation states and why? (a) What are the different oxidation states exhibited by the lanthanoids? (b) Explain the following observations: 6 d = 6 + 2 = 8. Transition elements show variable oxidation states because electrons from both s and d orbitals take part in bond formation. (iii) The members in the actinoid series exhibit larger number of oxidation states than the corresponding members in the lanthanoid series. Give reasons : Answer: (i) Metal-metal bonding is more extensive in the 4d and 5d series of transition elements than the 3d series. In d-block, oxidation states differ by one, whereas in p-block, it differs by two. Question 49: (i) Refer Ans. Question 5: (a) (i) Refer Ans. Higher the oxidation state, more will be acidic nature, e.g. The E 0 ( M 2+ / M ) value of a metal depends on the energy changes involved in the formation of the M 2+ ion: Why do transition elements show variable oxidation states? (c) Why do actinoids show a wide range of osidation states? to Q.16 (a) (ii). to Q.41 (ii). Answer: (i) Refer Ans. 16 (a) (ii). To reach a higher oxidation state, one obviously has to pay for it in the form of ionisation energy/energies. (i) It is because of presence of unpaired electrons in d-orbitals of transition metal. to Q.5 (ii). Question 37: Question 50: (ii) It is due to presence of unpaired electrons which undergo d-d transition by absorbing light from visible region and radiate complementary colour. (iii) Refer Ans. +2, +3 and +4 (out of which +3 is most common) because of large energy gap between 4f and 5d subshells. Answer: So, which do you mean? (a) Complete the following reactions in an aqueous medium: Answer: 46. Explain the following observations giving an appropriate reason for each: (i) It is due to lanthanoid contraction which is due to poor shielding effect of f-electrons. However, there is a compensatory effect in that elements in higher oxidation states generally get more out of bonding. (b) Account for the following: Answer: Answer: Question 14: On the other hand Actinoids show a large number of oxidation states because of small energy gap between 5f, 6d and 7s subshell. • The lanthanoids have similar ionic size and resemble with each other closely in their properties. Why Lanthanide show Variable Oxidation State? (iv) Out of Mn3+ and Cr3+, which is more paramagnetic and why? (i) Zn is not considered as a transition element. Assign reasons for the following: Question 71: Sc3+, V3+, Ti4+, Mn2+ Please enable Cookies and reload the page. (ii) Cr2O7-(aq) + Fe2+ (aq) + H+(aq) ——-> Manganese: Manganese has a very wide range of oxidation states in its compounds. Actinoids shows wide range of oxidation states due to the fact t hat the 5f,6d and 7s energy levels are of comparable energies.Therefore all these three subshells can participate. The radioactive nature of actinoids. The energies are decided on the basis of (n+l) rule. (ii) E°CU2+/CU *ias +ve value due to high ionisation enthalpies and sublimation energies and lower hydration energy. (iii) Refer Ans. 4f and 5d levels being close in energies. (ii) The lowest oxide of a transition metal is basic, the highest is amphoteric/ acidic. (ii) Why is E (Mn 2 + / Mn) 0 value highly negative as compared to other elements? Actinoids also show stable +3 oxidation state but show a number of oxidation states i.e. (ii) CrO4– + H+ ——-> (ii) It is because all of them are radioactive and some of them have short half life, therefore, the chemistry of actinoids is not smooth. (a) Complete the following chemical equations: (b) (i) Refer Ans. (iii) The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series. Answer: (iv) Europium shows +2 oxidation state. (v) Mn04 + 8H+ + 5e- ———> Mn2+ + 4H20. (a) Refer Ans. However, they are difficult to separate from one another. to Q.23 (a). Biology. Question 16: Manganese. (a) Copper exhibits +1 oxidation state frequently due to stable electronic configuration. (i) Mn04- (aq) + S2O32-(aq) + H20(l) ——-> (v) Cr2+ is a very good reducing agent. Question 32: 3) The principal O.S are + 3 and + 4 , ( + 3 oxidation state … (i) Cr2072- + 20H- ———-> (ii) Mn3+ gains one electron to form Mn2+ since 3dD is more stable, whereas Cr2+ loses one electron for Cr3+(i3p which is more stable, so, it acts as reducing agent. Oxidation states +3, +5 and +6 are typical for aqueous solutions, but also in the solid state. Write its one use. (iii) It is due to absence of unpaired electrons, they do not absorb light from visible region and do not radiate colour. All Actinides form oxides with different oxidation states. (b) Explain the following observations: The known oxidation states of actinoids are listed in Table 8.11. to Q.30 (iii). The actinoids exhibit more number of oxidation states in general than the lanthanoids. Those elements which either themselves or their ions have incompletely filled Characteristics: +2. This is because Option 1) the orbitals extend further from the nucleus than the orbitals Option 2) the orbitals are more buried than the orbitals Option 3) there is a similarity between and orbitals in their angular part of the wave function Option 4) the actinoids are more reactive than the lanthanoids. Performance & security by Cloudflare, Please complete the security check to access. It is converted to NpO 2 at high temperatures. For example, Cl – has an oxidation state of -1. (a) (i) Refer Ans. Answer: (ii) 5f orbitals have poor shielding effect than 4f orbitals, therefore, effective nuclear charge is more in actinoids than lanthanoids. (ii) The E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple or Fe3+/Fe2+ couple. (i) Transition metals and their compounds are generally found to be good catalysts. (ii) Transition metals form coloured compounds. (ii) Refer Ans. (b) 3MnO4– + 4H+ ——–> MnO2 + 2MnO4– + 2H2O. (iii) Refer Ans. Many of the transition metals can lose two or three electrons, forming cations with charges of +1 or +2. Answer the following: (a) What is the general electronic configuration of lanthanoids? View Answer. to Q.41 (ii). 2Cu+ —–> Cu2+ + Cu Answer: (a) Lanthanoids, mostly show +3 oxidation state but some of them show +2 and +4 oxidation states also due to the stability of electronic configuration (4f°, 4f7 and 4f14), e.g. to Q. Answer: (i) It is because 5f, 6d and 7s have comparable energy. (ii)Oxidation States. (b) (i) Refer Ans. (iii) It is because oxygen and fluorine are strong oxidising agents, highly electronegative, small size and can provide energy for formation of transition metal ion in higher oxidation state. Why? (iii) Refer Ans. Question 17: Question 53: to Q.62 (a) (i). (ii) Refer Ans. 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