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ENGLISH 1 d and f-block element(A.K.SAMAL,PGT(CHEM.) On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?
ANS: It is because Sc (21) has incompletely filled d-orbital, that is why it is transition element, whereas Zn(30) does not have incompletely filled d-orbitals, therefore, it is not regarded as transition element. 2 Why do transition metals show variable oxidation states? 1 ANS: It is because electrons from both ‘s’ and d-orbitals can take part in bond formation. 3 Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical. Give reason.
ANS: In Actinoids, 5f, 6d and 7s orbitals have comparable energies and electrons from these orbitals can take part to show higher oxidation states. 4 Among lanthanoids, Ln(III) compounds are predominant. However, occasionally in solutions or in solid compounds, +2 and +4 ions are also obtained. Give reason.
ANS: Lanthanoids show +3 oxidation state mostly as 2 electrons from outer 6s orbital and one electron from 5d orbital take part in bond formation. Some show +2 and +4 oxidation states due to stability of half filled and completely filled 4f orbitals. 5 Out of Cu 2 Cl 2 and CuCl 2 , which is more stable and why? 1 ANS: CuCl 2 is more stable due to more hydration energy. 6 Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them. Why?
ANS: It is due to almost same size (Zr = 160 pm, Hf = 159 pm) which is due to lanthanoid contraction. 7 E° of Cu is +0.34 V while that of Zn is – 0.76 V. Explain. 1 ANS: It is because Cu(s) is more stable than Cu2+^ due to high ionisation enthalpy which is not overcome by its hydration energy. In the case of Zn, after removal of 2 electrons from 4s orbtital, stable 3d^10 configuration is acquired. 8 Why do the transition metals have higher enthalpy of atomisation? In 3d series (Sc to Zn), which element has lowest enthalpy of atomisation and why?
ANS: It is due to the involvement of greater number of unpaired electrons from (n – 1)d as well as ns orbitals in the strong inter-atomic metallic bonding. Zinc has lowest enthalpy of atomisation due to larger size and in the absence of unpaired electrons, it forms weak metallic bond. 9 For the first row transition metals, the E° values are given below: Explain the irregularity in the above values.
ANS: It is due to irregular variation of sublimation enthalpies and ionisation enthalpies of elements of 3d transition series. 10 How would you account for the following? (i) Cr2+^ is reducing in nature while with the same d-orbital configuration (d^4 ), Mn3+^ is an oxidising agent. (ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series. or N ame the element showing maximum number of oxidation states among the first series of transition metal Sc (21) to Zn (30).
ANS: (i) It is because Cr2+^ loses electron to become Cr3+^ which is more stable due to half filled t2g orbitals, whereas Mn3+^ will gain electron to become Mn2+ which is more stable due to half filled
d-orbitals. (ii) Manganese. It is due to large number of unpaired electrons in d-orbitals in middle of the series. Mn (25) 4s^2 3d^5. 11 Explain the following observations giving an appropriate reason for each. (i) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e. 3rd series). (ii) Mn2+^ is much more resistant than Fe2+^ towards oxidation.
ANS: (i) Due to lanthanoid contraction, effective nuclear charge remains almost same therefore, metallic radii are nearly same, therefore, metal-metal bonding is more. (ii) Mn2+^ (3d^5 ) has stable electronic configuration, therefore, it does not get oxidised. Fe2+^ (3d^6 ) gets oxidised to form Fe3+(3d^5 ) which is more stable. 12 State reasons for the following: (i) Actinoids exhibit greater range of oxidation states than lanthanoids. (ii) Unlike Cr3+, Mn2+, Fe3+^ and the subsequent other M2+^ ions of the 3d series of elements, the 4d and the 5d series metals generally do not form stable cationic species.
ANS: (i) It is due to poor shielding effects of 4f and 5f electrons, more number of electrons take part in bond formation in actinoids. (ii) It is because energy required to remove electron is more due to greater effective nuclear charge which is due to lanthanoid contraction. 13 Assign reasons for each of the following: (i) T ransition metals generally form coloured compounds. (ii) Manganese exhibits the highest oxidation state of + 7 among the 3d series of transition elements.
ANS: (i) It is because transtion metals have unpaired electron in d-orbitals and undergo d-d- transitions by absorbing light from visible region and rediate complementary colour. (ii) Mn has electronic configuration (Ar)4s^2 3d^5 and all the electrons in ‘s’ as well as ‘d’ orbitals can take part in bond formation, therefore, it shows +7 (highest) oxidation state. 14 Explain the following observations: (i) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements. (ii) T ransition elements and their compounds are generally found to be good catalysts in chemical reactions.
ANS: (i) It is because atomic mass increases more than atomic volume, therefore, density increases from titanium (Z = 22) to copper (Z = 29). (ii) It is because they show variable oxidation states and have vacant d-orbitals forming unstable intermediates which readily change into products. 15 Explain the following observations: (i) Transition elements generally form coloured compounds. (ii) Zinc is not regarded as a transition element.
ANS: (i) It is due to presence of unpaired electrons in d-orbitals, therefore, they undergo d-d transitions by absorbing light from visible region and radiate complementary colour. (ii) It is because neither Zn nor Zn2+^ ion has incompletely filled d-orbital. 16 Complete the following equations:
ANS:
Complete the following equations:
whereas Ti3+^ is coloured due to presence of unpaired electrons, and undergoes d-d transition by absorbing light from visible region and radiate complementary colour. 25 How would you account for the following? (i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series. (ii) The E° value for the Mn3+/Mn2+^ couple is much more positive than that for Cr3+/Cr2+^ couple or Fe3+/Fe2+^ couple. (iii) The highest oxidation state of a metal is exhibited in its oxide or fluoride.
ANS: (i) It is due to lanthanoid contraction which is due to poor shielding effect of f-electrons. (ii) It is because Mn2+^ is more stable than Mn3+^ due to stable half filled 3d5 configuration, whereas Cr3+(t2g^3 ) and Fe3+(3d^5 ) are more stable than Cr2+^ and Fe2+^ respectively. (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. 26 Give reasons for each of the following: (i) Size of trivalent lanthanoid cations decreases with increase in the atomic number. (ii) T ransition metal fluorides are ionic in nature, whereas bromides and chlorides are usually covalent in nature. (iii) Chemistry of all the lanthanoids is quite similar.
ANS: (i) It is due to poor shielding effect of f-electrons, effective nuclear charge increases, so, ionic size decreases. (ii) F is more electronegative than Cl and Br, therefore, fluorides are ionic; whereas chlorides and bromides are covalent. (iii) It is due to similar ionic size which is due to lanthanoid contraction, they resemble in their properties. 27 A solution of KMnO 4 on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
ANS: Oxidising behaviour of KMnO 4 depends upon pH of solution. Different compounds with different colours are formed at different pH. 28 Identify A to E and also explain the reactions involved. 3
ANS:
29 When a chromite ore(A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallised from the solution. When compound (C) is treated with KCl, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.
ANS:
30 When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.
ANS: (a) (i) Oxygen can form double bond, therefore, it can form Mn 2 O 7 , whereas ‘F’ cannot form double bonds, so, it can form MnF 4. (ii) Transition metals show variable oxidation states, therefore, they and their compounds act as catalyst. (b) 3MnO 42 –^ + 4H+^ → MnO 2 + 2MnO 4 –^ + 2H 2 O 34 (a) How would you account for the following: (i) The chemistry of actinoids is more complicated as compared to lanthanoids. (ii) Transition metals form complex compounds. (b) Complete the following equation:
ANS: (a) (i) It is because they are radioactive and some of them have very short half life. (ii) It is due to small size, high charge and availability of d-orbitals of suitable energy. (b) 35 Explain the following: (i) The transition elements have great tendency for complex formation. (ii) There is a gradual decrease in the atomic sizes of transition elements in a series with increasing atomic numbers. (iii) Lanthanum and Lutetium do not show colouration in solutions. (At. No.: La = 57, Lu = 71)
ANS: (i) It is due to presence of vacant d-orbitals of suitable energy, smaller size of cations and higher charge. (ii) It is due to increase in effective nuclear charge gradually because unpaired electrons increases in the beginning with no repulsion. There is repulsion between paired electrons after middle of series, therefore, effective nuclear charge increases a little. (iii) It is due to absence of unpaired electrons, they do not absorb light from visible region and cannot undergo f-f transition. and do not radiate colour. 36 (a) Complete the following chemical equations for reactions in aqueous media : (i) Cr 2 O 72 –^ + H+^ + Fe2+^ → (ii) MnO 4 –^ + I–^ + H+^ → (b) How many unpaired electrons are present in Mn2+^ ion (At. no. of Mn = 25)? How does it influence magnetic behaviour of Mn2+^ ions?
ANS: (a) (b) Mn2+: 3d^5 4s^0 has 5 unpaired electrons. It is highly paramagnetic and attracted by magnet. 37 When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The gas taken in excess, reacts with NH 3 to give an explosive compound (C). Identify compounds A, B and C.
ANS: ‘A’ is MnO 2 which is brownish black. 38 (a) What are the different oxidation states exhibited by the lanthanoids? (b) Write two characteristics of the transition elements. (c) Which of the 3d-block elements may not be regarded as the transition elements and why?
ANS: (a) Lanthanoids, mostly show +3 oxidation state but some of them show +2 and + oxidation states also due to the stability of electronic configuration (4f 0 , 4f 7 and 4f 14 ). (b) (i) They show variable oxidation states. (ii) They form coloured ions. (c) Zn may not be regarded as transition metal because neither Zn nor Zn2+^ have incompletely filled d-orbital.
39 (a) Transition metals can act as catalysts, why? How does Fe(III) catalyse the reaction between iodide ion and persulphate ions? (b) Mention any three processes where transition metals act as catalysts.
ANS: (a) Transition metals act as catalyst because they show variable oxidation states as explained below: Reaction between iodide and persulphate ions is (b) 40 (a) Complete the following equations: (i) Cr 2 O 72 –^ + 2OH–^ → (ii) MnO 4 –^ + 4H+^ + 3 e–^ → (b) Account for the following: (i) Zn is not considered a transition element. (ii) T ransition metals form a large number of complexes. (iii) T he E° value for the Mn3+/Mn2+^ couple is much more positive than that for Cr3+/Cr2+^ couple.
ANS: (a) (b) (i) It is because neither Zn nor Zn2+^ has incompletely filled d-orbital. (ii) It is due to small size, higher charge and presence of vacant d-orbitals of suitable energy. (iii) It is because Mn2+^ is more stable than Mn3+^ due to half filled (3d^5 ) d-orbitals, whereas Cr3+^ is more stable than Cr2+^ due to half filled (t2g^3 ) orbitals. 41 (i) With reference to structural variability and chemical reactivity, write the differences between lanthanoids and actinoids. (ii) N ame a member of the lanthanoid series which is well known to exhibit +4 oxidation state. (iii) Complete the following equation: MnO 4 –^ + 8H+^ + 5 e–^ → (iv) Out of Mn3+^ and Cr3+, which is more paramagnetic and why? (Atomic nos.: Mn = 25, Cr = 24)
ANS: (i) (ii) Ce shows +4 oxidation state. (iii) MnO 4 –^ + 8H+^ + 5e–^ → Mn2+^ + 4H 2 O (iv) Mn3+^ (3d^4 ) has 4 unpaired electrons, therefore, it is more paramagnetic than Cr3+^ (3d^3 ) which has three unpaired electrons. 42 (a) Complete the following chemical equations: 5
Which of the following pairs has the same ionic size?
ANS: (a) have similar size due t o lanthanoid contraction. 48 Acidified K 2 Cr 2 O 7 solution turns green when SO 2 gas is passed through it due to formation of (a) Cr 2 (SO 4 ) 3 (b) CrO 42 – (c) Cr 2 (SO 3 ) 3 (d) CrSO 4
ANS: (a) It is due to formation of chromium sulphate. 49 The stability of Mn2+, Fe2+, Cr2+, Co2+^ is in order of (At No. of Mn = 25, Fe = 26, Cr = 24, Co = 27) (a) Mn2+^ > Fe2+^ > Cr2+^ > Co2+ (b) Fe2+^ > Mn2+^ > Co2+^ > Cr2+ (c) Co2+^ > Mn2+^ > Fe2+^ > Cr2+ (d) Cr2+^ > Mn2+^ > Co2+^ > Fe2+
ANS: (a) Mn2+^ (3d^5 ) is most stable, Fe2+^ (3d^6 ), Cr2+(3d^4 ), Co2+(3d^7 ) 50 Which of the following does not give O 2 on heating? (a) K 2 Cr 2 O 7 (b) (NH 4 ) 2 Cr 2 O 7 (c) KClO 3 (d) Zn(ClO 3 ) 2
ANS:
51 Which of the following lanthanoid ion is diamagnetic? (At No. of Ce = 58, Sm = 62, Eu = 63 Yb = 70) (a) Eu2+^ (b) Yb2+ (c) Ce2+^ (d) Sm2+
ANS: (b) Yb2+^ (4f^14 ) does not have unpaired election, therefore, diamagnetic. 52 The reaction of acidified KMnO 4 and H 2 O 2 gives (a) Mn4+^ and O 2 (b) Mn2+^ and O 2 (c) Mn2+^ and O 3 (d) Mn4+^ and MnO 2
ANS: (b) 2MnO 4 + 6H+^ + 5H 2 O 2 → 2Mn2+^ + 8H 2 O + 5O 2 53 Magnetic moment of 2.83 BM is given by which of the following ion? (a) Ti3+^ (b) Ni2+ (c) Cr3+^ (d) Mn2+
ANS: (b) Ni2+^ has 2 unpaired electrons. 54 The colour of KmnO 4 is due to (a) L → M charge transfer transition (b) σ → σ* transition (c) M → L charge transfer transition (d) d → d transition.
ANS: (a) It is due to L → M charge transfer transition by absorbing light from visible region and radiates purple colour. 55 KMnO 4 is not acidified by HCl instead of H 2 SO 4 because (a) H 2 SO 4 is stronger acid than HCl (b) HCl is oxidised to Cl 2 by KMnO 4 (c) H 2 SO 4 is dibasic acid (d) rate is faster in presence of H 2 SO 4
ANS: (b) 2KMnO 4 + 16 HCl → 2KCl + 2MnCl 2 + 5Cl 2 + 2H 2 O 56 Out of Mn 2 O 7 , V 2 O 3 , V 2 O 5 , CrO, Cr 2 O 3 , the basic oxides are (a) Mn 2 O 7 , V 2 O 3 (b) V 2 O 3 , V 2 O 5 (c) V 2 O 5 , CrO (d) V 2 O 3 and CrO
ANS: (d) V 2 O 3 and CrO are basic oxides due to lower, oxidation states. 57 The oxidation state of Cr in final product formed by reaction of KI and acidified dichromate solution is (a) +4 (b) + (c) +2 (d) +
ANS: (d) Cr3+^ is formed. 58 KMnO 4 gets reduced to (a) K 2 MnO 4 in neutral medium (b) MnO 2 in acidic medium (c) Mn2+^ in alkaline medium (d) MnO 2 in neutral medium
ANS:
59 The electronic configuration of Cu(II) is 3d^9 whereas that of Cu(I) is 3d^10. Which of the following is correct? [NCERT Exemplar Problem] (a) Cu(II) is more stable (b) Cu(II) is less stable (c) Cu(I) and Cu(II) are equally stable (d) Stability of Cu(I) and Cu(II) depends on nature of copper salts
ANS: (a) Cu(II) is more stable due to higher hydration energy. 60 Metallic radii of some transition elements are given below. Which of these elements will have highest density? [NCERT Exemplar Problem] Element Fe Co Ni Cu Metallic radii/pm 126 125 125 128 (a) Fe (b) Ni (c) Co (d) Cu
ANS: (d) Cu has highest density due to greater atomic mass. 61 Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state? (a) Ag 2 SO 4 (b) CuF 2 (c) ZnF 2 (d) Cu 2 Cl 2
ANS: (b) CuF 2 is coloured due top resence of unpaired electron in d-orbital 62 On addition of small amount of KMnO 4 to concentrated H 2 SO 4 , a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following. [NCERT Exemplar Problem] (a) Mn 2 O 7 (b) MnO 2 (c) MnSO 4 (c) Mn 2 O 3
ANS: (a) It is due to formation of Mn 2 O 7. 63 Which of the following reactions are disproportionation reactions? (i) Cu+^ → Cu2+^ + Cu
ANS: (a) (iii) (b) (iv) (c) (ii) (d) (v) (e) (i) 70 Match the compounds/elements given in Column I with uses given in Column II. Column I (Compound/element) Column II (Use) (a) Lanthanoid oxide (i) Production of iron alloy (b) Lanthanoid (ii) Television screen (c) Misch metal (iii) Petroleum cracking (d) Magnesium based alloy is constituent of (iv) Lanthanoid metal + iron (e) Mixed oxides of lanthanoids are employed (v) Bullets (vi) In X-ray screen
ANS: (a) (ii) (b) (i) (c) (iv) (d) (v) (e) (iii) 71 Match the properties given in Column I with the metals given in Column II. Column I (Property) Column II (Metal) (a) An element which can show +8 oxidation state (i) Mn (b) 3d block element that can show upto +7 oxidation state (ii) Cr (c) 3d block element with highest melting point (iii) Os (iv) Fe
ANS: (a) (iii) (b) (i) (c) (ii) Cr due to maximum number of unpaired electrons. 72 Match the statements given in Column I with the oxidation states given in Column II. Column I Column II (a) Oxidation state of Mn in MnO 2 is (i) + (b) Most stable oxidation state of Mn is (ii) + (c) Most stable oxidation state of Mn in oxides is (iii) + (d) Characteristic oxidation state of lanthanoids is (iv) + (v) + 7
ANS: (a) (iii) (b) (i) (c) (v) (d) (ii) 73 In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Both assertion and reason are true, and reason is the correct explanation of the assertion. (b) Both assertion and reason are true but reason is not the correct explanation of assertion. (c) Assertion is not true but reason is true. (d) Both assertion and reason are false. Assertion: Cu2+^ iodide is not known. Reason: Cu2+^ oxidises I–^ to iodine.
ANS: (a) Both assertion and reason are true, and reason is the correct explanation of the assertion. Cu2+^ 2I–^ → 2Cu+^ + I 2
ANS: 10 CO 2
ANS: 3I 2
ANS: 6 Fe3+ 77 Cu2+^ is reduced by CN–^ to Cu+^ which forms the complex [Cu(CN) 4 ]^3 –. [True/False] 1 ANS: True. 78 The number of moles of Mohr’s salt required per mole of dichromate ion are 6. [True/False] 1 ANS: True 79 The colour of light absorbed by an aqueous solution of CuSO 4 is orange red. [True/False] 1 ANS: True 80 The electronic configuration of Gd (64) is (a) [Xe] 4f^7 5d^1 6s^2 (b) [Xe] 4f^6 5d^2 6s^2 (c) [Xe] 4f^8 6s^2 (d) [Xe] 4f^9 5s^1
ANS: (a) because half-filled f-orbitals are more stable. 81 Which of the following statements related to lanthanoids is incorrect? (a) Eu shows +2 oxidation state (b) Pr(OH) 3 to Lu(OH) 3 , basicity decreases (c) All lanthanoids more reactive than Al (d) Ce4+^ is used as oxidising agent
ANS: (c) All are not more reactive than Al. 82 Name the gas that can readily decolourised by acidified KMnO 4 solution. (a) SO 2 (b) NO 2 (c) P 2 O 5 (d) CO 2
ANS: (a) SO 2 because it is good reducing agent. 83 The reason for greater range of oxidation state of actinoids is due to (a) actinoid contraction (b) 5f, 6d, 7s levels have comparable energies (c) 4f and 5d levels are close in energies (d) the radioactive nature of actinoids
ANS: (b) It is due to comparable energies of 5f, 6d, 7s, electron from these orbitals take part in bond formation. 84 The correct order of ionic radii Y3+, La3+, Eu3+^ and Lu3+^ is (a) Y3+^ < La3+^ < Eu3+^ < Lu3+ (b) Lu3+^ < Eu3+^ < La3+^ < Y3+
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92 Transition elements form binary compounds with halogens. Which of the following elements will form MF 3 type compounds? [NCERT Exemplar Problem] (a) Cr (b) Co (c) Cu (d) Ni
ANS: (a) and (b) CrF 3 , CoF 3 are easily formed ∵ Cr3+^ and Co3+^ are stable. 93 Which of the following will not act as oxidising agents? [NCERT Exemplar Problem] (a) CrO 3 (b) MoO 3 (c) WO 3 (d) CrO 42 –
ANS: (b) MoO 3 and (c) WO 3 because their +6 oxidation states are more stable. 94 Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows + oxidation state because _______________. [NCERT Exemplar Problem] (a) it has variable ionisation enthalpy (b) it has a tendency to attain noble gas configuration (c) it has a tendency to attain f^0 configuration (d) it resembles Pb4+
ANS: (b) and (c) It has stable electronic configuration. 95 Match the solutions given in Column I and the colours given in Column II. Column I (Aqueous solution of salt) Column II (Colour) (a) FeSO 4 .7H 2 O (i) Green (b) NiCl 2 .4H 2 O (ii) Light pink (c) MnCl 2 .4H 2 O (iii) Blue (d) CoCl 2 .6H 2 O (iv) Pale green (e) Cu 2 Cl 2 (v) Pink (vi) Colourless
ANS: (a) (iv) (b) (i) (c) (ii) (d) (v) (e) (vi) 96 Match the property given in Column I with the element given in Column II. Column I (Property) Column II (Element) (a) Lanthanoid which shows +4 oxidation state (i) Pm (b) Lanthanoid which can show +2 oxidation state (ii) Ce (c) Radioactive lanthanoid (iii) Lu (d) Lanthanoid which has 4f^7 electronic configuration in +3 oxidation state (iv) Eu (e) Lanthanoid which has 4f^14 electronic configuration in +3 oxidation state (v) Gd (vi) Dy
ANS: (a) (ii) (b) (iv) (c) (i) (d) (v) (e) Lu3+(iii)
97 Match the properties given in Column I with the metals given in Column II. Column I (Property) Column II (Metal) (a) Element with highest second ionisation enthalpy (i) Co (b) Element with highest third ionisation enthalpy (ii) Cr (c) M in M (CO) 6 is (iii) Cu (d) Element with highest heat of atomisation (iv) Zn (v) Ni
ANS: (a) (iii) (b) (iv) (c) (ii) (d) (i) 98 In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Both assertion and reason are true, and reason is the correct explanation of the assertion. (b) Both assertion and reason are true but reason is not the correct explanation of assertion. (c) Assertion is not true but reason is true. (d) Both assertion and reason are false. Assertion: Actinoids form relatively less stable complexes as compared to lanthanoids. Reason: Actinoids can utilise t heir 5f orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbital for bonding.
ANS: (c) Assertion is not true but reason is true. 99 Cr3+^ is _________ stable than Mn2+. 1 ANS: more 100 The general molecular formula of compounds formed by heating lanthanoids with sulphur is ________.
ANS: Ln 2 S 3 101 Cr in CrO 42 –^ is sp^3 hybridised and tetrahedral shape. [True/False] 1 ANS: True 102 MnO 4 –^ and MnO 42 –^ have tetrahedral structure. [True/False] 1 ANS: True
