GOVERNMENT OF KARNATAKA DEPARTMENT OF PRE UNIVERSITY EDUCATION, BANGALORE II PU 2014 VIKASANA TELE PROGRAME SUBJECT : CHEMISTRY TOPIC : The d- and f- block Elements MANJUNATH BHAT LECTURER IN CHEMISTRY GOVT. PRE UNIVERSITY COLLEGE, BRAHMAVAR, UDUPI DIST. MOBILE NO. 9448416987 EMAIL :[email protected] 1 The transition elements (d block) and inner transition elements (f block) in the periodic table The d-block of the periodic table contains the elements of the groups 3-12 in which the (n-1)d - orbital's are progressively filled with electrons in each of the four long periods. There are mainly three series of the transition metals, 3d series - Sc to Zn 4d series - Y to Cd 5d series - La to Hg (omitting Ce to Lu). 6d series - starting from Ac is still incomplete The d–block elements are also known as Transition metals , they possess properties that are in between the s– and p– . block elements All transition elements are metals, whereas main-group elements in each period change from metal to nonmetal Characteristic properties of transition elements are; Variable oxidation states - due to the availability of both ns & (n – 1 ) d electrons for bond formation. Paramagnetic behavior – due to the presence of one or more unpaired electrons. Colored compound formation - due to presence of unpaired electron & d-d transition of these electrons. Catalytic activity - due to their ability to adopt multiple oxidation states and to form complexes and provide a suitable surface for the reaction to take place. Alloy formation – due to similar radii . Complex formation - due to Small size and high nuclear charge of cations The properties of transition metal compounds are related to the electron configuration of the metal ion. The electrons in (n-1) d orbitals are very singificant , because they influence their properties , such as variable o.s, coloured compounds , magnetic character, catalytic property , complex formation etc. Element Sc Ti V Cr Mn Fe Co Ni Cu Zn Z 21 22 23 24 25 26 27 28 29 30 Electronic Arrangement 3d [Ar] [Ar] [Ar] [Ar] [Ar] [Ar] [Ar] [Ar] [Ar] [Ar]                                          4s           Which one of the following pairs of ions have the same electronic configuration; a) Ni2+, Co3+ 8 4s 0 , Co3+ [Ar] 3d64s0 2+ 2+ Ni2+ [Ar] 3d b) Fe , Mn Fe2+ - [Ar] 3d64s0 , Mn2+ [Ar]3d54s0 c) Fe 3+, Mn 2+ Ans: c Fe3+ - [Ar] 3d54s0 ,Mn2+[Ar]3d54s0 d) Sc+3 , Ti+3 Sc+3 - [Ar] 3d04s0 , Ti+3 - [Ar]3d24s0 Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d 5 4s 0 .What is its atomic number? a.25 b. 26 X +3 - [ Ar]3d 5  X - [Ar]3d 6 4s 2 i.e atomic no. is 26 c. 27 Ans. b d. 24 Which of the following do not have valence electron in 3d sub shell? a) Fe (III) Fe3+ - [Ar] 3d54s0 Cr+ - [Ar] 3d54s0 c) Cr (I) Ans. d 2+ - [Ar] 3d54s0 MnMn b) (II) Sc3+ - [Ar] 3d04s0 d)valence Sc (III) Fe (III), Mn (II) , Cr (I) all have electron in 3d sub shell, but Sc (III) has no electron in 3d sub shell The transitional elements elements show paramagnetism, form alloyspair effect a.Transition Show diamagnetism b.Undergo inert and do not undergo inert pair effect . But they Show variable oxidation states. In addition to the electrons of ns electrons of incompletely filled (n1)d involve in bonding. c. Do not form alloys Ans. d d. Show variable oxidation states Which one of the following statement is not true regarding variable oxidation state of transition metals. dblock elements contain partially filled penultimate orbitals i.e d a.Theorbital. variable oxidation state is due to incomplete filling of (n-1) d orbitals Oxidation state of a transition metal varies , in which they differ from other by unity . E.g Vanadium b. Oxidation stateeach differ from each other by unity shows o.s. of +2, +3, +4 and +5 Generallystate non differ transition o.s differ by of a unit c. Oxidation fromelements each other by unit twoof two. E.g S shows o.s of -2 , +2, +4 and +6. d. It isIt due to to involvement inbonding bonding is due involvementofof ns nsand and(n-1) (n-1) d electrons electrons in Ans. c Transitional elements exhibit variable valences because they release electrons from the following orbital's. a.ns orbitals b. (n - 1)d and ns orbitals In transition elements , ns and (n-1)d orbital's have very little difference in energy , hence they release electrons from both these orbital's and exhibit variable valences c. ns and np orbital's Ans. b d. (n - 1)d orbital's Out of Cr2+ and Cr3+,which one is stable in aqueous solution? a.Cr2+ b.Cr3+ Cr3+ is more stable in aqueous solution due to higher hydration energy which is due to smaller size and higher charge. c. Both Cr2+ & Cr3+ Ans. b. d. None of the two In transition elements , the incoming electron occupies (n-1)d orbital in preference to. a.ns b. np In transition elements , (n-1)d orbital's are filled with electrons after filling ns orbital. (n-1)d orbital has lower energy than np orbital , hence (n-1)d is filled in preference to np orbital. c. (n-1)p Ans.b. d. (n-1)s Zn and Cd do not show variable valence , because : a. They have only two electrons in outermost sub shells b. Their d – sub shells are incomplete c. Their d – sub shells are complete d. They are relatively soft metals Zn - [Ar] 4s2 3d10 Cd - [Kr] 5s2 4d10 Since their d – sub shells are complete , they show fixed valence of 2. Ans. c Which of the following has the maximum number of unpaired ‘d’ electrons? a.Co3+ b. Zn2+ c. Cu+ d. Ni3+ Cu+ - [Ar] 3d104s0 – 0 Co3+ - [Ar] 3d64s0 - 4 Zn2+ - [Ar] 3d104s0 - 0 Ans. a Ni3+ - [Ar] 3d7 4s0 – 3 In which of the following compounds the radius of chromium is smallest? a. CrO2 c. K2Cr2O7 Ans. c The ionic radii decreases as charge on cation b. Cr 2 O3 increases ( i.e higher oxidation state) In CrO2 o.s of Cr is + 4, Cr2O3 and CrCl3 - + 3 and in K2Cr2O7 -- +6 Hence radius of chromium is smallest in d.CrCl 3 K2Cr2O7 8. The outer most electronic configuration of an ion in +3 oxidation state is [Ar] 3d54s0, the element belong to The electronic configuration of an ion in +3 54s0 oxidation state is [Ar] 3d a. III group b. 18th group The electronic configuration of the element is [Ar] 3d64s2. c. 5th group d. 8th group It belong to 8th group Ans. d The minimum oxidation state shown by transition element is +2 which of the following is an exception? a. Fe Fe- [Ar] 3d64s2 c. Ni Ni- [Ar] 3d84s2 b. Co Co- [Ar] 3d74s2 d. Cu Cu- [Ar] 3d104s1 Oxidation States of the 1st Row Transition Metals Iron , cobalt and nickel shows common minimum oxidation state of +2, copper shows +2 as well as +1 oxidation states. Ans. d 22.1 Within each transition series the oxidation state, a.Decreases regularly in moving from left to right across a period . b. First increases till the middle of the period & then decreases c. . First decreases till the middle of the period & then increases d. Increases regularly in moving from left to right across a period Oxidation States of the 1st Row Transition Metals (most stable oxidation numbers are shown in red) O.S first increases till the middle decreases . Ans. b of the period & then 22.1 Which of the following transition metal does not show the variable oxidation state ? a.Cu b. Zn Cu- [Ar] 3d104s1 c. Mn Mn- [Ar] Zn- [Ar] 3d104s2 d. V 3d54s2 V- [Ar] 3d34s2 Mn, V and Cu shows variable oxidation state Zn is the last element in the 3d series which shows +2 o.s. Ans. b First ionisation enthalpy of Cr is lower than that of Zn, because, a. Cr is smaller atom than Zn b. Removal of an electron from 4s level needs less energy. c. Removal of one electron from 4s orbital of Cr does not change the d configuration. d. Removal of one electron from 4s orbital of Cr change the d configuration. IE1 of Cr is lower, because removal of one electron from 4s orbital Cr does not change the d configuration . Cr (z= 24 , 3d5 4s1)  Cr+ (3d5 4s0) ------ IE1 IE1 value for Zn is higher, because removal of electron from 4s level needs more energy. Zn (3d10 4s2)  Zn+ (3d10 4s1)  Ans. c IE1 (Zn) > IE1 (Cr) ------ IE1 Species having same number Zn+2 ion is isoelectronic with of electrons are known as iso electronic species . Zn+2 - [Ar] 3d10 4s0 a. Cu+ Cu+ b. Cu+2 - [Ar]3d10 4s0 Cu+2 c. Ni+2 Ni+2 - [Ar]3d8 Ans. a -[Ar] 3d9 4s0 d. Fe+2 4s0 Fe+2 - [Ar]3d6 4s0 Which of the following shows positive E0 value? a.Mn/Mn2+ c.Cu/Cu2+ Ans. c b. Ni/Ni2+ Mn2+, Zn2+ shows negative E0 value due stability of half filled and completely filled configurations, but for Ni2+ it is due to highest –ve enthalphy of hydration Cu2+ shows +ve E0 value. d. Zn/Zn2+ Standard reduction potential of transition metals is a.Negative c. Zero Ans .d b. positive T.M ions like Mn2+, Zn2+ shows negative E0 value due stability of half filled and completely filled 2+ it is due to highest –ve configurations. Ind.Nimay be positive or negative enthalphy of hydration. Cu2+ shows +ve E0 value due to higher + hydration ethalpy. Metallic radii of some transition elements are given below. Which of these elements will have highest density? Element Fe Co Ni Cu Metallic radii/pm 126 125 a.Fe b.Ni c. Co d. Cu 124 128 As atomic radius decreases, density increases along the period, hence Cu has highest density Ans. d Which one of the following transition metal has lowest density? a. Copper b. Nickel As atomic radius decreases density increases along the period, Scandium has larger atomic radius , hence it has lowest density (d = 3.0g/cm3) c. Scandium Ans. c d. Zinc The melting points and boiling points up to middle of 3d- series a.Increases b. decreases As the no. of unpaired electrons increases, metallic bond becomes strong due involvement of ns & (n-1)d electrons, increases up to middle , they become max. in the element having d5 confn (Cr). Then they show down word trends. c. Remains same d. no regular trend The m.p. of Mn is lower because it has stable half filled conf. so electrons are held tightly, delocalization is less & metallic bond is weak. Ans . a Which one of the following first row transition metal is expected to have the highest 3rd IE? V - [Ar] 3d34s2 Mn - [Ar] 3d54s2 a. V (z =Removal 23) of b.Mn 3rd electron from Mn+2 requires large amount(z of = energy due its stable d5 configuration. Cr - [Ar] 3d54s1 c.Cr (z = 24) Ans. b 25) Fe - [Ar] 3d64s2 d. Fe (z = 26) Which of the following oxide of vanadium is most basic ? a. VO 2 When metal in oxide has higher o.s itb.VO is acidic and when metal in oxide has lower o.s it is basic. V2O5 - acidic oxide. VO, V2O3 - basic oxides. c. V2O5, VO2 - is amphoteric Ans. a d.V2O3 Which of the following is not an acidic oxide? a. Mn2O7 b.CrO3 In Mn2O7 ,CrO3 , V2O5, are acidic oxides. Fe2O3 is amphoteric c. V2O5, Ans. d d.Fe2O3 Which of the following compound of manganese is amphoteric? a. MnO b. Mn2O3 Soln. When metal in oxide has higher o.s it is acidic, metal in oxide has lower o.s it is basic and metal in oxide has intermidiate o.s , it is amphoteric. MnO and Mn2O3 are in lower o.s are basic oxides Mn2O7 is an acidic oxide . Mn2O5 is amphoteric c. Mn2O7 Ans. d d. Mn2O5 The “spin-only” magnetic moment [µ] of Ni 2+ in aqueous solution would be (Ni ,z = 28) . a. 2.84 b. 4.90 Ni +2 - [Ar]3d8 4s0 it contains 2 unpaired electrons If n = 2 c. 0 Ans. a µ = n(n+2) = 2.84 d. 1.73 The spin only magnetic moment of an ion is 3.87 B.M the unpaired electrons present in it are, a. 2 b. 3 µ = n(n+2) (3.87) 2 = n(n+2) 15 = n 2 + 2n By approx n = 3 c. 4 Ans = b d. 5 Titanium shows magnetic moment of 1.73 BM in its compound. What is the oxidation state of Ti in the compound? a. +1 b.+4 µ = n(n+2) (1.73) 2 = n(n+2) 3.0 = n 2 + 2n By apporx n = 1 No. of unpaired electron = 1 Ti (z = 22) is in +3 o.s Ti +3 - [Ar]3d1 4s0 c.+3 Ans. c d. +2 The magnetic moment of Manganese is 5.92 BM in its compound. What is the symbol of the ion? a. Mn+ b. Mn 2+ µ = n(n+2) (5.92) 2 = n(n+2) 35 = n 2 + 2n By apporx n = 5 No. of unpaired electron = 5 Mn (z = 25) in +2 o.s has 5 unpaired electrons. Mn+2 - [Ar]3d5 4s0 c. Mn3+ Ans. b d. Mn 7+ Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behavior? a. [Cr(CN) 6] 3Cr+3 -[Ar]d3- 3 c.[Fe(CN) 6] 3- b. [Mn(CN) 6] 3Mn+3 -[Ar]d4 - 4 d.[Co(CN) 6] 3Co+3 -[Ar]d4- 4 5- 5 Fe+3 -[Ar]d (At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27) Paramagnetic behavior depends on the no. of unpaired electrons Ans. a Which one of the following shows highest magnetic moment ? (a) Fe2+ (b) Co2+ Fe+2 -[Ar]d6 - 4 Co+2 -[Ar]d7 - 3 (c) Cr3+ Cr+3 -[Ar]d3 - Ans.a (d) Ni2+ 3 Ni+2 -[Ar]d2 - 2 Which pair of compounds is expected to show similar colour in aqueous medium? a. FeCl2 and CuCl2 b. FeCl2 and MnCl2 Fe+2 (FeCl2 ) - 4 unpaired electrons Cu+2 (CuCl2) electron. - 1 unpaired  VOCl2 and CuC l2 expected to show similar colour in aqueous medium . c. VOCl2 and FeCl2 Mn+2 (MnCl2 ) - 5 unpairedAns: electrons d d. VOCl2 and CuCl2 V+4 (VOCl2 ) - 1 unpaired electron. Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds are coloured ? a.TiCl4 b. CuF2 c. ZnF2 d. Cu2Cl2 Ti4+ has 3d0 Configuration Cu+ and Zn2+ both have 3d10 Configuration Cu2+ also has 3d9 Configuration. Ans. b CuF2 Ti(H2O)6]3+ is purple in complimentary colour of colour because a.Greenish Yellow b.Red c.Green d.Blue is Each color has a complementary color; the one opposite it on the artist’s wheel. The color an object exhibits depends on the wavelengths of light that it absorbs. Purple colour is complimentary colour of Greenish Yellow. The color of + [Ti(H2O)6]3 Green and yellow light are absorbed while other wavelengths are transmitted. This gives a purple color. The hydrated Ti3+ ion is purple. Ans. a When a compound of transition element is dissolved in a solution of salt then it produces a.Simple ions b. complex ions Eg .4KCN + Fe(CN)2 ----- [Fe(CN)6 ] 4 - + 4 K + c. double salts Ans. b d.strong anions d-block element form complexes due to a.Large size and high nuclear charge of cations b. Small size and high nuclear charge of cations d-block elements and their ions show strong tendency to form complexes due to Small size and high nuclear charge of cations c. Small size and low nuclear charge d. Availabilty of 4s electrons Ans.b The catalytic activity of the transition metals and their compounds is due to a.The chemical reactivity b. Presence of unpaired electrons Their ability adopt multiple oxidation states (form c.Their unfilledtod-orbitals unstable intermediates) and availability of adsorption sites d.Their ability to adopt multiple oxidation states and availability of adsorption sites Ans d. V2O5 acts as catalyst , because a.it has large surface area. b.It can form unstable intermediates . c.It can form unstable intermediates .which readily change into products. d. All the above statement are correct It has large surface area. It can form unstable intermediates . It can form unstable intermediates .which readily change into products. Ans. d. When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning, but becomes instantaneous after some time because a. CO 2 5C is formed as the -product. O 2− + 2MnO + 16H+ —> 2Mn2+ + 8H O + 10CO 2 4 4 2 b. Reaction is exothermic. c. MnO4– catalyses the reaction. d. Mn2+ acts as autocatalyst. Ans. d 2 In the reaction, 2 MnO4- + 6H+ + 5NO2-  2Mn 2+ + 3H2O + 5NO3 a.MnO4- is reduced b.Oxidation number of Mn decreases from + 7 to + 2 c. Oxidation number of N increases from + 3 to + 5 - + 6H+ + 5NO -  2Mn 2+ + 3H O + 5NO – 2 MnO d. All the4 above statements are correct 2 2 3 Ans. d KMnO4 acts as an oxidising agent in acidic medium. The number of moles of KMnO4 that will be needed to react with one mole of sulphide ions in acidic solution is a 2/5 b 3/5 S2− + 2/5MnO4− + 16/5H+ —> 2/5Mn2+ + 8/5H2O + S c 4/5 Ans. a d 1/5 Which one of the following does not decolourise an acidified KMnO4 solution? a.SO2 b. FeCl3 2KMnO4 + 5SO2 + 2H2 O  2MnSO4 + K2SO4 +2H2 SO4 2KMnO4 + 8H2 SO4 + 10FeSO4  2MnSO4+ K2SO4+8H2O+5Fe 2(SO4 )3 c. H2 O2 2KMnO4 + 3H2 SO4 +5 H2 O2  2MnSO4 + K2SO4 +8H2 O+ 5O 2 Ans. b d.FeSO4 Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4? a.Both HCl and KMnO4 act as oxidising agents. b.KMnO4oxidises HCl into Cl2 which is also an oxidizing c. KMnO4 is a weaker oxidising agent than HCl. d.KMnO4 acts as a reducing agent in the presence of HCl Ans. b 2KMnO4 + 16HCl 2 MnCl2 + 2KCl + 8H2O + 5Cl2 The oxidation state of chromium in the reaction between KI and acidified potassium dichromate solution changes from a. +4 to +2 Cr 2 O 7 2− + 14H + + 6I −  3 I2 + 2Cr c. +2 to +4 Ans. b b. +6 to +3 3+ + 7H 2 O d. +3 to +6 When acidified K2Cr2O7 solution is added to Sn2+ salts, then Sn2+ changes to a. Sn b. Sn 3+ Cr2O72- + 14H+ + 3Sn2+  3Sn 4+ + 7H2O + 2Cr3+ c. Sn 4+ Ans. c d. Sn + Potassium chromate is dichromate by adding; a. Potassium hydroxide 2− 2Cr O + 2 H +  Cr 4 2O7 converted 2− into potassium b. Sulphuric acid +H O 2 chromates and dichromate ions exist in equilibrium and are inter convertible by altering the pH of the solution. c.acetic acid Ans b d. ammonium hydroxide When hydrogen peroxide is treated with cold acidified K2Cr2O7 solution containing ether a blue colour is obtained this is due to, a. Chromium sulphate b.Potassium chromate K2Cr 2O7 + H2 SO4 +2H2O2 2K2SO4 +CrO5 + 5H2O c. Perchromic acid Ans d d.Chromium pentoxide Which of the following statements is not correct? a. K2Cr2O7 solution in acidic medium is orange color K2Cr2O7 solution in acidic medium is orange colour b.K2Cr2−2O7 solution becomes yellow on increasing the pH of the + 2− 2Cr O 4 +2H  Cr 2 O 7 + H2O solution beyond 7 . K2Cr2O7 solution becomes yellow on increasing the pH of the solution beyond 7 2−passing On H 2S acidified Crc.2 O + 2 OH -  2Crthrough O 4 2− + Hthe 7 2O . K2Cr2O7 solution a milky is observed . K2Crcolour 2O7 + 4H2 SO4 +3H2SK2SO4 + Cr 2 (SO4) 3 +7H2O + 3S gives milky colour . d. Na Cr O is prefered over K Cr OSulphur in volumetric analysis 2 2 7 2 Ans. d 2 7 KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidised to ____________ b. IO– a. I2 2 MnO4- + 16H+ +10 I -  2Mn 2+ + 8H2O + 5I2 c. IO3 – Ans. a d. IO4 – Non-stoichiometric compounds of transition elements are called-----------a.Hydrates b.Hydrides Hydrates : In which water molecules are Hydrides : A binary compound of hydrogen chemically bound to a compound. E.g with metal. E.g NaH CoCl2.6H2O. c.Interstitial compounds d.Binary compounds Interstitial compounds are non stoichiometric compds formed when small atoms (like H,C or N ) are trapped inside the crystal lattice of T. metals. E.g VH0.56 ,TiH1.7 Ans. c Which one of the following is not the characteristic property of interstitial compounds? Interstitial compounds are formed when small atoms (like H,C or a.They have high melting points in comparison to pure metals. N ) are trapped inside the crystal lattice of T. metals b.They are very hard. They have highmetallic melting points in comparison to pure metals. They are very hard. c.They retain conductivity. . They retain metallic conductivity. They have chemical properties as the parent metals d.They aresame chemically very reactive Ans.d Lanthanides and Actinides The lanthanides are also called the rare earth elements. The atomic properties of the lanthanides vary little across the period, and their chemical properties are also very similar. General electron configuation of Ln - [Xe] 4f 1-14 5d 0-16s2 The +3 oxidation state is common for both lanthanides and actinides. General electron configuation of Ac - [Rn] 5f 1-14 6d 0-17s2 All actinides are radioactive, and have very similar physical and chemical properties. The element with electronic configuration [Xe] 4f 75d16s2 is a. Representative element b.Transition element [Xe] 4f 75d16s2 z = 64 i. e lanthanoid , Gd c. Lanthanoids Ans c. d. Actinoids Which of the following oxidation state is common for all lanthanoids? a. +2 b.+3 [Xe] 4f 1-14 5d 0-16s2 common oxidation state of lanthanides is +3 c. +4 Ans. b d. +5 Which one of the following ion is not a good reducing agent in solution ? a.Sm2+ b.Eu2+ c. Ce4+ d.Yb 2+ The most stable oxidation state of lanthanides is +3. Hence the ions in +2 oxidation state tend to change +3 state by loss of electron acting as reducing agents Ln2+  Ln3+ + e (e.g Sm2+, Eu2+, Yb2+ ) Those in +4 oxidation state tend to change to +3 oxidation state by gain of electron acting as a good oxidising agent in aqueous solution. Ce4+ + e  Ce3+ Ans c Identify the incorrect statement among the following a. d-Block elements show irregular and erratic chemical properties among themselves b. La and Lu have partially filled d orbitals and no other partially filled orbitals c. The chemistry of various lanthanoids is very similar d. 4f and 5f orbitals are equally shielded Sol. 4f and 5f belongs to different energy levels, hence the shielding effect is on them is not the same. Shielding of 4f is more than 5f. Ans. d Number of elements belonging to 4f series are --------a. 6 b. 10 4f series from Ce to Lu contains 14 elements. c. 28 Ans. d d. 14 Lanthanoid contraction is caused due to, a. The appropriate shielding on outer electrons by 4f electrons from the nuclear charge b.The aprpriate shielding on outer electrons by 5d electrons from the nuclear charge c. The same effective nuclear charge from Ce to Lu . d.The imperfect shielding on outer electrons by 4f electrons from the nuclear charge. Ans. c Although Zirconium belongs to 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because___________. Both similar atomic radius Zr 160pm a. both belong to have d-block and Hf 159pm b.both have same number of electrons. c.both have similar atomic radius. d.both belong to the same group of the periodic table Ans. d Lanthanoids and actinoids resemble in a.Electronic configuration b. Ionisation energy Ln - [Xe] 4f 1-14 5d 0-16s2 , Ac- [Rn] 5f 1-14 6d 0-17s2 c. Oxidation state Both show a dominant o.s of +3 Ans. c d. Formation of complexes Ln – Show lesser tendency for complex formation Ac- Show greater tendency for complex formation Which one of the following set correctly represents the increase in the paramagnetic property of the ion? a.Mn2+ < +2 -[Ar]d5 - 5 Mn 2+ V < Cr2+ < Cu2+ +2 -[Ar]d Crb. - 4V2+ Cu2+4 < < Cr2+ < Mn2+ Paramagnetic behavior depends on the no. of unpaired electrons c. Cu2+ < Cr2+ < V2+ < Mn2+ Cu+2-[Ar]d9 - Ans. b 1 d.Mn2+ < Cu2+ < Cr2+ < V2+ V+2 -[Ar]d3 - 4 The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr3+ ion is a. 2.84 B.M. b. 3.87 B.M. Cr +3 - [Ar]3d3 4s0 it contains 3 unpaired electrons If n = 3 µ = n(n+2) = 3.87 c. 3.47 B.M. Ans. b d. 3.57 B.M. KMnO 4 can be prepared from K2MnO4as per the reaction ; 3MnO4 2− + 2H2 O  2MnO4− + MnO2 + 4OH − The reaction goes to completion by removing OH − ion by adding 4KMnO + 4KOH 2 K MnO + O + 2H O 4 a. KOH 2 4 2 2 b. CO 2 3K2MnO4 + 2CO2 2 KMnO4 +MnO2 + 2K2CO3 HCl c.SO react with OH − ion and gives water hence reaction goes d.HCl 2 to completion Ans. d Which of the following reactions are disproportionation reactions? i. Cu+ → Cu2+ + Cu ii. 3MnO42- + 4H+ →2MnO4– + MnO2 + 2H2O iii. 2KMnO4 →K2MnO4 + MnO2 + O2 iv. 2MnO4– + 3Mn2+ + 2H2O →5MnO2 + 4H+ a. i, ii b. i, ii, iii A particular oxidation state, which is relatively less stable compared to other oxidation states, undergoes disproportionation. Manganese(VI) which is relatively less stable changes over to manganese(VII) and manganese(IV) in acid solution. 3MnO42- + 4H+  2MnO4- + MnO2 + 2H2O c. ii, iii, d. i, iv Similarly Cu+ iv → Cu2+ + Cu Ans. a Which of the following statements is not correct? a. Copper liberates hydrogen from acids. b.In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine. c.Mn3+ and Co3+ are oxidising agents in aqueous solution. d. Ti2+ and Cr2+ are reducing agents in aqueous solution Ans.a