Explore topic-wise InterviewSolutions in Current Affairs.

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1.	Square planar complexes with coordination number 4 exhibit geometrical isomerism whereas tetrahedral complexes do not. Why ?
Answer» SOLUTION :TETRAHEDRAL complexes do not exhibit geometrical isomerism because the RELATIVE positions of the ligands with respect to each other is the same. The SQUARE planar complexes on the other hand show geometrical isomerism. If same kind of ligands occupy positions adjacent to each other, it is called cis- form and if these are OPPOSITE to each other, it is called trans- form.

2.	Square planar complexes with coordination number of four exhibit geometrical isomerism whereas tetrahedral complexes do not. Why ?
Answer» Solution :TETRAHEDRAL complexes do not GEOMETRICAL isomerism because the RELATIVE positions of the ligands attached to the central metal atom are same with respect to each other.

3.	Square planar complexes of the type MABXL (where A, B, X and L are unidentates) show
Answer» TWO CIS and ONE TRANS isomer two trans and one cis isomer two cis and two trans isomers one cis and one trans isomer Solution :

4.	Square planar complexes are generally formed by d^(8) ions with strong field ligands. The crystal field splitting is larger for second and third row transition elements and for more highly charged species. All the complexes having 4d^(8)" and "5d^(8) configuration are mostly square including those with weak field ligands such halide ions. Square planar complexes with coordination number four can show diastereoisomerism. Which one of the following square planar complexes will show geometrical isomerism ?
Answer» `[Pt(EN)_(2)]^(2+)` `[Pt(gly)_(2)]` `[Pt(NH_(3))_(2)CL(NH_(2)CH_(3))]Cl` (b) and (C ) both ANSWER :D

5.	Square planar complexes are generally formed by d^(8) ions with strong field ligands. The crystal field splitting is larger for second and third row transition elements and for more highly charged species. All the complexes having 4d^(8)" and "5d^(8) configuration are mostly square including those with weak field ligands such halide ions. Square planar complexes with coordination number four can show diastereoisomerism. Which of the following staements is true for the complex [Ni(PPh_(3))_(2)Br_(2)]?
Answer» Hybridisation is the same as found in NICKEL `(+2)` complex with STRONG field ligands like CN. Hybridisation is the same as found in nickel (0) complex with strong field ligands like CO. Hybridisation is the same as found in nickel `(+2)` complex with weak field ligands like halide ions. (b) and (c ) both SOLUTION :`[Ni(PPh_(3))_(2)Br_(2)] implies sp^(3)`,tetrahedral `[Ni(CO)_(4)]""sp^(3)` `[NiCl_(4)]^(2-) "" implies sp^(3)`

6.	Square planar complexes are generally formed by d^(8) ions with strong field ligands. The crystal field splitting is larger for second and third row transition elements and for more highly charged species. All the complexes having 4d^(8)" and "5d^(8) configuration are mostly square including those with weak field ligands such halide ions. Square planar complexes with coordination number four can show diastereoisomerism. Amongst the following complexes which has square planar geometry ?
Answer» `[RhCl(CO)(PPh_(3))_(2)]` `K_(3)[Cu(CN)_(4)]` `K_(2)[ZN(CN)_(4)]` `[Ni(CO)_(4)]` Answer :A

7.	Square planar complexes have………… type of hybridisation.
Answer» `SP^(3)` `dsp^(2)` `sp^(3)d` `sp^(3)d^(2)` Answer :B::D

8.	Square planar complexes are generally formed by d^(8) ions with strong field ligands. The crystal field splitting is larger for second and third row transition elements and for more highly charged species. All the complexes having 4d^(8)" and "5d^(8) configuration are mostly square including those with weak field ligands such halide ions. Square planar complexes with coordination number four can show diastereoisomerism. Which of the ions do not form the square planar complexes amongst the following ions ?
Answer» `NI(+II)" and "Rh(+I)` with strong field ligands `Rh(+I)" and "Au(+III)` with strong as well as weak field ligands. `Pd(+II), Pt(+II)" and "Au(+III)` with strong as well as weak field ligands. None of the above Solution :`Ni^(2+),Rh^(+),Au^(3+),Pd^(2+),Pt^(2+)` form SQUARE PLANAR COMPLEXES.

9.	Square planar complex of the type MAXBL (where A, B, X and L are unidentate ligands) shows following set of isomers
Answer» two cis and ONE trans two trans and one cis two cis and two trans three cis and one trans Solution :Square PLANAR complex of TYPE `M_("AXBL")`will show two-cis and one-trans isomers. These above three isomers are formed by fixing the position of one LIGAND, A.

10.	Spring's reaction is / are
Answer» `Na_(2)S+Na_(2)SO_(3)+I_(2) to Na_(2)S_(2)O_(3)+2NaI` `Na_(2)SO_(3)+S to Na_(2)S_(2)O_(3)` `2Na_(2)S+SO_(2)+Na_(2)CO_(3) to 3Na_(2)S_(2)O+CO_(2)` `2Na_(2)S+3O_(2) to 2Na_(2)S_(2)O_(3)+6S` SOLUTION :Spring.s reaction is are `Na_(2)S+Na_(2)SO_(3)+I_(2) to Na_(2)S_(2)O_(3)+2NaI`

11.	sqrt2.h/(2pi)
Answer» -6 V -2.1 V -3.9 V -8.1 V Answer :C

12.	Sprouting barley is the main source of ___ sugar.
Answer» GLUCOSE FRUCTOSE CELLULOSE MALTOSE SOLUTION :maltose

13.	Spraying of DDT on crops produces pollution of
Answer» AIR only Air and SOIL only Air soil and WATER Air and water only Answer :C

14.	Spontanelity of any process can be predicated withthe help of DeltaS_("total").But this requires calculation of changes in system as will as surrounding. If some criteria (depanding upon the system only) can be developed for checking spontaneity unde specific conditions, then that would be a more useful parameter.The criteria can be derived from Clausius inequality. T ds ge q,(gt sign for irreversible process= sign for rev. process) or T dsgtdU-W for an irreversible process or T dS_(sys)gt dU_(sys)+p dV [consider no non-PV work] If U and V are constant, T dS_(sys)gt0 or (dS)_(u.v)gt0 for spontaneous process if V and T are constant T dS gt dU or dU-T dS lt 0 As temperature is constant, dU-d(TS)lt0 or d(U-TS)lt0 Another state function A(Helmholtz's function)=U-TS A decrease in Helmholtz function (A) under constant volume and temperature is the criteria of spontaneity of a process. For an irreversible cyclic process oint (dq)/T of system is :
Answer» EQUAL to zero Greater than zero Less than zero Equal to CHANGE in GIBB's energy. Solution :For a system in cyclic process `oint(dq)/Tle0` `oint(dq)/T=0` for REVERSIBLE `oint(dq)/Tgt0` for reversible

15.	Spontaneous adsorption of a gas on solid surface is an exothermic process because
Answer» `DELTAH` increases for SYSTEM `DeltaS` increases for gas `DeltaS` DECREASES for gas `DELTAG` increases for gas Answer :C

16.	Spontanelity of any process can be predicated withthe help of DeltaS_("total").But this requires calculation of changes in system as will as surrounding. If some criteria (depanding upon the system only) can be developed for checking spontaneity unde specific conditions, then that would be a more useful parameter.The criteria can be derived from Clausius inequality. T ds ge q,(gt sign for irreversible process= sign for rev. process) or T dsgtdU-W for an irreversible process or T dS_(sys)gt dU_(sys)+p dV [consider no non-PV work] If U and V are constant, T dS_(sys)gt0 or (dS)_(u.v)gt0 for spontaneous process if V and T are constant T dS gt dU or dU-T dS lt 0 As temperature is constant, dU-d(TS)lt0 or d(U-TS)lt0 Another state function A(Helmholtz's function)=U-TS A decrease in Helmholtz function (A) under constant volume and temperature is the criteria of spontaneity of a process. If a state function G is defined such as G=H-TS, then dG must decrease for a spontaneou process (involving only PV work ) occuring at
Answer» Constant VOLUME and TEMPERATURE Constant pressure and temperature Constant volume and pressure Constant entropy and volume Solution :For a spontaneous PROCESS (at constant T & P) `G=H-TS("at" dG lt 0)`

17.	Spontanelity of any process can be predicated withthe help of DeltaS_("total").But this requires calculation of changes in system as will as surrounding. If some criteria (depanding upon the system only) can be developed for checking spontaneity unde specific conditions, then that would be a more useful parameter.The criteria can be derived from Clausius inequality. T ds ge q,(gt sign for irreversible process= sign for rev. process) or T dsgtdU-W for an irreversible process or T dS_(sys)gt dU_(sys)+p dV [consider no non-PV work] If U and V are constant, T dS_(sys)gt0 or (dS)_(u.v)gt0 for spontaneous process if V and T are constant T dS gt dU or dU-T dS lt 0 As temperature is constant, dU-d(TS)lt0 or d(U-TS)lt0 Another state function A(Helmholtz's function)=U-TS A decrease in Helmholtz function (A) under constant volume and temperature is the criteria of spontaneity of a process. For a spontaneous process, if entropy and volume are constant, the internal energy of system must
Answer» Increases Decreases Remain constant Be zero Solution :`dU-T dSlt0(dS=0)` or `dU LT 0` So internal ENERGY of SYSTEM decreases

18.	splitting of spectral lines under the influences of magnetic field is called
Answer» PHOTO ELECTRIC effect stark effect Crompton effect Zeeman effect Answer :D

19.	Splitting of spectral lines when atoms are subjected to strong electric field is called :
Answer» ZEEMAN effect Stark effect Decay Disintegration Answer :B

20.	Splitting energy (Delta_(0)) can usually be measured form the absorption spectra of the complex ions. In simple cases when light is absorbed by a complex ion, an electron in one of the lower energy orbitals is excited to one of the higher energy orbital. The energy corresponding to the frequency of absorbed light is equal to Delta_(0). If value of Delta_(0) for the complex is in visible region, the complex is coloured and the value of Delta_(0) lies in ultraviolet or infrared region, the complex is colourless. For octahedral complexes, the cyrstal field stabilisation energy is given by CFSE= +[-0.4t_(2g)N+0.6e_(g)N_(1)]Delta_(0) Where N and N_(1) are number of electrons in t_(2g)" and "e_(g) orbitals respectively. The value of CFSE can be used for the correction of the experimental values of heats of hydration of divalent can be obtained by substracting the calculated CFSE values form the experimental values. The heat of hydration of Cr^(2+) ion is -450k" "cal//mol. For [Cr(H_(2)O)_(6)]^(2+), Delta_(0)=13,900" "cm(-1). What heat of hydration would be, if there were no crystal field stabilisation energy ?
Answer» `-426 k" "CAL//"MOLE"` `-245 k" "cal//"mole"` `-4.84 k" "cal//"mole"` NONE of these Answer :A

21.	Splitting energy (Delta_(0)) can usually be measured form the absorption spectra of the complex ions. In simple cases when light is absorbed by a complex ion, an electron in one of the lower energy orbitals is excited to one of the higher energy orbital. The energy corresponding to the frequency of absorbed light is equal to Delta_(0). If value of Delta_(0) for the complex is in visible region, the complex is coloured and the value of Delta_(0) lies in ultraviolet or infrared region, the complex is colourless. For octahedral complexes, the cyrstal field stabilisation energy is given by CFSE= +[-0.4t_(2g)N+0.6e_(g)N_(1)]Delta_(0) Where N and N_(1) are number of electrons in t_(2g)" and "e_(g) orbitals respectively. The value of CFSE can be used for the correction of the experimental values of heats of hydration of divalent can be obtained by substracting the calculated CFSE values form the experimental values. The value of crystal field splitting (Delta_(0)) for [Ti(H_(2)O)_(6)^(3+) is 243 kj mol^(-1). The crystal field stabilisation energy (CFSE) in this complex is : (in kj mol^(-1))
Answer» `-(3)/(5)xx243` `-(2)/(5)xx243` `-3xx(2)/(5)xx243` `-243` SOLUTION :`[Ti(H_(2)O)_(6)]^(3+) implies d^(1),(t_(2g)^(1,0,0)EG^(0,0))` CFSE`=-0.4xx1xxDelta_(0)`

22.	Splitting energy (Delta_(0)) can usually be measured form the absorption spectra of the complex ions. In simple cases when light is absorbed by a complex ion, an electron in one of the lower energy orbitals is excited to one of the higher energy orbital. The energy corresponding to the frequency of absorbed light is equal to Delta_(0). If value of Delta_(0) for the complex is in visible region, the complex is coloured and the value of Delta_(0) lies in ultraviolet or infrared region, the complex is colourless. For octahedral complexes, the cyrstal field stabilisation energy is given by CFSE= +[-0.4t_(2g)N+0.6e_(g)N_(1)]Delta_(0) Where N and N_(1) are number of electrons in t_(2g)" and "e_(g) orbitals respectively. The value of CFSE can be used for the correction of the experimental values of heats of hydration of divalent can be obtained by substracting the calculated CFSE values form the experimental values. Which of the following statements is correct ?
Answer» ZINC (II) ION has a zero CFSE for any geometry. A solution of `[Ti(H_(2)O)_(6)]^(3+)` is purple as the value of `Delta` for the `H_(2)O` complex is in the visible region. Solution of `[Fe(CN)_(6)]^(4-)" and "[Fe(H_(2)O)_(6)]^(2+)` appear colourless in dilute solutions. All of these Solution :`Cr^(2+) impliesd^(4),t_(2g)^(1,1,1)eg^(1,0)` CFSE`=-1.2 Delta_(0)+0.6Delta_(0)=-0.6 Delta_(0)` NOTE :`349.75 cm^(-1)=1 KCAL//"mol"` `implies CFSE=-0.6xx(13900)/(349.75)"kcal"//"mol"` `=-23.84 kcal//mol~=-24` Heat of hydration withoutCFSE will be : `-450+24=-426 "kcal"//"ml"`

23.	Spirit of salt is:
Answer» HBr HCl `H_2SO_4` HI Answer :B

24.	Spinel is a important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has one-eight of the tetrahedral holes occupied by one type of metal ion and one half of the octahedral hole occupied by another type of metal ion. Such a spinel is formed by Zn^(2+), Al^(3+) and O^(2–), with Zn^(2+) in the tetrahedral holes. Give the formulae of spinel.
Answer» ANSWER :`ZnAl_(2)O_(4)`

25.	Spirit is 95% solution of
Answer» Methanol Acetone Benzene Ethanol. Answer :D

26.	Spinel is a important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has one-eight of the tetrahedral holes occupied by one type of metal ion and one half of the octahedral hole occupied by another type of metal ion. Such a spinel is formed by Zn^(2+), Al^(3+)andO^(2–), with Zn^(2+) in the tetrahedral holes. Give the formulae of spinel.
Answer» Solution :`ZN^(2+)=8//8=1` `Al^(3+)=4//2=2` `O^(2-)`=CCP(4) `ZnAl_(2)O_(4)` `Zu^(2+)=(1)/(8)xx8` octahedral voids `O^(2)`=CCParrangment(1+3)=4

27.	Spinels are denoted with the formula MAl_2 O_4, where 'M' is
Answer» Mg in + 2 state or Mn in + 2 state FE in + 2 state or Mn is + 4 state Fe in + 3 state or Mg in + 2 state Mg in + 2 state or Mn in + 4 state Answer :A

28.	Spin only magnetic moment of the compound Hg(Co(SCN)_(4)is
Answer» `sqrt3` `sqrt15` `sqrt24` `sqrt8` Solution : m`n=3` mu_(8)=SQRT(n(n+2))=sqrt(3xx5))=sqrt15`

29.	Spin only magnatic moment can be calculated by using mu=sqrt(n(n+2)) where 'n' represents
Answer» PRINCIPAL QUANTUM No Magnetic Quantum No. Number of unpaired ELECTRONS SPIN Quantum No. Answer :C

30.	Spin only' magnetic moment of Ni in [Ni(dmg)_(2)] is same as that found in:
Answer» `NI " in " [NiCl_(2)(PPh_(3))_(2)]` `Mn" in " [MnO_(4)]^(-)` `Co " in " [CoBr_(4)]^(2-)` `Pt " in " [Pt(H_(2)O)_(2)Br_(2)]` Answer :B::D

31.	Spin only' magnetic moment of Ni in [Ni(dmg)_2] is same as that found in
Answer» Ni in `[NiCl_2(PPh_3)_2]` Mn in `[MnO_4]^(2-)` Co in `[CoBr_4]^(2-)` Pt in `[Pt (H_2O)_2 Br_2]` SOLUTION :Ni`(dmg)_2` COMPLEX is a square PLANAR and dimagnetic. `Pt(II)-5d^8` configuration .complex is square planar and THEREFORE, dimagnetic. So , `mu=0`

32.	Spin only' magnetic moment of Ni in [Ni(dmg)_(2)] is same as that found in :
Answer» NI in `[NiCl_(2)(PPh_(3))_(2)]` Mn in `[MnO_(4)]^(-)` Co in `[CoBr_(4)]^(2-)` PT in `[Pt(H_(2)O)_(2)Br_(2)` Solution :`[overset(+2)NI(DMG)_(2)] implies d^(8)" config".(dsp^(2))'O' u.p.`` E^(-)`. `{MnO_(4)^(-),[Pt(H_(2)O)_(2)Br_(2)]" are diamagetic"}`

33.	Spinof elctron is
Answer» ratation of electron about it 's own axis clockwise and anticlockwise ratation of electron a quntum number which depends UPON direction and SPEEDOF ratation of electron an intrisiccharacteristic of electron con nected with its MAGNETIC field ANSWER :D

34.	Spiegeleisen is on alloy of
Answer» FE, MN C Fe, MG, C Mn, C, Zn Fe, Mn, Mg Answer :A

35.	Spiegeleisen is an alloy of:
Answer» FE and MN Fe,Mn and CFE,Mn and Cr Fe and Cr Answer :B

36.	Spiegeleisen is an alloy of
Answer» `Fe, CU and Mn` `Fe, MG and C` `Fe, Co and CR` `Fe, Cu and NI` Answer :A

37.	Sphalerite is ore of :
Answer» IRON CALCIUM COPPER Zinc ANSWER :B

38.	Spherical gold paricles of radius 10nm, then the number of gold particles per mm^(3) of sol will be
Answer» `1.9xx10^(12)` `6.3xx10^(14)` `6.3xx10^(10)` `2.4xx10^(6)` Solution :Volume of the gold dispresed in one LITRE water `=("mass")/("density") = (1.9xx10^(-4)g)/(19g cm^(-3)) = 1 xx 10^(5)cm^(-3)` Radius of gold sol particle `=10NM = 10 xx 10^(-9)m = 10xx10^(-7) cm = 10^(-6)cm` Volume of gold sol particle= `4/3 pi r^(3)` `4/3 xx 22/7 xx (10^(-6))^(3) = 4.19xx10^(-18) cm^(3)` No. of gold sol particle in `1 xx 10^(-5)cm^(3)` No, of gold sol particle in `1 xx 10^(-5) cm^(3)` `=(1xx10^(-5))/(4.19xx10^(-18)) = 2.38xx10^(12)` No. of gold sol particle in one `mm^(3)` `=(2.38xx10^(12))/(10^6) = 2.38 xx 10^(6)`.

39.	Sphaleriteis concentratedby
Answer» GRAVITYSEPARATION frothfloatation magneticsepartion hydraulicwashing Solution :SPHALERITE (ZNS) beinga sulphide ORE isconcernedby FROTHFLOATATIONPROCESS.

40.	Sphalerite and siderite are the ores of the metals
Answer» AI and ZN Fe and Cu Cu and Zn Fe and Zn ANSWER :D

41.	Sperrylite is :
Answer» `AGCL` `PtAs_2` `Fe_2O_3` `SnO_2` ANSWER :B

42.	Spermaceti is commonly used in :
Answer» Fermentation of cane sugar Preparation of ACETIC ACID BIRTH control Cosmetics and soaps ANSWER :D

43.	Speeds of diffusion of carbon dioxide and unknown gas under identical conditions were 0.29 and 0.271 mL ("minute")^- respectively. Calculate the density of the gas if that of carbon dioxide is 1.964 gL^(-1) .
Answer» ANSWER :`2.249L^(-1)`

44.	Speigel, used in the manufacture of steel by the Bessemer process, is an alloy of.....
Answer» iron, CHROMIUM and carbon. iron, MANGANESE and carbon. iron, tungstan and carbon. iron, nickel and carbon. Answer :B

45.	Spelter is:
Answer» IMPURE CU Impure ZINC ZnO CuO Answer :B

46.	Specify the oxidation numbers of the metals in the following coordination entities : [PtCl_(4)]^(2-)
Answer» Solution :Oxidation numbers can be obtained by adding the oxidation numbers of CONSTITUENTS and equating them with the overall charge on the ION, or zero, if it is a MOLECULAR complex. Assuming x as the oxidation number of the METAL, we have `x - 4 = - 2` or `x = +2`

47.	Spectomycin is
Answer» an TRANQUILIZER an antibitoic an antiseptic an antipyretic Answer :B

48.	Specify the oxidation numbers of the metals in the following coordination entities : (i) [Co(H_(2)O)(CN)(en)_(2)]^(2+) (ii) [PtCl_(4)]^(2-) (iii) [Cr(NH_(3))_(3)Cl_(3)] (iv) [CoBr_(2)(en)_(2)]^(+) (v) K_(3)[Fe(CN)_(6)]
Answer» Solution :(i) x+(0)+(-1)+(0)=+2orx=+3 (ii) x-4=-2or x=+2 (III) x+0+3(-1)=0 or x=+3 (iv) x+2(-1)+0=+1 or x=+3 (V) `[Fe(CN)_(6)]^(3-)`, x+6(-1)=-3 or x=+3

49.	Specify the oxidation numbers of the metals in the following coordination entities : K_(3)[Fe(CN)_(6)]
Answer» Solution :Oxidation numbers can be OBTAINED by adding the oxidation numbers of constituents and equating them with the OVERALL charge on the ion, or zero, if it is a MOLECULAR COMPLEX. Assuming x as the oxidation number of the metal, we have `x + 6(-1) = - 3` or `x = +3.`

50.	Specify the oxidation numbers of the metals in the following coordination entities : [Co(H_(2)O)(CN)(en)_(2)]^(2+)
Answer» Solution :Oxidation numbers can be obtained by adding the oxidation numbers of constituents and equating them with the overall CHARGE on the ion, or ZERO, if it is a molecular COMPLEX. Assuming x as the oxidation NUMBER of the METAL, we have `x + (0) + (-1) + (0) = +2` or `x = +3`