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Q.1. The IUPAC name of the complex [Pt(NH3)2Cl(NH2CH3)]Cl is (2020)
The correct IUPAC name for the given complex [Pt(NH3)2 Cl(NH2 CH3 )]Cl is diamminechlorido(methanamine)platinum(II) chloride.
Q.2. The theory that can completely/properly explain the nature of bonding in [Ni(CO)4] is (2020)
(a) Werner’s theory.
(b) Molecular orbital theory.
(c) Crystal field theory.
(d) Valence bond theory
The carbonyl (CO) ligand is a -acceptor ligand, so, it is involved in synergic bonding, which is explained by molecular orbital theory.
Q.3. The number of possible optical isomers for the complexes MA2B2 with sp3 and dsp2 hybridized metal atom, respectively, is
Note: A and B are unidentate neutral and unidentate monoanionic ligands, respectively. (2020)
(a) 0 and 2
(b) 2 and 2
(c) 0 and 0
(d) 0 and 1
The isomers of the complex compound of type MA2B2 with sp3 hybridization (tetrahedral) is zero. Since, the geometry is symmetrical, thus, no optical isomer is possible. For dsp2 hybridization (square planar) is also zero, due to symmetrical geometry.
Q.4. Among the statements (I)-(IV), the incorrect ones are (2020)
(I) Octahedral Co(III) complexes with strong field ligands have very high magnetic moments.
(II) When Δo < P, the d-electron configuration of Co(III) in an octahedral complex is
(III) Wavelength of light absorbed by [Co(en)3]3+ is lower than that of
(IV) If the Δo for an octahedral complex of Co(III) is 18,000 cm−1, the Δt for its tetrahedral complex with the same ligand will be 16,000 cm−1.
(a) (I) and (IV) only
(b) (III) and (IV) only
(c) (I) and (II) only
(d) (II) and (III) only
(I) In presence of strong field ligands, electrons get paired up, so, number of unpaired electrons decreases, thus, the magnetic moment decreases.
The relation between CFSE for octahedral and tetrahedral complexes is:
(IV) For,Δo = 18000 cm-1,the CFSE for tetrahedral complex is
Δt = 4/9 x 18000 = 8000 cm-1
Q.5. The complex that can show fac- and mer- isomers is (2020)
Complex of type Ma3b3 shows fac and mer geometrical isomers. So, complex [Co(NH3)3 (NO3)3] will shows fac and mer geometrical isomers.
Q.6. The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3 g of [Co(NH3)6]Cl3 is __________. (2020)
M[Co(NH3)6]Cl2 = 267.46 g mol−1
MAgNO3 = 167.87 g mol-1
For the reaction,
M1 V1 = M2 V2
Q.7. Among (I) – (IV), the complexes that can display geometrical isomerism are (2020)
(I) [Pt(NH3)2 Cl]+
(III) [Pt(NH3)2 Cl(NO2)]
(a) (II) and (III)
(b) (IV) and (I)
(c) (III) and (IV)
(d) (I) and (II)
The complexes of type Ma3b or Ma5b can never displaced geometrical isomerism. The geometrical isomerism exhibit by complex [Pt(NH3)2 Cl(NO2)]
The geometrical isomerism exhibit by complex [Pt(NH3)4 ClBr]2+
Q.8. The correct order of the calculated spin-only magnetic moments of complexes (I) to (IV) is (2020)
(a) (I) ≈ (III) < (II) ≈ (IV)
(b) (III) < (IV) < (II) < (I)
(c) (III) ≈ (IV) < (II) < (I)
(d) (I) ≈ (III) ≈ (IV) < (II)
For the given complex, the spin-only magnetic moment is:
For [Ni(CO)4], the oxidation state of Ni is zero. Ni0 : [Ar]3d84s04p2
Thus, the magnetic moment for (I) is 0.
For [Ni(H2O)6]Cl2, the oxidation state of Ni is +2. Ni2+: [Ar]3d84s0
The number of unpaired electrons (n) = 2
Thus, spin only magnetic moment for (II) is =
For Na2[Ni(CN)4], the oxidation state of Ni is +2. Ni2+: [Ar]3d84s0
Thus, the magnetic moment for (III) is 0
For [PdCl2(PPh3)4], the oxidation state of Pd is +2. Pd2+: [Kr]4d84s0
Thus, the magnetic moment for (IV) is 0.
Hence, the correct order for spin-only magnetic moment is: (I) ≈ (III) ≈ (IV) < (II).
Q.9. Complexes (ML5) of metals Ni and Fe have ideal square pyramidal and trigonal bipyramidal geometries, respectively. The sum of the 90°, 120° and 180° L–M–L angles in the two complexes is_______. (2020)
The square pyramidal structure for Ni complex is
This complex has eight 90° L–Ni–L and two 180° L–Ni–L bonds.
The trigonal bipyramidal structure for Fe complex is
This complex has six 90° L–Fe–L, three 120° L–Fe–L and one 180° L–Fe–L bonds.
Thus, total of 8 + 2 + 6 + 3 + 1 = 20 L–M–L bonds.
Q.10. Complex X of composition Cr(H2O)6Cln has a spin only magnetic moment of 3.83 BM. It reaches with AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of X is (2020)
(a) hexaaqua choromium(III) chloride.
(b) tetraaquadichlorido chromium (IV) chloride dihydrate.
(c) dichloridotetraaqua chromium(IV) chloride dihydrate.
(d) tetraaquadichlorido chromium (III) chloride dehydrate.
The spin only magnetic moment = 3.83 BM, that is, there are 3 unpaired electrons in d-orbital of Cr. So, the oxidation state of Cr is +III. This complex can show geometrical isomerism which complex type of Ma4b2 or Ma3b3. But this complex reacts with AgNO3, so the complex must of type Ma4b2. Hence, the complex is [Cr(H2O)4Cl2]Cl. 2H2O and its IUPAC name is: tetraaquadichloridochromium(III)chloride dihydrate.
Q.11. [Pd(F)(Cl)(Br)(I)]2− has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilization energy [CFSE] of [Fe(CN)6]n – 6, respectively, are
[Note: Ignore the pairing energy] (2020)
(a) 2.84 BM and −1.6 Δo
(b) 5.92 BM and 0
(c) 1.73 BM and −2.0 Δo
(d) 0 BM and −2.4 Δo
The possible geometrical isomers of the complex of type Mabcd are three.
Thus, the coordination complex of iron
The oxidation state of Fe in above complex is Fe3+ = [Ar] 3d54s0.
Due to the strong field ligand, pairing will occur, hence, the electronic configuration will beSo, the magnetic moment is
The CFSE of the complex is
CFSE 2.0 = −2.0Δ0
Q.12. The correct order of the spin-only magnetic moments of the following complexes is (2020)
(III) Na3[Fe(C2O4)3] (Δo > P)
(a) (III) > (I) > (IV) > (II)
(b) (III) > (I) > (II) > (IV)
(c) (I) > (IV) > (III) > (II)
(d) (II) (I) > (IV) > (III)
For the given complex, the spin-only magnetic moment is:
For [Cr(H2O)6]Br2, the oxidation state of Cr is +2. Cr2+:[Ar]3d44s0
Since, the ligand is weak filed ligand, so no pairing will take place, thus, electronic configuration is .
The number of unpaired electrons (n) = 4
For Na4[Fe(CN)6], the oxidation state of Fe is +2. Fe2+ : [Ar] 3d64s0
Since, the ligand is strong filed ligand, so pairing of electrons will take place, thus, electronic configuration is . The number of unpaired electrons (n) = 0
For Na3[Fe(C2O3)3], the oxidation state of Fe is +3. Fe3+: [Ar]3d54s0
Since, it is given, Δ0 > P, so pairing will take place, thus, electronic configuration is . The number of unpaired electrons (n) = 1
For (Et4N)2[CoCl4], the oxidation state of Co is +2. Co2+: [Ar]4d7 4s0.
Since, the ligand is weak filed ligand, so no pairing will take place, thus, electronic configuration is .
The number of unpaired electrons (n) = 3
Hence, the correct order for spin-only magnetic moment is: (I) (IV) (III) (II).
Q.13. 5 g of zinc is treated separately with an excess of
(I) dilute hydrochloric acid and
(II) aqueous sodium hydroxide
The ratio of the volumes of H2 evolved in these two reactions is (2020)
(a) 1 : 2
(b) 1 : 1
(c) 1 : 4
(d) 2 : 1
The moles of zinc = 0.076 mol.
The reaction of zinc with HCl and NaOH is
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Zn(s) + 2NaOH(aq) + 2H2O(l) → Na2n(OH)4(aq) + H2(g)
1 mol of Zn react with HCl and NaOH respectively, 1 mol of hydrogen gas is liberated from each reaction. Thus, the ratio of hydrogen gas evolved when 5 g of Zn reacts with dil. HCl and aq. NaOH solutions is 1:1.
Q.14. The isomer(s) of [Co(NH3)4Cl2] that has/have a Cl–Co–Cl angle of 90°, is/are (2020)
(a) meridional and trans
(b) cis and trans
(c) trans only
(d) cis only
The geometrical isomers of [Co(NH3 )4Cl2 ] is:
Hence, only cis-isomer have Cl–Co–Cl bond angle equal to 90°.
Q.15. The atomic radius of Ag is closest to (2020)
Due to the lanthanoid contraction, the atomic radius of Ag is closest to Au.
Q.16. Consider the following reactions:
NaCl + K2Cr2O7 + conc.H2SO4 → (A) + side products
(A) + NaOH → (B) + Side products
(B) + dil.H2SO4 + H2O2 → (C) + Side products
The sum of the total number of atoms in one molecule each of (A), (B) and (C) is________. (2020)
The reaction involved is
NaCl + K2Cr2O7 + conc. H2SO4 → + 2KHSO4(aq) + 4NaHSO4(aq) + 3H2O (l)
+ NaOH (aq) → + NaCl + H2O
+ H2SO4 +H2O2 → + Na2 SO4 + H2O
Thus, the total number of electrons in compound (A), (B) and (C) is 5 + 7 + 6 = 18
Q.17. The third ionization enthalpy is minimum for (2020)
The electronic configuration of Fe (Z = 26) is [Ar] 3d6 4s2. After losing three electrons, the electronic configuration will be Fe3+ : [Ar] 3d5 4s0 due to the degenerate d-orbital configuration, the third ionization energy of Fe is minimum among the given options.
Q.18. Assertion: For hydrogenation reactions, the catalytic activity increases from Group 5 to Group 11 metals with maximum activity shown by Group 7-9 elements. (2020)
Reason: The reactants are most strongly adsorbed on Group 7-9 elements.
(a) Both assertion and reason are true, and the reason is the correct explanation for the assertion.
(b) Both assertion and reason are false.
(c) Both assertion and reason are true, but the reason is not the correct explanation for the assertion.
(d) Assertion is not true, but reason is true.
The d-block elements show catalytical activity in hydrogenation reaction, due to the large surface area, elements of Group 7 – 9, the reactant more strongly absorbed on them.
Q.19. The electronic configurations of bivalent europium and trivalent cerium are
(Atomic number: Xe = 54, Ce = 58, Eu = 64) (2020)
(a) [Xe] 4f2 and [Xe] 4f7
(b) [Xe] 4f7and [Xe] 4f1
(c) [Xe] 4f7 6s2 and [Xe] 4f26s2
(d) [Xe] 4f4 and [Xe] 4f9
The electronic configuration of Ce3+ is [Xe]4f1 and that of Eu2+ is [Xe]4f7.
Q.20. The sum of the total number of bonds between chromium and oxygen atoms in chromate and dichromate ions is _______. (2020)
The structure of chromate and dichromate is
Thus, total number of bonds between Cr and O in chromate is 4 and in dichromate is 8. Hence, the sum of total bonds in chromate and dichromate is 4 + 8 = 12.
Q.21. The transition element that has lowest enthalpy of atomisation is: (2019)
As zinc has no unpaired of electrons to take part in the bond, it has least enthalpy of atomisation amongst the given transition elements.
Q.22. The effect of lanthanoid contraction in the lanthanoid series of elements by and large means: (2019)
(a) increase in both atomic and ionic radii
(b) decrease in atomic radii and increase in ionic radii
(c) decrease in both atomic and ionic radii
(d) increase in atomic radii and decrease in ionic radii
Due to lanthanoid contraction, both atomic radii and ionic radii decrease gradually in the lanthanoid series. The Lanthanide contraction is caused by a poor shielding effect of the 4f electrons.
Q.23. The electrolytes usually used in the electroplating of gold and silver, respectively, are: (2019)
(a) [Au (CN)2] and [Ag(CN)2]-
(b) [Au(CN)2]- and [AgCl2]-
(c) [Au(OH)4]- and [Ag(OH)2]-
(d) [Au(NH3)2]+ and [Ag(CN)2]-
The electrolytes used in the electroplating of Au and Ag are [Au(CN)2]- and [Ag(CN)2]- respectively.
Q.24. The element that usually does NOT show variable oxidation states is: (2019)
Sc shows oxidation state of +3 only.
Q.25. Mn2(CO)10 is an organometallic compound due to the presence of: (2019)
(a) Mn-C bond
(b) Mn-Mn bond
(c) Mn-O bond
(d) C-O bond
Compounds having atleast one carbon metal (M - C) bond are known as organometallic compounds. It contains Mn-C bond.
Q.26. The correct order of atomic radii is: (2019)
(c) Ce> Eu>Ho>N
Atomic radii follows the order
Q.27. The lanthanide ion that would show colour is: (2019)
Sm = 4f6 6s2
Sm3+ = 4f5 = Partially filled/orbital
∴ Sm3+ will be coloured
Lu3+ = 4f14 = colourless.
Q.28. The statement that is INCORRECT about the interstitial compounds is: (2019)
(a) they are chemically reactive.
(b) they are very hard.
(c) they have metallic conductivity.
(d) they have high melting points.
Interstitial compounds are inert, i.e., they are chemically non-reactive.
Q.29. The maximum number of possible oxidation states of actinoides are shown by: (2019)
(a) Nobelium (No) and lawrencium (Lr)
(b) Actinium (Ac) and thorium (Th)
(c) Berkelium (Bk) and californium (Cf)
(d) Neptunium (Np) and plutonium (Pu)
∴ Maximum oxidation state is shown by Np and Pu.
Q.30. Consider the hydrated ions of Ti2+, V2+, Ti3+, and Sc3+. The correct order of their spin-only magnetic moments is: (2019)
(a) V2+ < Ti2+ < Ti3+ < Sc3+
(b) Sc3+ < Ti3+ < Ti2+ < V2+
(c) Ti3+ < Ti2+ < Sc3+ < V2+
(d) Sc3+ < Ti3+ < V2+ < Ti2+
Electronic configuration of the given transition metal ions are:
Since, magnetic moment is directly proportional to the number of unpaired electrons. The correct increasing order of magnetic moment is Sc3+ < Ti3+ < Ti2+ < V2+ because they have 0, 1, 2 and 3 unpaired electrons respectively.
Q.31. The highest possible oxidation states of uranium and plutonium, respectively, are: (2019)
(a) 6 and 7
(b) 6 and 4
(c) 7 and 6
(d) 4 and 6
Maximum oxidation state shown by Uranium is + 6 and Plutonium is 7.
Q.32. The correct order of the first ionization enthalpies is: (2019)
(a) Ti < Mn < Zn < Ni
(b) Ti < Mn < Ni < Zn
(c) Mn < Ti < Zn < Ni
(d) Zn < Ni < Mn < Ti
I.E. increases on moving left to right in a period.
∴ Ti < Mn < Ni < Zn
Q.33. The pair that has similar atomic radii is: (2019)
(a) Mn and Re
(b) Ti and Hf
(c) Sc and Ni
(d) Mo and W
Mo and W belong to group 6 and period 5 (4d series) and 6 (5d series) respectively.
Due to lanthanoid contraction, radius of Mo and W are almost same i.e. 0.140 nm and 0.141 nm respectively.
Q.34. Two complexes [Cr (H2O)6] Cl3 (A) and [Cr (NH3)6] Cl3 (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is: (2019)
(a) Δ0 values of (A) and (B) are calculated from the energies of violet and yellow light, respectively.
(b) both are paramagnetic with three unpaired electrons.
(c) both absorb energies corresponding to their complementary colors.
(d) Δ0 value for (A) is less than that of (B).
Here, both the complexes (A) and (B) are paramagnetic with 3 unpaired electrons each. Also H2O is a weak field ligand which causes lesser splitting than NH3 which is comparatively stronger field ligand. Hence, the (Δ0) value of (A) and (B) are calculated from the wavelengths of light absorbed and not from the wavelengths of light emitted.
Q.35. The highest value of the calculated spin only magnetic moment (in BM) among all the transition metal complexes is: (2019)
Magnetic moment, (where, n = no. of unpaired electrons)
As transition metal atom/ion in a complex may have unpaired electrons ranging from zero to 5. So, maximum number of unpaired electrons that may be present in a complex is 5.
∴ Maximum value of magnetic moment among all the transition metal complexes is
Q.36. The complex that has highest crystal field splitting energy (A) is: (2019)
In case of similar metal atom or ion, the value of coordination number and the strength of the ligands determine the value of crystal field splitting energy.
Greater the co-ordination number and strength of value of the ligand, greater will be the value of CFSE.
Strength of ligands : CN- > NH3 > H2O > Cl-
∴ K3[Co(CN)6] has the highest crystal field splitting energy.
Q.37. Homoleptic octahedral complexes of a metal ion 'M3+' with three monodentate ligands L1, L2 and L3 absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is: (2019)
(a) L3 < L1 < L2
(b) L3 < L2 < L1
(c) L1 < L2 < L3
(d) L2 < L1 < L3
Lesser the wavelength of light absorbed (more energy) greater will be ligand strength.
Q.38. The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO2)] is: (2019)
Q.39. The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is: (2019)
Q.40. A reaction of cobalt(III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two isomeric products A (violet coloured) and B (green coloured). A can show optical activity, but, B is optically inactive. What type of isomers does A and B represent? (2019)
(a) Geometrical isomers
(b) Coordination isomers
(c) Linkage isomers
(d) Ionisation isomers
Reaction for the given condition can be written as:
A = optically active (cis-isomer), violet
B = optically inactive (trans-isomer), green
Q.41. Match the metals (column I) with the coordination compound(s)/enzyme(s) (column II): (2019)
(a) (A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)
(b) (A)-(i); (B)-(ii); (C)-(iii); (D)-(iv)
(c) (A)-(ii); (B)-(i); (C)-(iv); (D)-(iii)
(d) (A)-(iv); (B)-(iii); (C)-(i); (D)-(ii)
Wilkinson catalyst: [Rh(PPh)3Cl]
Chlorophyll : C55H72O5N4Mg
Vitamin B12 contains Co.
Carbonic anhydrase contains a Zn ion.
Q.42. The coordination number of Th in K4[Th(C2O4)4(H2O)2] is:
(C2O2-4 = oxalato) (2019)
C2O2-4 (oxalato) : bidentate ligand
H2O (aqua): monodcntatc
∴ Co-ordination no. of Th = 2 x 4 + 2 = 10
Q.43. The number of bridging CO Iigand(s) and Co-Co bond(s) in Co2(CO)8, respectively are: (2019)
(a) 2 and 1
(b) 2 and 0
(c) 0 and 2
(d) 4 and 0
The structure of Co2(CO)8 is represented as
It contains two bridging CO ligands and one metal -metal (Co - Co) bond.
Q.44. The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the compex [M(H2O)6]Cl2, is: (2019)
(a) V2+ and Co2+
(b) V2+ and Fe2+
(c) Co2+ and Fe2+
(d) Cr2+ and Mn2+
So, the central metal ion has 3 unpaired electrons.
∴ Configuration is either d3 or d7
As H2O is a weak field ligand. V2+ and Co2+ will have 3 unpaired electrons.
V2+ has d3 configuration; Co2+ has d7 configuration.
Q.45. The metal d-orbitals that are directly facing the ligands in K3[Co(CN)6] are: (2019)
K3[Co(CN)6] is an octahedral complex.
During splitting of d orbitals in octahederal complexes, orbitals point towards the direction of ligands (i.e. they experience more repulsion and will be raised in energy by 3/5 Δ0).
Q.46. The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 BM. The suitable ligand for this complex is: (2019)
Electronic configuration of Mn is,
Presence of 5 unpaired e- shows that the complex of Mn2+ has only weak field ligand (NCS-).
Q.47. The following ligand is: (2019)
It has four atoms containing lone pair of e-, therefore, it will be able to donate these lone pairs and acts as a tetradentate ligand.
Q.48. The compound that inhibits the growth of tumors is: (2019)
CM-Platin is used as an anti-cancer drug.
Q.49. The calculated spin-only magnetic moments (BM) of the anionic and cationic species of [Fe(H2O)6]2 and [Fe(CN)6], respectively, are: (2019)
(a) 0 and 4.9
(b) 2.84 and 5.92
(c) 4.9 and 0
(d) 0 and 5.92
Where n = no. of unpaired electrons
Q.50. The one that will show optical activity is:
(en = ethane 1, 2-diamine) (2019)
No plane of symmetry or centre of symmetry Hence it is optically active.
Q.51. The degenerate orbitals of [Cr(H2O)6]3+ are: (2019)
(a) dxz and dyz
(b) dyz and dz2
Cr3+ has (P configuration and forms an octahedral inner orbitals complex.
The set of degenerate orbitals are
Q.52. The correct statements among I to III are: (2019)
(I) Valence bond theory cannot explain the color exhibited by transition metal complexes.
(II) Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes.
(III) Valence bond theory cannot distinguish ligands as week and strong field ones.
(a) (II) and (III) only
(b) (I), (II) and (III)
(c) (I) and (III) only
(d) (I) and (II) only
Valence bond theory cannot distinguish between weak field ligands and strong field ligands. Therefore, it cannot predict quantitatively the magnetic properties of transition metal complexes.
Q.53. The maximum possible denticities of a ligand given below towards a common transition and inner-transition metal ion, respectively, are: (2019)
(a) 8 and 6
(b) 6 and 8
(c) 6 and 6
(d) 8 and 8
The maximum possible denticites of the given ligand towards transition metal ion is 6 and towards inner transition metal ion (due to greater ionic radii and more atomic orbitals) is 8.
Q.54. Three complexes, [CoCl(NH3)5]2+(I), [Co(NH3)5H2O]3+(II) and [Co(NH3)6]3+ (III) absorb light in the visible region. The correct order of the wavelength of light absorbed by them is: (2019)
(a) (III) > (I) > (II)
(b) (III) > (II) > (I)
(c) (II) > (1) > (III)
(d) (I) > (II) > (III)
Wavelength of the energy absorbed by the coordination compound is inversely proportional to ligand field strength of the given co-ordination compound. The strong field ligand causes higher splitting of the d-orbitals. The decreasing order of ligand field strength is NH3 > H2O > Cl. Therefore decreasing order of absorbed wavelength is (I) > (II) > (III).
Q.55. The crystal field stabilization energy (CFSE) of [Fe(H2O)6] Cl2 and K2[NiCl4], respectively are: (2019)
(a) -0.6Δo and -0.8Δt
(b) -0.4Δo and -0.8Δt
(c) -2.4Δo and -1.2Δt
(d) -0.4Δo and -1.2Δt
Q.56. The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to +3 state is: (2019)
So, only Fe2+ will lose crystal field stabilisation energy upon oxidation to +3, others will gain crystal field stabilisation energy.
Q.57. The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and K3[A1(C2O4)3], respectively, are:
(en = ethane-1,2-diamine) (2019)
(a) 5 and 3
(b) 3 and 3
(c) 6 and 6
(d) 5 and 6
Q.58. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting [3Ca3 (PO4)2 . Ca(OH)2] to (2018)
Q.59. Hydrogen peroxide oxides [Fe(CN)6 ]4- to [Fe(CN)6]3- in acidic medium but reduces [Fe(CN)6 ]3- to [Fe(CN)6]4- in alkaline medium. The other products formed are, respectively: (2018)
(a) (H2O + O2) and H2O
(b) (H2O + O2) and (H2O+OH-)
(c) H2O and (H2O + O2)
(d) H2O and (H2O+OH-)
Q.60. Consider the following reaction and statements:
[Co(NH3)4Br2]+ +Br- → [Co(NH3)3Br3] + NH3
(I) Two isomers are produced if the reactant complex ion is a cis-isomer.
(II) Two isomers are produced if the reactant complex ion is a trans-isomer.
(III) Only one isomer is produced if the reactant complex ion is a trans-isomer.
(IV) Only one isomer is produced if the reactant complex ion is a cis-isomer.
The correct statements are: (2018)
(a) (I) and (II)
(b) (I) and (III)
(c) (III) and (IV)
(d) (II) and (IV)
Q.61. The correct combination is: (2018)
(a) [NiCl4]2 - paramagnetic ; [Ni(CO)4] — tetrahedral
(b) [NiCl4]2— — square-planar ; [Ni(CN)4]2 - paramagnetic
(c) [NiCl4]2- — diamagnetic ; [Ni(CO)4] —square-planar
(d) [Ni(CO)4]2- — tetrahedral; [Ni(CN)4 — paramagnetic
(1) [Ni (Cl)4]2 → d8 (Ni)2+
Cl− is weak
Field ligand → So due to unpraised is paramagnetic
Ni(CO)4 → Tetrachedral
Q.62. When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3 ; a dark green product is formed which disproportioates in acidic solution to afford a dark purple solution. X is: (2018)
Q.63. In a complexometric titration of metal ion with ligand M (Metal ion) + L (Ligand) → C (Complex) end point is estimated spectrophotometrically (through light absorption). If 'M' and 'C' do not absorb light and only 'L' absorbs, then the titration plot between absorbed light (A) versus volume of ligand 'L' (V) would look like: (2018)
Initially ligand consumed by metal due to formation of complex. So absorbed light (A) remain constant, after complex formation is completed, extra volume of ligand solution increases ligand concentration and also increases absorbed light.
Q.64. In Wilkinson's catalyst, the hybridization of central metal ion and its shape are respectively: (2018)
(a) sp3d, trigonal bipyramidal
(b) d2sp3, octahedral
(c) dsp2, square planar
(d) sp3, tetrahedral
Q.65. Which of the following complexes will show geometrical isomerism? (2018)
(a) Potassium tris(oxalato)chromate(III)
(c) Aquachlorobis(ethylenediamine)cobalt(II) chloride
(d) Potassium amminetrichloroplatinate(II)
Q.66. On treatment of 100 mL of 0.1 M solution of CoCl3 × 6H2O with excess AgNO3; 1.2 × 1022 ions are precipitated. The complex is (2017)
(a) [Co(H2O)4Cl2]Cl • 2H2O
(b) [Co(H2O)3Cl3] • 3H2O
(d) [Co(H2O)5Cl]Cl2 • H2O
Millimoles of AgNO3 =
Millimoles of CoCl3·6H2O = 0.1 × 100 = 10
∴ Each mole of CoCl3·6H2O gives two chloride ions.
Q.67. The pair of compounds having metals in their highest oxidation state is: (2017)
(a) MnO2 and CrO2Cl2
(b) [Fe(CN)6]3- and [Cu(CN)4]2-
(c) [NiCl4]2- and [CoCl4]2-
(d) [FeCl4]- and Co2O3
MnO2 = + 4
CrO2Cl2= + 6
Q.68. Which of the following ions does not liberate hydrogen gas on reaction with dilute acids? (2017)
Q.69. [Co2(CO)8] displays: (2017)
(a) no Co–Co bond, six terminal CO and two bridging CO
(b) no Co–Co bond, four terminal CO and four bridging CO
(c) one Co–Co bond, six terminal CO and two bridging CO
(d) one Co-Co bond, four terminal CO and four bridging CO
Q.70. The reaction of zinc with dilute and concentrated nitric acid, respectively produces: (2016)
(a) N2O and NO2
(b) NO2 and NO
(c) NO and N2O
(d) NO2 and N2O
Zn reacts with dil. HNO3 (20%) to form nitrous oxide (N2O)
Zn reacts with conc. HNO3 (70%) to form nitrogen dioxide (NO2)
4 Zn + 10HNO3(dil.) → 4Zn(NO3)2 + N2O + 5H2O
Zn + 4HNO3 (conc.) → Zn(NO3)2 + 2NO2 + 2H2O
Q.71. The pair having the same magnetic moment is: (2016)
[At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
(a) [Cr(H2O)6 ]2+ and [CoCl4 ]2-
(b) [Cr(H2O)6]2+ and [Fe(H2O)6 2+
(c) [Mn(H2O)6 ]2+ and [Cr(H2O)6 ]2+
(d) [CoCl4 ]2- and [Fe(H2O)6]2+
Cr+2 → (Ar) 3d4 4s0
Fe+2 → (Ar) 3d64s0
Q.72. Which one of the following complexes shows optical isomerism? (2016)
(c) trans[Co(en)2Cl2]Cl (en = ethylenediamine)
(d) [Co(NH3)4Cl2] Cl
Q.73. Which one of the following complexes will consume more equivalents of aqueous solution of Ag(NO3)? (2016)
Complex [Cr(H2O)6]Cl3 will consume more equivalents of aqueous solution of Ag(NO3).
Q.74. Which one of the following species is stable in aqueous solution? (2016)
MnO2-4 disproportionates in neutral or acidic solution.
3MnO2-4 + 4H+ → 2MnO2-4 + MnO2 + 2H2O
(3) Many Cu+ compounds are unstable in aqueous solution and undergo disproportionation as follows
2Cu+ → Cu2+ + Cu
Q.75. Identify the correct trend given below:
(Atomic No.: Ti = 22, Cr = 24 and Mo = 42) (2016)
The splitting is affected by the oxidation state of the central metal ion. A higher oxidation state leads to larger splitting hence
Δ0 of [Ti(H2O)6]3+ < Δ0 of [ Ti(H2O)6]2+
Further Δ0 also depends of Zeff and Zeff of 4d series if more than 3d series.
Δ0 of [Cr(H2O)6]2+ < Δ0 of [Mo(H2O)6]2+
Q.76. The transition metal ions responsible for color in ruby and emerald are, respectively: (2016)
(a) Cr3+ and Cr3+
(b) Co3+ and Co3+
(c) Co3+ and Cr3+
(d) Cr3+ and Co3+
Q.77. Which of the following is an example of homoleptic complex? (2016)
Complex having only 1 type of ligands are examples of homoleptic complex.