JEE Chemistry Coordination Compounds MCQs Set 25

Answer: (b) bidentate

Explanation:
1. H2N-CH2-CH2-NH2 has two nitrogen donor atoms
2. Both can bind the same metal, so it is bidentate and forms a five-membered chelate ring

Teacher's Note:
1. Count the donor atoms that can bind one metal ion at the same time.

Answer: (c) two nitrogen atoms and four oxygen atoms

Explanation:
1. EDTA has two tertiary amine nitrogens and four carboxylate groups
2. It is hexadentate, using 2 N and 4 O donors

Teacher's Note:
1. EDTA is the standard hexadentate chelating ligand used in titrations and in treating metal poisoning.

Answer: (a) complex salts

Explanation:
1. In a complex salt one metal is held inside the coordination sphere and does not release its ions
2. Example: K4[Fe(CN)6] gives a test for K+ but not for Fe2+

Teacher's Note:
1. A double salt such as alum gives tests for all of its ions, a complex salt does not.

Answer: (b) Lewis bases

Explanation:
1. A ligand donates a lone pair to the central metal ion
2. So it behaves as a Lewis base and the metal ion as a Lewis acid

Teacher's Note:
1. Ligands can be neutral or anionic, but they are always electron-pair donors.

Answer: (b) +3

Explanation:
1. O is -2 (four of them: -8); C2H4 and H2O are neutral
2. 2x - 8 = -2, so 2x = +6 and x = +3

Teacher's Note:
1. Neutral ligands do not change the oxidation-state sum.

Answer: (d) 6

Explanation:
1. Four NH3 and two Cl- are each bonded through one donor atom
2. Coordination number = 4 + 2 = 6

Teacher's Note:
1. Coordination number counts the donor atoms attached to the metal, not the number of ligand species.

Answer: (b) [Cu(CN)4]3-

Explanation:
1. In [Cu(CN)4]3- the sum is x + 4(-1) = -3, so x = +1
2. The other two ions are Cu(II) because water and ammonia are neutral and the ion charge is +2

Teacher's Note:
1. Divide the overall charge among the ligands first and what remains is the metal oxidation state.

Answer: (a) 6 and +3

Explanation:
1. Two Cl- and two bidentate en give 2 + 4 = 6 donor atoms
2. The complex ion is [CoCl2(en)2]+, so x - 2 + 0 = +1 and x = +3

Teacher's Note:
1. A bidentate ligand contributes two to the coordination number.

Answer: (d) -1

Explanation:
1. Charge = +3 + 2(-2) + 0 = -1

Teacher's Note:
1. Oxalate carries a charge of -2.

Answer: (d) [Co(en)3]2(SO4)3

Explanation:
1. The cation is [Co(en)3]3+ and sulphate is 2-
2. Two cations (+6) balance three sulphates (-6), giving [Co(en)3]2(SO4)3

Teacher's Note:
1. Balance the charge using the criss-cross rule between cation and anion.

Answer: (d) tetraamminechloridonitrito-O-cobalt(III) chloride

Explanation:
1. Ligands are named in alphabetical order: ammine, chlorido, nitrito
2. ONO- binds through oxygen, so it is nitrito-O
3. The cation charge is +1, so cobalt is +3

Teacher's Note:
1. The binding atom of an ambidentate ligand is written after its name.

Answer: (c) [Co(NH3)3(H2O)2Cl]Cl2

Explanation:
1. The coordination sphere has 3 NH3, 2 H2O and 1 Cl-, six donors in all
2. Charge on the cation = +3 - 1 = +2, so two Cl- remain outside

Teacher's Note:
1. Ligand formulae are written in the order: anionic, neutral, with the metal first.

Answer: (c) 5

Explanation:
1. EAN = Z - oxidation state + 2 x number of ligands
2. 26 - 0 + 2x = 36, so x = 5

Teacher's Note:
1. Fe(CO)5 follows the EAN rule exactly and is a stable carbonyl.

Answer: (b) 86

Explanation:
1. Pt(IV) has 78 - 4 = 74 electrons
2. Six Cl- donate 6 x 2 = 12 electrons
3. EAN = 74 + 12 = 86 (the radon configuration)

Teacher's Note:
1. EAN = Z - oxidation number + 2 x number of ligands.

Answer: (d) 25

Explanation:
1. Only the ionisable Cl- outside the coordination sphere precipitates
2. That is 1 of the 4 chlorine atoms, which is 25 percent

Teacher's Note:
1. Chloride inside the coordination sphere is not free and does not react with Ag+.

Answer: (c) [Pt(NH3)5Cl]Cl3

Explanation:
1. [Pt(NH3)5Cl]Cl3 gives one complex cation and three Cl- ions, four in all
2. [Pt(NH3)4Cl2]Cl2 gives three ions and [Pt(NH3)4]Cl4 gives five
3. [Pt(NH3)2Cl4] gives none

Teacher's Note:
1. Count the complex ion as one particle and add the counter ions.

Answer: (a) [Co(NH3)6]Cl3

Explanation:
1. Molar conductance increases with the number of ions: [Co(NH3)6]Cl3 gives 4 ions
2. [Co(NH3)5Cl]Cl2 gives 3, [Co(NH3)4Cl2]Cl gives 2 and [Co(NH3)3Cl3] is a non-electrolyte

Teacher's Note:
1. Four ions (1:3 electrolyte) conduct best of the four.

Answer: (b) Cl- ions

Explanation:
1. Only Cl- is outside the coordination sphere
2. Br- and Cr3+ are inside the complex ion and do not give their usual tests

Teacher's Note:
1. Dissociation separates only the counter ion.

Answer: (a) [MnO4]-

Explanation:
1. Mn in MnO4- is +7 with configuration 3d0
2. Co3+ is d6, Fe3+ is d5 and Cr3+ is d3

Teacher's Note:
1. Subtract the oxidation state from the group number to count d electrons.

Answer: (c) d2sp3

Explanation:
1. Cr(0) is d5 s1 and CO is a strong field ligand, so the electrons pair in the 3d orbitals
2. Two 3d, one 4s and three 4p orbitals hybridise to d2sp3

Teacher's Note:
1. Strong field ligands give inner-orbital (d2sp3) octahedral complexes.

Answer: (a) [Pt(en)2]2+

Explanation:
1. Pt(II) (5d8) forms square planar complexes
2. Ni(CO)4 (Ni0, sp3), [Zn(NH3)4]2+ (d10) and [NiCl4]2- (weak field Cl-) are tetrahedral

Teacher's Note:
1. d8 ions of the 5d series are always square planar.

Answer: (a) sp3d2 and 4

Explanation:
1. Fe2+ is d6 and water is a weak field ligand, so no pairing occurs
2. The outer 4d orbitals are used, giving sp3d2 with 4 unpaired electrons (t2g4 eg2)

Teacher's Note:
1. Weak field ligands give high spin outer-orbital complexes.

Answer: (d) 0

Explanation:
1. Low spin d6 is t2g6 eg0
2. All six electrons are paired in the t2g orbitals

Teacher's Note:
1. Low spin d6 complexes such as [Co(NH3)6]3+ are diamagnetic.

Answer: (d) [FeF6]3-

Explanation:
1. F- is a weak field ligand, so Fe3+ (d5) in [FeF6]3- stays t2g3 eg2
2. NH3 and CN- are strong enough to pair electrons in the other octahedral complexes, and [Ni(CN)4]2- is low spin square planar

Teacher's Note:
1. Compare the position of the ligand in the spectrochemical series.

Answer: (c) [Mn(H2O)6]2+

Explanation:
1. Mn2+ is d5 and water is weak, so there are 5 unpaired electrons
2. Cu2+ has 1 and [Fe(CN)6]4- (low spin d6) has 0

Teacher's Note:
1. Spin-only moment is sqrt(n(n+2)) with n unpaired electrons, here 5.92 BM.