JEE Chemistry Coordination Compounds MCQs Set 17

Answer: (a) 4 water molecules

Explanation:
1. Four of the water molecules are bound to \( \ce{Cu^{2+}} \) in a square plane, forming \( \ce{[Cu(H2O)4]^{2+}} \).
2. The fifth water molecule is held by hydrogen bonds between the sulphate ion and the coordinated water.

Teacher's Note:
1. The hydrated salt is better written as \( \ce{[Cu(H2O)4]SO4.H2O} \).
2. Only one water molecule is not coordinated.

Answer: (a) \( \ce{[Ni(CN)4]^{2-}} \) is square planar, and \( \ce{[NiCl4]^{2-}} \) and \( \ce{Ni(CO)4} \) are tetrahedral

Explanation:
1. Cyanide is a strong field ligand and pairs the electrons of \( \ce{Ni^{2+}} \) (\( d^8 \)), giving \( dsp^2 \) and a square planar shape.
2. Chloride is a weak field ligand, giving \( sp^3 \) and a tetrahedral shape.
3. In \( \ce{Ni(CO)4} \) nickel is \( 3d^{10} \) (after pairing), hybridised \( sp^3 \), also tetrahedral.

Teacher's Note:
1. Strong field ligands with d8 ions give square planar complexes.
2. Zero-valent nickel (d10) is always tetrahedral with four ligands.

Answer: (c) \( \ce{[CrCl2(H2O)4]Cl.2H2O} \)

Explanation:
1. One-third of the chlorine means one of the three chloride ions is ionisable (outside the coordination sphere).
2. That corresponds to \( \ce{[CrCl2(H2O)4]Cl.2H2O} \), which has two chlorides inside the bracket and one outside.

Teacher's Note:
1. Only ionisable chloride reacts with silver nitrate.
2. Match the fraction precipitated to the number of free chloride ions.

Answer: (b) excitation of electrons from d to d

Explanation:
1. Ti(III) is \( d^1 \). Visible light promotes the electron from the \( t_{2g} \) to the \( e_g \) level.
2. This d-d transition absorbs in the blue-green region, so the complex appears violet.

Teacher's Note:
1. Colour in most transition metal complexes comes from d-d transitions.
2. No d electrons means no d-d colour.

Answer: (d) All of the above are correct

Explanation:
1. Ammonia is a strong enough ligand to pair the electrons of \( \ce{Co^{3+}} \) (\( d^6 \)) in the \( t_{2g} \) orbitals.
2. Two inner d orbitals are then free for \( d^2sp^3 \) hybridisation, giving an octahedral, diamagnetic, inner orbital complex.
3. Statements (a), (b) and (c) are all true, so (d) is correct.

Teacher's Note:
1. Low spin d6 octahedral complexes are diamagnetic.
2. Inner orbital means the (n-1)d orbitals take part in hybridisation.

Answer: (a) \( \ce{[Ni(H2O)6]^{2+}} \)

Explanation:
1. \( \ce{Ni^{2+}} \) is \( d^8 \); in an octahedral field it keeps two unpaired electrons in the \( e_g \) set.
2. \( \ce{[Fe(CN)6]^{4-}} \) is low spin \( d^6 \), \( \ce{Ni(CO)4} \) is \( d^{10} \) and square planar \( \ce{[Ni(CN)4]^{2-}} \) is low spin \( d^8 \); all three are diamagnetic.

Teacher's Note:
1. Draw the d-electron configuration to count the unpaired electrons.
2. Strong ligands such as CN- and CO pair the electrons.

Answer: (d) all of these

Explanation:
1. Ions of the 4d and 5d series have large crystal field splitting, so their complexes are nearly always low spin.
2. \( \ce{Rh^{3+}} \), \( \ce{Pt^{2+}} \) and \( \ce{Ru^{2+}} \) all belong to these series.

Teacher's Note:
1. Splitting increases down a group.
2. Heavier transition metals form low spin complexes even with weaker ligands.

Answer: (a) \( \ce{Cl-} < \ce{F-} < \ce{C2O4^{2-}} < \ce{NO2-} < \ce{CN-} \)

Explanation:
1. In the spectrochemical series, chloride is weaker than fluoride, which is weaker than oxalate.
2. Nitrite and then cyanide are strong field ligands, so the order is \( \ce{Cl-} < \ce{F-} < \ce{C2O4^{2-}} < \ce{NO2-} < \ce{CN-} \).

Teacher's Note:
1. Halides come at the weak end, with Cl weaker than F.
2. Option (b) is not even a consistent increasing order.

Answer: (b) \( sp^3d^2 \)

Explanation:
1. Fluoride is a weak field ligand, so it does not pair the d electrons of \( \ce{Co^{3+}} \).
2. The six ligands use the outer 4s, 4p and 4d orbitals, giving \( sp^3d^2 \) hybridisation (an outer orbital complex).

Teacher's Note:
1. High spin means outer orbital complex.
2. Count 1 s, 3 p and 2 d orbitals for six bonds.

Answer: (b) trigonal bipyramidal

Explanation:
1. Iron in \( \ce{Fe(CO)5} \) is \( 3d^8 4s^2 \) and the electrons pair to leave one d orbital free.
2. Hybridisation is \( dsp^3 \), giving a trigonal bipyramidal shape.

Teacher's Note:
1. Five ligands around a central atom give a trigonal bipyramid.
2. This molecule obeys the E.A.N. rule (26 + 10 = 36).

Answer: (b) \( \ce{Ru(CO)5} \)

Explanation:
1. Ruthenium has atomic number 44. To reach the 54 electrons of xenon, it needs 10 more electrons from carbonyl ligands.
2. Each CO donates 2 electrons, so five CO are needed: \( \ce{Ru(CO)5} \).

Teacher's Note:
1. Use the E.A.N. rule to predict the formula of a simple carbonyl.
2. Number of CO = (noble gas atomic number - metal atomic number) / 2.

Answer: (d) all of these

Explanation:
1. Carnallite (\( \ce{KCl.MgCl2.6H2O} \)), Mohr's salt (\( \ce{FeSO4.(NH4)2SO4.6H2O} \)) and alum (\( \ce{K2SO4.Al2(SO4)3.24H2O} \)) are all double salts.
2. All of them dissociate completely into simple ions in water.

Teacher's Note:
1. A double salt is stable only in the solid state.
2. A complex ion retains its identity in solution.

Answer: (c) \( \ce{[Fe(C2O4)3]^{3-}} \)

Explanation:
1. Cyanide is a carbon donor strong field ligand, so both cyanide complexes are low spin.
2. Oxalate is an oxygen donor and not strong enough to force pairing in \( \ce{Fe^{3+}} \) (\( d^5 \)), so \( \ce{[Fe(C2O4)3]^{3-}} \) is high spin.

Teacher's Note:
1. Oxygen donor ligands are comparatively weak.
2. Compare the donor atom of each ligand.

Answer: (c) optical isomers

Explanation:
1. None of the ligands is ambidentate and the complex is a single ion, so linkage and coordination isomerism are impossible.
2. The cis arrangement of the two chloride ligands (with the two ammonia molecules also cis) has no mirror plane, so that form is optically active.

Teacher's Note:
1. Ambidentate ligands such as NO2- and SCN- are needed for linkage isomerism.
2. A bidentate chelate with cis monodentate ligands often gives a chiral complex.

Answer: (b) H.O.M.O. to L.U.M.O. transition

Explanation:
1. Halogen molecules have no d electrons.
2. Visible light promotes an electron from the highest occupied molecular orbital (a pi antibonding orbital) to the lowest unoccupied one (a sigma antibonding orbital), which gives them their colour.

Teacher's Note:
1. The colour deepens down the group as the gap narrows.
2. d-d transitions need partially filled d orbitals.

Answer: (d) all of the above

Explanation:
1. Chromium is in the +6 state (\( d^0 \)) in all three species, so no d-d transition is possible.
2. Their colours arise from charge transfer from oxygen or chlorine to chromium.

Teacher's Note:
1. Look at the d-electron count first.
2. Charge transfer gives intense colours.

Answer: (c) \( \ce{CuSO4.5H2O} \)

Explanation:
1. \( \ce{Cu^{2+}} \) is \( d^9 \) with one unpaired electron, and its hydrated salt is blue because of a d-d transition.
2. \( \ce{KMnO4} \) (Mn(VII), \( d^0 \)) is diamagnetic, AgI (Ag(I), \( d^{10} \)) is diamagnetic, and the square planar \( \ce{[Ni(CN)4]^{2-}} \) is diamagnetic.

Teacher's Note:
1. Paramagnetism needs unpaired electrons.
2. Count the d electrons before judging colour or magnetism.

Answer: (c) \( \ce{[Pt(NH3)2Cl2Br2]} \)

Explanation:
1. Square planar \( \ce{[Pt(NH3)2(NO2)2]} \) has a plane of symmetry.
2. \( \ce{[Co(en)(NH3)3Cl]+} \) (Ma3b(AA)) and \( \ce{[Pt(NH3)4ClBr]^{2+}} \) (Ma4bc) always have a mirror plane.
3. The octahedral Ma2b2c2 type \( \ce{[Pt(NH3)2Cl2Br2]} \) has an all-cis arrangement with no mirror plane, which is chiral.

Teacher's Note:
1. Octahedral Ma2b2c2 is the first simple type with enantiomers.
2. Check each geometry for a plane of symmetry.

Answer: (b) sodium nitroprusside

Explanation:
1. Sulphide ions react with sodium nitroprusside to give the violet complex \( \ce{Na4[Fe(CN)5(NOS)]} \), used as a test for sulphide.

Teacher's Note:
1. Remember the test: S2- with nitroprusside gives violet.
2. Nitroprusside is \( \ce{Na2[Fe(CN)5NO]} \).

Answer: (d) \( \ce{KFe[Fe(CN)6]} \)

Explanation:
1. Prussian blue is made from \( \ce{Fe^{3+}} \) and \( \ce{[Fe(CN)6]^{4-}} \), and Turnbull's blue from \( \ce{Fe^{2+}} \) and \( \ce{[Fe(CN)6]^{3-}} \).
2. Both are in fact the same soluble compound \( \ce{KFe[Fe(CN)6]} \) when potassium ions are present.

Teacher's Note:
1. Each of the older formulas applies to only one of the two.
2. Modern studies show the two blues are identical.

Answer: (d) \( \ce{C6H4(OH)2 + 2AgBr -> C6H4O2 + 2HBr + 2Ag} \)

Explanation:
1. Reaction (a) prepares the emulsion, (c) is the exposure and (b) is fixing, where hypo dissolves unchanged silver bromide as a complex.
2. Developing uses a mild reducing agent such as hydroquinone to reduce the exposed silver bromide to metallic silver.

Teacher's Note:
1. Developer reduces exposed AgBr; hypo removes the unexposed AgBr.
2. Fixing is a complex formation reaction.

Answer: (a) Cis-platin

Explanation:
1. Cis-platin, \( \ce{cis-[Pt(NH3)2Cl2]} \), binds to the DNA of cancer cells and blocks their division.
2. Zeise's salt has no such medical use.

Teacher's Note:
1. Only the cis isomer is active.
2. Platinum complexes are important anticancer drugs.