JEE Chemistry Coordination Compounds MCQs Set 15

Answer: (a) Diamminedichloridoplatinum (II)

Explanation:
1. Ligands are named in alphabetical order (ammine before chlorido), with prefixes di- not counted in the order.
2. The neutral complex has two neutral NH3 and two Cl-, so the oxidation state of Pt is +2.

Teacher's Note:
1. Alphabetical order ignores the multiplying prefixes.
2. Work out the oxidation state from the charges of the ligands.

Answer: (d) \( \ce{[Cr(H2O)6]Cl3} \)

Explanation:
1. Only chloride ions outside the coordination sphere are precipitated by silver nitrate.
2. Three moles of AgCl mean that all three chloride ions are free, so all water molecules are inside the coordination sphere: \( \ce{[Cr(H2O)6]Cl3} \).

Teacher's Note:
1. The number of moles of AgCl formed equals the number of ionisable chloride ions.
2. Chloride inside the square bracket is not precipitated.

Answer: (a) \( d^2sp^3 \) hybridisation, diamagnetic

Explanation:
1. Fe is in the +2 state (d6). CN- is a strong field ligand, so the six electrons pair up in the t2g orbitals.
2. Two inner d orbitals are free for \( d^2sp^3 \) hybridisation and there are no unpaired electrons, so the complex is diamagnetic.

Teacher's Note:
1. Strong field ligands pair the electrons and give inner orbital complexes.
2. Count the d electrons of the metal ion first.

Answer: (d) no isomerism

Explanation:
1. The two compounds have the same complex cation but different counter ions, Br- and Cl-.
2. They have different molecular formulae, so they are two different compounds and not isomers.

Teacher's Note:
1. Isomers must have the same molecular formula.
2. Ionisation isomers need an exchange between a ligand and the counter ion.

Answer: (b) \( \ce{NH4+} \)

Explanation:
1. A ligand must have a lone pair to donate to the metal.
2. \( \ce{NH4+} \) has no lone pair on nitrogen, whereas NO, ethylenediamine and CO can donate electrons.

Teacher's Note:
1. Ligands are electron pair donors.
2. The nitrogen of ammonium has used its lone pair to bind a proton.

Answer: (d) (a) and (c) both

Explanation:
1. \( \ce{[Co(en)3]^3+} \) has a propeller shape and no plane of symmetry, so it has two enantiomers.
2. The cis form of \( \ce{[Co(en)2Cl2]+} \) also lacks a plane of symmetry and exists as a non-superimposable mirror image pair. The trans form is symmetrical and optically inactive.

Teacher's Note:
1. Complexes with two or three bidentate ligands arranged in a cis manner are chiral.
2. Look for a plane of symmetry.

Answer: (b) tetrahedral, square planar and octahedral

Explanation:
1. \( \ce{Cl-} \) is a weak field ligand and gives tetrahedral \( \ce{[NiCl4]^2-} \).
2. \( \ce{CN-} \) is a strong field ligand and gives square planar \( \ce{[Ni(CN)4]^2-} \).
3. Water gives octahedral \( \ce{[Ni(H2O)6]^2+} \).

Teacher's Note:
1. The shapes depend on the field strength of the ligand for d8 ions.
2. Strong field ligand means dsp2 square planar for Ni(II).

Answer: (b) 2, 2, 3

Explanation:
1. The complex is of the type \( \ce{M(AA)2B2} \). It has two geometrical isomers, cis and trans.
2. The trans form is optically inactive and the cis form exists as a pair of enantiomers, so there are 2 optical isomers.
3. The total number of stereoisomers is 1 (trans) + 2 (cis pair) = 3.

Teacher's Note:
1. Only the cis isomer of M(AA)2B2 is chiral.
2. Total isomers are counted by listing all different structures.

Answer: (d) \( \ce{[CoBr2(NH3)4]Cl} \)

Explanation:
1. AgCl forms immediately, so chloride is outside the coordination sphere. AgBr does not form, so both bromides are inside.
2. The formula is \( \ce{[CoBr2(NH3)4]Cl} \), which gives two ions (the complex cation and Cl-) in solution.

Teacher's Note:
1. Ions outside the bracket give an immediate precipitate with silver nitrate.
2. Two ions per formula unit means one complex ion and one counter ion.

Answer: (c) 3

Explanation:
1. The complex dissociates as \( \ce{[Co(NH3)6]Cl2 -> [Co(NH3)6]^2+ + 2Cl-} \).
2. That gives 1 complex ion and 2 chloride ions, a total of 3 ions.

Teacher's Note:
1. The coordination sphere does not break up in water.
2. Count the complex ion as one particle.

Answer: (b) CO

Explanation:
1. A pi-acid ligand accepts electron density from filled metal d orbitals into its empty pi* orbitals.
2. CO has empty pi* orbitals and does this back bonding, whereas NH3, F- and en have no such orbitals.

Teacher's Note:
1. CO, CN-, NO+ and phosphines are typical pi-acid ligands.
2. Back bonding strengthens the metal-carbon bond.

Answer: (d) \( dsp^2 \)

Explanation:
1. \( \ce{Cu^2+} \) is d9. The unpaired electron of the 3d orbital is promoted to the 4p orbital so that a 3d orbital is vacated.
2. One d, one s and two p orbitals hybridise to give \( dsp^2 \) and the complex is square planar.

Teacher's Note:
1. Cu(II) forms square planar ammine complexes.
2. The promoted electron makes the complex paramagnetic with one unpaired electron.

Answer: (a) 4

Explanation:
1. Each Co atom needs 36 electrons. Co has 27, the Co-Co bond contributes 1 and each CO donates 2.
2. \( 27 + 1 + 2n = 36 \), so \( n = 4 \). The compound is \( \ce{Co2(CO)8} \).

Teacher's Note:
1. Use the EAN rule with the metal-metal bond counted.
2. Each metal-metal bond adds one electron to each metal.

Answer: (a) \( \ce{[Cu(NH3)4]^2+} \)

Explanation:
1. \( \ce{Cu^2+} \) has 27 electrons and four NH3 add 8, giving 35, which is not 36.
2. \( \ce{Zn^2+} \) (28 + 8), \( \ce{Hg^2+} \) (78 + 8) and Fe in Fe(CO)5 (26 + 10) all reach the noble gas number of electrons.

Teacher's Note:
1. Compute EAN for each option.
2. Cu(II) complexes are the common exceptions.

Answer: (c) \( \ce{[Co(NH3)3Cl3]} \) : 2

Explanation:
1. \( \ce{[Co(NH3)3Cl3]} \) (MA3B3) has two geometrical isomers, facial and meridional.
2. \( \ce{[Co(en)3]^3+} \) has 2, \( \ce{[Co(en)2Cl2]+} \) has 3 (trans plus cis pair) and square planar \( \ce{[Pt(NH3)2Cl2]} \) has 2, so the other matches are wrong.

Teacher's Note:
1. MA3B3 gives fac and mer isomers.
2. Count each enantiomer as a separate stereoisomer.

Answer: (d) \( \ce{[Co(NH3)3Cl3]} \)

Explanation:
1. Both isomers of \( \ce{[Co(NH3)3Cl3]} \), fac and mer, have a plane of symmetry, so neither is chiral.
2. The complexes with one or more bidentate en ligands in a cis arrangement (options a, b and c) can have non-superimposable mirror images.

Teacher's Note:
1. A plane of symmetry rules out optical activity.
2. Complexes without a chelating ligand are rarely chiral.

Answer: (d) square planar and paramagnetic

Explanation:
1. The deep blue complex is \( \ce{[Cu(NH3)4]^2+} \). Cu2+ is d9 and uses \( dsp^2 \) hybridisation, so the shape is square planar.
2. One unpaired electron remains, so the complex is paramagnetic.

Teacher's Note:
1. d9 complexes always keep one unpaired electron.
2. Square planar means dsp2 hybridisation.

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

Explanation:
1. The oxalate ion is a bidentate ligand and forms three five-membered chelate rings with Fe3+.
2. The chelate effect makes \( \ce{[Fe(C2O4)3]^3-} \) the most stable among the options.

Teacher's Note:
1. Look for the chelating ligand first.
2. Chelates beat complexes of monodentate ligands.

Answer: (b) Zeise's salt

Explanation:
1. An organometallic compound has at least one metal-carbon bond to an organic group.
2. Zeise's salt, \( \ce{K[PtCl3(\eta^2-C2H4)]} \), has a bond between platinum and the carbon atoms of ethene. The others have no metal-carbon bond of an organic group.

Teacher's Note:
1. Cyanide and carbonyl complexes are normally not called organometallic when the carbon is not part of an organic group.
2. Zeise's salt is the classic example of a pi complex.

Answer: (d) \( \ce{CoCl3.3NH3} \)

Explanation:
1. \( \ce{CoCl3.3NH3} \) is the neutral complex \( \ce{[Co(NH3)3Cl3]} \), so it does not ionise.
2. The other three give 4, 3 and 2 ions respectively.

Teacher's Note:
1. A neutral complex has no counter ions.
2. The formula unit with all chlorides inside gives no ions.

Answer: (c) +1

Explanation:
1. The brown ring is \( \ce{[Fe(H2O)5(NO)]^2+} \), and water is neutral.
2. In this complex NO is present as \( \ce{NO+} \), so Fe is \( +2 - 1 = +1 \).

Teacher's Note:
1. NO behaves as NO+ in the brown ring complex.
2. The charge of the complex minus the charge of the ligands gives the oxidation state.

Answer: (b) 5

Explanation:
1. Fe3+ is d5, and fluoride is a weak field ligand, so the electrons do not pair.
2. The complex \( \ce{[FeF6]^3-} \) uses \( sp^3d^2 \) hybridisation and has 5 unpaired electrons.

Teacher's Note:
1. Outer orbital means the weak field ligand case.
2. Weak field ligands leave the d electrons unpaired.

Answer: (c) both tetrahedral

Explanation:
1. In \( \ce{[Ni(CO)4]} \), Ni is d10 (4s2 3d8 rearranged), so the hybridisation is \( sp^3 \) and the shape is tetrahedral.
2. \( \ce{[Ni(PPh3)2Cl2]} \) contains the bulky ligands PPh3 and Cl-, and Ni2+ forms the tetrahedral \( sp^3 \) complex.

Teacher's Note:
1. Zero-valent nickel carbonyl is tetrahedral.
2. Bulky ligands favour tetrahedral over square planar.

Answer: (c) \( \ce{[PtCl4I2]^2-} \)

Explanation:
1. \( \ce{[PtCl4I2]^2-} \) is an octahedral complex of type \( \ce{MA4B2} \) and has cis and trans isomers.
2. A linear complex (a), a tetrahedral complex (b) and the cation and anion pair (d), each with only one arrangement, do not show geometrical isomerism.

Teacher's Note:
1. Linear and tetrahedral complexes never show geometrical isomerism.
2. Check for MA4B2 octahedral or MA2B2 square planar patterns.

Answer: (b) \( \ce{[Co(en)2Cl2]Cl} \)

Explanation:
1. Enantiomorphs are non-superimposable mirror images.
2. Only the cis form of \( \ce{[Co(en)2Cl2]+} \) is chiral. The other options are symmetrical and have no mirror image isomers.

Teacher's Note:
1. Enantiomorph means optical isomer.
2. Chiral complexes contain chelating ligands in a cis arrangement.

Answer: (c) \( \ce{CuF2} \)

Explanation:
1. Colour in these compounds arises from d-d transitions, which need unpaired d electrons.
2. Cu+ (d10) in options (a), (b) and (d) cannot have d-d transitions, so these are colourless. Cu2+ (d9) in \( \ce{CuF2} \) has a vacancy and is coloured.

Teacher's Note:
1. d10 and d0 ions are colourless in the absence of charge transfer.
2. Cu(I) compounds are colourless, Cu(II) compounds are coloured.