JEE Chemistry Coordination Compounds MCQs Set 23

Answer: (a) 1 and 3

Explanation:
1. Co\(^{3+}\) (d\(^6\)) and Fe\(^{2+}\) (d\(^6\)) with strong field ligands pair all six electrons in the t\(_{2g}\) orbitals, so these complexes are diamagnetic.
2. Mn\(^{3+}\) (d\(^4\)) has two unpaired electrons and Fe\(^{3+}\) (d\(^5\)) has one unpaired electron even in the low spin state.

Teacher's Note:
1. Low spin d\(^6\) is diamagnetic.

Answer: (a) 1 and 3

Explanation:
1. Mn\(^{3+}\) (d\(^4\)) in a weak field has 4 unpaired electrons, Fe\(^{3+}\) (d\(^5\)) has 5, Co\(^{3+}\) (d\(^6\)) has 4 and Ni\(^{2+}\) (d\(^8\)) has 2.
2. So \( \ce{[MnCl6]^{3-}} \) and \( \ce{[CoF6]^{3-}} \) both have 4.

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

Answer: (a) 1 and 3

Explanation:
1. Optical activity needs no plane or centre of symmetry.
2. \( \ce{[Co(en)3]^{3+}} \) and cis-\( \ce{[Co(en)2Cl2]^+} \) are chiral.
3. The trans isomer has a plane of symmetry and \( \ce{[Cr(NH3)5Cl]} \) is also symmetric.

Teacher's Note:
1. Tris-chelate and cis bis-chelate octahedral complexes are chiral.

Answer: (a) 1, 2 and 3

Explanation:
1. Ethylenediamine \( \ce{H2N-CH2-CH2-NH2} \) carries no charge, and the two nitrogen atoms each donate a lone pair (bidentate).
2. Bonding through two donor atoms forms a ring, so it is chelating. It is not unidentate.

Teacher's Note:
1. Chelating ligands form ring structures.

Answer: (a) 1 and 3

Explanation:
1. Linkage isomerism needs an ambidentate ligand.
2. \( \ce{NO2^-} \) (N or O donor) and \( \ce{SCN^-} \) (S or N donor) are ambidentate, so complexes 1 and 3 show it.

Teacher's Note:
1. Ambidentate ligands: NO\(_2^-\), SCN\(^-\), CN\(^-\).

Answer: (c) \( \Delta_o[\ce{CrCl6}] < \Delta_o[\ce{Cr(H2O)6}] < \Delta_o[\ce{Cr(NH3)6}] \)

Explanation:
1. The energy absorbed is \( \Delta_o \propto 1/\lambda_{max} \).
2. The longest wavelength (758 nm, CrCl\(_6^{3-}\)) means the smallest \( \Delta_o \), then H\(_2\)O (694 nm), and the shortest wavelength (465 nm, NH\(_3\)) means the largest \( \Delta_o \).

Teacher's Note:
1. Larger absorption wavelength means smaller splitting.

Answer: (c) Ethylenediamine triacetate ion

Explanation:
1. Ethylenediamine triacetate has 2 nitrogen donors and 3 carboxylate oxygen donors, i.e. 5.
2. Triethylene tetramine has 4, EDTA has 6 and diethylene triamine has 3.

Teacher's Note:
1. EDTA is hexadentate.

Answer: (d) \( \ce{[Co(NH3)5Br]SO4} \)

Explanation:
1. Ionisation isomers exchange a ligand in the coordination sphere with the counter ion.
2. \( \ce{[Co(NH3)5Br]SO4} \) and \( \ce{[Co(NH3)5SO4]Br} \) are such a pair; the chlorides listed have only Cl\(^-\) outside the sphere and no other counter ion to exchange.

Teacher's Note:
1. Look for a ligand that can also be a counter ion.

Answer: (d) 1, 2 and 3

Explanation:
1. Statement 1 describes a complex ion correctly.
2. Statement 2 is the definition of ligands.
3. Statement 3 is the effective atomic number (EAN) rule.
4. So all three are correct.

Teacher's Note:
1. EAN rule is satisfied by many carbonyls.

Answer: (d) \( \ce{[Cu(NH3)4]^{2+}} \) has a tetrahedral structure and is paramagnetic

Explanation:
1. Water ligands are replaced by ammonia (substitution), and the colour changes from light blue to deep blue. NH\(_3\) is a stronger field ligand than H\(_2\)O.
2. \( \ce{[Cu(NH3)4]^{2+}} \) is square planar (dsp\(^2\)) with one unpaired electron, so it is not tetrahedral.

Teacher's Note:
1. Cu(II) d\(^9\) complexes with four ligands are normally square planar.

Answer: (c) dichlorobis(methylamine)copper(II)

Explanation:
1. The ligands are named alphabetically: chlorido and methylamine.
2. The complex is neutral, so the name is dichloridobis(methylamine)copper(II).

Teacher's Note:
1. Polydentate ligands names with complex prefixes use bis, tris.

Answer: (b) \( \ce{[Cr(H2O)4Cl2](NO2)} \)

Explanation:
1. Ionisation isomers give different ions in solution because a ligand and a counter ion are exchanged.
2. Exchanging the coordinated NO\(_2^-\) with the outer Cl\(^-\) gives \( \ce{[Cr(H2O)4Cl2](NO2)} \).

Teacher's Note:
1. Choice (c) is a linkage isomer and (d) is a hydrate isomer.

Answer: (c) measuring the solid state magnetic moment

Explanation:
1. Fe(II) with NO\(^+\) (both diamagnetic as d\(^6\) low spin) gives a diamagnetic complex.
2. Fe(III) with NO (odd electron) would be paramagnetic, so the magnetic moment distinguishes the two forms.

Teacher's Note:
1. Magnetic measurements show the number of unpaired electrons.

Answer: (b) \( d_{xy}, d_{yz} \) and \( d_{zx} \) orbitals

Explanation:
1. Fe\(^{3+}\) is d\(^5\) and CN\(^-\) is a strong field ligand.
2. In the octahedral field the electrons fill the lower t\(_{2g}\) set (\( d_{xy}, d_{yz}, d_{zx} \)), as \( t_{2g}^5 \).

Teacher's Note:
1. The e\(_g\) orbitals stay empty in low spin d\(^5\).

Answer: (c) A is diamagnetic and B is paramagnetic with two unpaired electrons

Explanation:
1. A is \( \ce{[Ni(CN)4]^{2-}} \): strong field CN\(^-\) pairs the d\(^8\) electrons (square planar), so it is diamagnetic.
2. B is \( \ce{[NiCl4]^{2-}} \): weak field Cl\(^-\) gives a tetrahedral complex with two unpaired electrons.

Teacher's Note:
1. Ni(II) is d\(^8\).

Answer: (b) \( L_1 < L_3 < L_2 < L_4 \)

Explanation:
1. Energy \( E \propto 1/\lambda \), and the absorbed energy equals \( \Delta_o \).
2. The order of wavelengths is red > yellow > green > blue, so the order of energy is L\(_1\) < L\(_3\) < L\(_2\) < L\(_4\).
3. A larger \( \Delta_o \) means a stronger ligand.

Teacher's Note:
1. Blue light has the highest energy in the visible range among these.

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

Explanation:
1. \( \ce{[Co(NH3)3Cl3]} \) has facial and meridional isomers, both of which have a plane of symmetry, so neither is optically active.
2. The other three have chiral forms.

Teacher's Note:
1. No plane of symmetry means optical activity.

Answer: (b) \( \ce{[Cr(en)2Br2]Br} \)

Explanation:
1. Two Br, two en and one Cr(III) are in the complex ion: \( 3 + 2(-1) + 0 = +1 \), so the ion is \( \ce{[Cr(en)2Br2]^+} \).
2. Its counter ion is bromide, so the formula is \( \ce{[Cr(en)2Br2]Br} \).

Teacher's Note:
1. The cation takes bromide as the counter ion.

Answer: (b) \( \ce{[CuCl2\{O=C(NH2)2\}2]} \)

Explanation:
1. The name shows two chloride ligands and two urea ligands on copper.
2. With Cu(II) and 2 Cl\(^-\) the complex is neutral, so the formula is \( \ce{[CuCl2\{O=C(NH2)2\}2]} \).

Teacher's Note:
1. The name lists the ligands alphabetically.

Answer: (b) \( \ce{NH3} \) < en < \( \ce{CN^-} \) < CO

Explanation:
1. From the spectrochemical series: \( \ce{NH3} < \text{en} < \ce{CN^-} < \text{CO} \).
2. CO is a strong \( \pi \)-acceptor, so it has the highest field strength.

Teacher's Note:
1. Spectrochemical series: I\(^-\) < Br\(^-\) < Cl\(^-\) < F\(^-\) < OH\(^-\) < H\(_2\)O < NH\(_3\) < en < CN\(^-\) < CO.

Answer: (c) \( \ce{[Co(NH3)2(en)2]^{3+}} \)

Explanation:
1. \( \ce{[Co(NH3)2(en)2]^{3+}} \) is of the type M(AA)\(_2\)b\(_2\) and exists as cis and trans forms.
2. \( \ce{[Co(en)3]^{3+}} \), \( \ce{[Ni(NH3)5Br]^+} \) and \( \ce{[Cr(NH3)4(en)]^{3+}} \) have no geometrical isomers.

Teacher's Note:
1. M(AA)\(_2\)b\(_2\) gives cis and trans.

Answer: (b) \( \ce{[Co(en)3]^{3+}} \)

Explanation:
1. \( \ce{[Co(en)3]^{3+}} \) has no plane or centre of symmetry, so it is optically active.
2. The Zn(II) complexes are tetrahedral and symmetric, and \( \ce{[Co(H2O)4(en)]^{3+}} \) also has a plane of symmetry.

Teacher's Note:
1. Tris-chelate octahedral complexes are chiral.

Answer: (d) \( \ce{[Co(en)2(NH3)2]^{3+}} \)

Explanation:
1. \( \ce{[Co(en)2(NH3)2]^{3+}} \) is of type M(AA)\(_2\)b\(_2\), and its cis isomer has no symmetry plane, so it shows optical isomerism.
2. The other species have a plane of symmetry or are not octahedral isomers with chirality.

Teacher's Note:
1. Chiral octahedral complexes need a non-superimposable mirror image.

Answer: (d) 6 and 3

Explanation:
1. Each en is bidentate (2 donors each) and oxalate is bidentate, so the coordination number is \( 2 \times 2 + 2 = 6 \).
2. The complex ion has charge +1 (nitrite is outside): \( x + 0 + (-2) = +1 \), so x = +3.

Teacher's Note:
1. Count donor atoms, not ligands.

Answer: (a) \( \ce{[Co(CN)6]^{3-}} \)

Explanation:
1. All the complexes have Co(III), so \( \Delta_o \) depends on the ligand field strength.
2. The spectrochemical series gives CN\(^-\) > NH\(_3\) > H\(_2\)O > oxalate, so CN\(^-\) gives the highest \( \Delta_o \).

Teacher's Note:
1. Same metal ion and charge means compare the ligands.

Answer: (d) \( \ce{[PtCl4]^{2-}} \)

Explanation:
1. Pt(II) (5d\(^8\)) forms square planar complexes even with the weak field ligand Cl\(^-\), because the large 5d orbitals give a large splitting.
2. The 3d metal complexes with Cl\(^-\) are tetrahedral.

Teacher's Note:
1. Pd(II) and Pt(II) are nearly always square planar.

Answer: (b) both \( \sigma \) and \( \pi \) character

Explanation:
1. CO donates a lone pair to Fe (a \( \sigma \) bond).
2. Fe also donates d electrons back into the \( \pi^* \) orbitals of CO (\( \pi \) back bonding), so the bond has both \( \sigma \) and \( \pi \) character.

Teacher's Note:
1. Synergic bonding strengthens the metal-carbon bond.