JEE Chemistry Coordination Compounds MCQs Set 24

Answer: (d) pentamminenitrito-N-cobalt(III) chloride

Explanation:
1. The complex ion is \( \ce{[Co(NO2)(NH3)5]^{2+}} \), so Co is +3 (x + (-1) = +2).
2. Ligands are named alphabetically: ammine then nitrito-N, giving pentamminenitrito-N-cobalt(III) chloride.

Teacher's Note:
1. Two chloride ions outside imply a +2 complex ion.

Answer: (b) two, tetrahedral

Explanation:
1. Ni\(^{2+}\) is d\(^8\). A paramagnetic four-coordinate complex has unpaired electrons, which means the electrons are not paired.
2. This happens in a tetrahedral (sp\(^3\)) complex with 2 unpaired electrons, while square planar d\(^8\) complexes are diamagnetic.

Teacher's Note:
1. d\(^8\) square planar is diamagnetic.

Answer: (d) d\(^4\) (in strong ligand field)

Explanation:
1. \( \mu = 2.84 \) BM corresponds to \( \sqrt{n(n+2)} = \sqrt{8} \), so n = 2.
2. Strong field d\(^5\) has 1, d\(^3\) has 3, weak field d\(^4\) has 4, and strong field d\(^4\) has 2 unpaired electrons.

Teacher's Note:
1. Count the unpaired electrons for each case.

Answer: (c) potassium hexacyanoferrate(III)

Explanation:
1. The complex is anionic, so the metal is named with the suffix -ate (ferrate).
2. The oxidation state of Fe: \( 3(+1) + x + 6(-1) = 0 \), so x = +3.

Teacher's Note:
1. Iron in an anionic complex is ferrate.

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

Explanation:
1. The unpaired electrons are: Co\(^{3+}\) (d\(^6\), low spin) 0, Fe\(^{3+}\) (d\(^5\)) 1, Mn\(^{3+}\) (d\(^4\)) 2, Cr\(^{3+}\) (d\(^3\)) 3.
2. The lowest magnetic moment (zero) is for the Co complex.

Teacher's Note:
1. CN\(^-\) is a strong field ligand.

Answer: (a) the number of ligands around a metal ion bonded by sigma bonds

Explanation:
1. The coordination number is the number of ligand donor atoms attached to the metal through sigma bonds.
2. Pi bonds (back bonding) do not add to the count.

Teacher's Note:
1. Coordination number is counted as donor atoms, not ligands.

Answer: (a) \( \ce{[MnCl4]^{2-}} > \ce{[CoCl4]^{2-}} > \ce{[Fe(CN)6]^{4-}} \)

Explanation:
1. \( \ce{[MnCl4]^{2-}} \): Mn\(^{2+}\) d\(^5\), 5 unpaired electrons (5.92 BM).
2. \( \ce{[CoCl4]^{2-}} \): Co\(^{2+}\) d\(^7\), 3 unpaired electrons (3.87 BM).
3. \( \ce{[Fe(CN)6]^{4-}} \): Fe\(^{2+}\) d\(^6\) low spin, 0 unpaired electrons.

Teacher's Note:
1. Compare the numbers of unpaired electrons.

Answer: (d) \( \ce{[Ni(NH3)6]^{2+}} \)

Explanation:
1. Ni\(^{2+}\) is d\(^8\) and has no vacant 3d orbitals even with NH\(_3\), so it uses outer 4d orbitals (sp\(^3\)d\(^2\)).
2. The other three are low spin d\(^2\)sp\(^3\) inner orbital complexes.

Teacher's Note:
1. d\(^8\) octahedral complexes are always outer orbital.

Answer: (a) \( \ce{[Co(NH3)5Cl]Cl2} \)

Explanation:
1. Three ions in solution means one complex cation and two chloride ions.
2. Two moles of AgCl precipitate, so two Cl\(^-\) are ionisable and one is in the coordination sphere, giving \( \ce{[Co(NH3)5Cl]Cl2} \).

Teacher's Note:
1. Ionisable chloride precipitates with silver nitrate.

Answer: (b) 0

Explanation:
1. The complex ion has charge -4 (four K\(^+\)).
2. \( x + 4(-1) = -4 \), so x = 0.

Teacher's Note:
1. CN\(^-\) carries a charge of -1.

Answer: (d) linkage

Explanation:
1. The nitro ligand is ambidentate and can bond through N (nitro) or O (nitrito).
2. So the complex shows linkage isomerism.

Teacher's Note:
1. No geometrical or optical isomers exist for this type.

Answer: (c) d, s, p\(_x\), p\(_y\)

Explanation:
1. Square planar geometry arises from dsp\(^2\) hybridisation.
2. This uses one d orbital (\( d_{x^2-y^2} \)), the s orbital and the p\(_x\) and p\(_y\) orbitals.

Teacher's Note:
1. dsp\(^2\) is square planar and sp\(^3\) is tetrahedral.

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

Explanation:
1. Ni\(^{2+}\) is d\(^8\). With the strong field ligand CN\(^-\) the electrons pair up and one 3d orbital is vacated.
2. This gives dsp\(^2\) hybridisation and a square planar diamagnetic complex.

Teacher's Note:
1. Strong field plus d\(^8\) means dsp\(^2\).

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

Explanation:
1. \( \ce{CoCl3.3NH3} \) is \( \ce{[Co(NH3)3Cl3]} \), where all three chlorides are inside the coordination sphere.
2. So no free Cl\(^-\) is available to precipitate with AgNO\(_3\).

Teacher's Note:
1. Only ionisable chloride gives a precipitate.

Answer: (c) It has no unpaired electrons and is in a low-spin configuration

Explanation:
1. Co\(^{3+}\) is d\(^6\), and CN\(^-\) is a strong field ligand, so all the electrons pair up in t\(_{2g}\).
2. The complex has no unpaired electron and is low spin.

Teacher's Note:
1. d\(^6\) low spin is \( t_{2g}^6 e_g^0 \).

Answer: (b) \( \ce{[Fe(H2O)6]^{3+}} \)

Explanation:
1. Fe\(^{3+}\) is d\(^5\) and water is a weak field ligand, giving \( t_{2g}^3 e_g^2 \).
2. CFSE = \( 3(-0.4) + 2(+0.6) = 0 \).
3. The other ions have non-zero CFSE.

Teacher's Note:
1. High spin d\(^5\) and d\(^{10}\) have zero CFSE.

Answer: (b) cis-\( \ce{[PtCl2(NH3)2]} \)

Explanation:
1. cis-Diamminedichloroplatinum(II), known as cisplatin, is used in cancer chemotherapy.

Teacher's Note:
1. The trans isomer is inactive.

Answer: (b) \( -0.6\Delta_o \)

Explanation:
1. High spin d\(^4\) is \( t_{2g}^3 e_g^1 \).
2. CFSE = \( 3(-0.4\Delta_o) + 1(+0.6\Delta_o) = -0.6\Delta_o \).

Teacher's Note:
1. t\(_{2g}\) electron contributes -0.4 and e\(_g\) electron +0.6.

Answer: (b) cis-isomer

Explanation:
1. When the two chloride ligands are adjacent (at 90 degrees), the isomer is the cis form.
2. In the trans isomer they are at 180 degrees.

Teacher's Note:
1. Cis means adjacent, trans means opposite.

Answer: (c) six membered

Explanation:
1. Each acac anion binds through its two oxygen atoms.
2. The ring contains Co, O, C, C, C and O, which is six atoms.

Teacher's Note:
1. Acetylacetonate gives six-membered chelate rings.

Answer: (c) \( \ce{[CoBr4]^{2-}} \)

Explanation:
1. Co\(^{3+}\) (d\(^6\), low spin with NH\(_3\)), Pt(II) (square planar d\(^8\)) and Mo(0) (d\(^6\) with CO) are all diamagnetic.
2. \( \ce{[CoBr4]^{2-}} \) is tetrahedral Co(II) d\(^7\) with 3 unpaired electrons.

Teacher's Note:
1. Tetrahedral complexes of Co(II) are high spin.

Answer: (c) The red complex has a tetrahedral geometry

Explanation:
1. The nickel dimethylglyoximate complex is square planar with two intramolecular O-H...O hydrogen bonds.
2. Dimethylglyoxime acts as a bidentate (didentate) ligand, so the statement that the complex is tetrahedral is not true.

Teacher's Note:
1. Ni(II) with a strong field ligand gives a square planar complex.

Answer: (d) \( 3d^1_{xy}, 3d^1_{yz}, 3d^1_{xz} \)

Explanation:
1. Cr\(^{3+}\) is d\(^3\) with 3 unpaired electrons, giving \( \mu = \sqrt{15} = 3.87 \) BM.
2. The electrons go singly into the lower energy t\(_{2g}\) orbitals: \( d_{xy}, d_{yz}, d_{xz} \).

Teacher's Note:
1. In an octahedral field the t\(_{2g}\) orbitals fill first.

Answer: (a) \( \ce{Al(OC2H5)3} \)

Explanation:
1. In aluminium ethoxide the ethyl groups are attached to aluminium through oxygen (Al-O-C), not through carbon.
2. The other three compounds have direct metal-carbon bonds (Mg-C, Pt-C, Ni-C).

Teacher's Note:
1. An organometallic compound has at least one M-C bond.

Answer: (d) \( \ce{(CH3)4Sn} \)

Explanation:
1. Zeise's salt, ferrocene and dibenzenechromium have the metal bonded to the \( \pi \) electrons of the unsaturated ligand.
2. In tetramethyltin the methyl groups are bonded by ordinary \( \sigma \) bonds, so it is not a \( \pi \)-bonded compound.

Teacher's Note:
1. Eta (hapticity) notation indicates pi bonding.

Answer: (b) 3

Explanation:
1. The complex is of type M(AA)\(_2\)bc. The trans form has a plane of symmetry (1 isomer).
2. The cis form is chiral and exists as a pair of enantiomers (2 isomers).
3. The total is 3 stereoisomers.

Teacher's Note:
1. Stereoisomers include geometrical and optical isomers.

Answer: (c) 4

Explanation:
1. Fe\(^{2+}\) is d\(^6\). In the high spin state it has 4 unpaired electrons (\( t_{2g}^4 e_g^2 \)).
2. In the low spin state all electrons pair (\( t_{2g}^6 \)), so there are 0 unpaired electrons.
3. The difference is 4.

Teacher's Note:
1. d\(^6\): high spin 4, low spin 0.