JEE Chemistry Coordination Compounds MCQs Set 16

Answer: (d) all of the above are possible

Explanation:
1. Ionisation isomers arise when the counter ion \( \ce{Cl-} \) exchanges places with a ligand inside the coordination sphere.
2. Exchanging Cl with I gives \( \ce{[Pt(NH3)3(Br)(NO2)Cl]I} \) (option a), with NO2 gives \( \ce{[Pt(NH3)3(Br)(I)Cl]NO2} \) (option b), and with Br gives \( \ce{[Pt(NH3)3(NO2)(I)Cl]Br} \) (option c).
3. Ligands are written in alphabetical order in each name, so all three names are correct.

Teacher's Note:
1. Swap the free anion with each ligand in turn to list the ionisation isomers.
2. Ligand names are arranged alphabetically, ignoring numerical prefixes.

Answer: (b) \( \ce{[Fe(CN)6]^{3-}} \)

Explanation:
1. Cyanide is a stronger field ligand than water, so the two cyanide complexes have larger splitting than the two aqua complexes.
2. For the same ligand, a metal ion with a higher oxidation state gives larger splitting. \( \ce{Fe^{3+}} \) in \( \ce{[Fe(CN)6]^{3-}} \) therefore splits the d orbitals more than \( \ce{Fe^{2+}} \) in \( \ce{[Fe(CN)6]^{4-}} \).

Teacher's Note:
1. Splitting depends on ligand strength and on the charge of the metal ion.
2. Compare the ligands first and the oxidation states second.

Answer: (b) A = Q, B = P, C = P, D = P

Explanation:
1. \( \ce{[Co(en)2Cl2]+} \) has a trans form and a cis pair of enantiomers, so 3 stereoisomers (Q).
2. \( \ce{[Co(en)3]^{3+}} \) exists only as a pair of enantiomers (P).
3. Square planar \( \ce{[Pt(NH3)2Cl2]} \) has cis and trans forms (P).
4. Octahedral \( \ce{[Pt(NH3)3Cl3]+} \) has facial and meridional forms, both achiral (P).

Teacher's Note:
1. Count geometrical isomers first and then check each for a mirror image that cannot be superimposed.
2. Bis-chelate cis forms are chiral; tris-chelate complexes are chiral.

Answer: (a) \( \ce{[Fe(H2O)5(NO)]^{2+}} \)

Explanation:
1. Nitrate is reduced to nitric oxide by iron(II) in concentrated sulphuric acid.
2. The nitric oxide displaces one water molecule of \( \ce{[Fe(H2O)6]^{2+}} \) to give the brown complex \( \ce{[Fe(H2O)5(NO)]^{2+}} \).

Teacher's Note:
1. The brown colour comes from NO bound to iron in a complex ion.
2. Only one water molecule is replaced.

Answer: (d) \( \ce{[CrCl2(CN)2(NH3)2]^{-}} \)

Explanation:
1. Chromium is +3, each chloride and cyanide is -1 and ammonia is neutral.
2. Charge = +3 + 2(-1) + 2(-1) + 0 = -1, so the complex ion is \( \ce{[CrCl2(CN)2(NH3)2]^-} \), consistent with the suffix "chromate".

Teacher's Note:
1. The ending "-ate" signals an anionic complex.
2. Add the charges of the metal and ligands to get the overall charge.

Answer: (c) \( \ce{[Cr(CO)6]^-} > \ce{[Cr(CO)6]} > \ce{[Cr(CO)6]^+} \)

Explanation:
1. In neutral \( \ce{[Cr(CO)6]} \): E.A.N. = 24 + 12 = 36.
2. Adding one electron gives the anion an E.A.N. of 37, and removing one gives the cation an E.A.N. of 35.
3. Hence anion > neutral > cation.

Teacher's Note:
1. E.A.N. is the metal's electron count plus two electrons per ligand, adjusted for the charge on the complex.
2. A negative charge raises the count by one.

Answer: (a) (\( \eta^6 \)-benzene)tricarbonylchromium(0)

Explanation:
1. The complex is neutral, so the metal keeps the name chromium and the oxidation state is 0.
2. Ligands are cited alphabetically: benzene (b) comes before carbonyl (c), so the name is (\( \eta^6 \)-benzene)tricarbonylchromium(0).

Teacher's Note:
1. Alphabetical order ignores numerical prefixes such as tri-.
2. Write the hapticity as a superscript on eta before the ligand name.

Answer: (c) \( \ce{K[Pt(\eta^2-C2H4)Cl3]} \)

Explanation:
1. Zeise's salt is the first organometallic complex of an alkene, potassium trichlorido(ethene)platinate(II).
2. Its anion contains one ethene molecule and three chloride ions around platinum.

Teacher's Note:
1. Ferrocene and dibenzenechromium are sandwich compounds, not Zeise's salt.
2. Remember the formula as potassium, platinum, one ethene and three chlorides.

Answer: (a) \( \ce{[Ca(EDTA)]^{2-}} \)

Explanation:
1. EDTA is a hexadentate ligand and uses all six donor atoms (two nitrogen and four oxygen).
2. So calcium has coordination number 6 in \( \ce{[Ca(EDTA)]^{2-}} \); oxalate (bidentate) gives only four in option (c).

Teacher's Note:
1. Coordination number equals the number of donor atoms attached to the metal.
2. Multiply the denticity by the number of such ligands.

Answer: (c) \( \ce{V(CO)6} \) is not very stable and is easily reduced to \( \ce{[V(CO)6]^-} \)

Explanation:
1. \( \ce{V(CO)6} \) has 17 valence electrons in the metal-based shell (E.A.N. 35), so it readily gains one electron to reach 36 in \( \ce{[V(CO)6]^-} \).
2. Option (a) is wrong because the iron here is Fe(III) low spin d5 and has one unpaired electron; (b) is wrong because the chelated \( \ce{[Ni(dmg)2]} \) is far more stable; (d) is wrong because these ligands are pi-acceptors using empty antibonding orbitals.

Teacher's Note:
1. Metal carbonyls prefer the 18-electron configuration.
2. CO, CN- and NO+ accept electron density from the metal into empty orbitals.

Answer: (d) 1.150 Angstrom

Explanation:
1. In metal carbonyls the metal donates electrons into the antibonding pi orbitals of CO (back bonding).
2. This lowers the C-O bond order, so the bond becomes longer than in free CO (1.128 Angstrom); the observed value in \( \ce{Fe(CO)5} \) is about 1.15 Angstrom.

Teacher's Note:
1. Back bonding strengthens the M-C bond and weakens the C-O bond.
2. A longer C-O bond means a lower stretching frequency.

Answer: (c) \( \ce{H2O} < \ce{NH3} < \ce{NO2-} < \ce{CN-} \)

Explanation:
1. In the spectrochemical series, water is weaker than ammonia, and ammonia is weaker than the pi-acceptor ligands nitrite (N-bonded) and cyanide.
2. Cyanide is stronger than nitrite, so the order is \( \ce{H2O} < \ce{NH3} < \ce{NO2-} < \ce{CN-} \).

Teacher's Note:
1. Pi-acceptor ligands sit at the strong end of the series.
2. Remember the sequence I, Br, Cl, F, OH, oxalate, H2O, NH3, en, NO2, CN, CO.

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

Explanation:
1. \( \ce{[Co(NH3)3Cl3]} \) exists as facial and meridional isomers, and both have a plane of symmetry, so neither is optically active.
2. \( \ce{[Co(en)3]^{3+}} \) and cis-\( \ce{[Co(en)(NH3)2Cl2]+} \) lack a mirror plane and are chiral.

Teacher's Note:
1. A complex with a plane of symmetry cannot be optically active.
2. Octahedral Ma3b3 is never chiral.

Answer: (b) Chlorophyll

Explanation:
1. Chlorophyll is a magnesium complex of a porphyrin-type ring.
2. Haemoglobin contains iron and vitamin B12 contains cobalt.

Teacher's Note:
1. Chlorophyll is a Mg(II) coordination compound.
2. Match the metal with the biomolecule: Mg - chlorophyll, Fe - haemoglobin, Co - B12.

Answer: (b) \( sp^3 \), tetrahedral

Explanation:
1. Iron is in the -2 state with configuration \( 3d^{10} \), so no d orbital is available for inner hybridisation.
2. Four carbonyl ligands use one 4s and three 4p orbitals, giving \( sp^3 \) hybridisation and a tetrahedral shape.

Teacher's Note:
1. A d10 metal ion with four ligands is tetrahedral.
2. Work out the oxidation state first.

Answer: (b) Dimethylglyoxime

Explanation:
1. Nickel(II) reacts with dimethylglyoxime in ammoniacal solution to give a bright red precipitate of \( \ce{[Ni(dmg)2]} \), a square planar chelate.

Teacher's Note:
1. This red chelate is a standard qualitative test for Ni(II).
2. Dimethylglyoxime is also called Tschugaev's reagent.

Answer: (b) \( 4\Delta_o = 9\Delta_t \)

Explanation:
1. For the same metal and ligands, \( \Delta_t = \dfrac{4}{9}\Delta_o \).
2. Cross-multiplying gives \( 9\Delta_t = 4\Delta_o \), which is option (b).

Teacher's Note:
1. The tetrahedral splitting is less than half of the octahedral splitting.
2. Only four ligands and no direct overlap with the d orbitals lower it.

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

Explanation:
1. \( \ce{[Cr(C2O4)3]^{3-}} \) is an octahedral tris-chelate that lacks a plane of symmetry, so it exists as two non-superimposable mirror images.
2. The other three ions have all identical ligands and are symmetric.

Teacher's Note:
1. Tris-chelate octahedral complexes are chiral.
2. Complexes with only one kind of monodentate ligand are never chiral.

Answer: (b) \( \ce{[Co(NH3)3(NO2)3]} \)

Explanation:
1. \( \ce{[Co(NH3)3(NO2)3]} \) is a neutral complex with no counter ions, so it gives no ions in solution.
2. The others release 3, 5 or 2 ions per formula unit.

Teacher's Note:
1. Neutral complexes are non-electrolytes.
2. Count the ions given by each formula.

Answer: (c) \( \ce{K3[Cu(CN)4]} \)

Explanation:
1. Copper(II) is reduced by cyanide to copper(I) with the formation of cyanogen.
2. Excess cyanide then dissolves the copper(I) cyanide as the very stable tetracyanidocuprate(I), \( \ce{K3[Cu(CN)4]} \).

Teacher's Note:
1. Cu(II) with cyanide gives a Cu(I) complex and cyanogen gas.
2. The complex is colourless as Cu(I) is d10.

Answer: (b) 5

Explanation:
1. This double salt (Mohr's salt) dissociates completely into its simple ions.
2. The ions are one \( \ce{Fe^{2+}} \), two \( \ce{NH4+} \) and two \( \ce{SO4^{2-}} \), a total of 5.

Teacher's Note:
1. A double salt loses its identity in water.
2. Count each ion according to its subscript.

Answer: (b) hexadentate

Explanation:
1. EDTA has two nitrogen atoms and four carboxylate oxygen atoms that can bind a metal ion, so it is hexadentate.

Teacher's Note:
1. It forms five-membered chelate rings with the metal.
2. It is used to estimate the hardness of water.