JEE Chemistry Coordination Compounds MCQs Set 26

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Multiple Choice Questions

Question 1: The number of unpaired electrons in the complex ion [FeF6]3- is:
(a) 0
(b) 5
(c) 6
(d) 4
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. Fe3+ is d5 and F- is a weak field ligand
2. The configuration is t2g3 eg2 with 5 unpaired electrons

Teacher's Note:
1. Five unpaired electrons give 5.92 BM.

Question 2: Which of the following complexes has a geometry different from the others?
(a) [Zn(NH3)4]2+
(b) Ni(CO)4
(c) [NiCl4]2-
(d) [Ni(CN)4]2-
Show Answer & Explanation

Answer: (d) [Ni(CN)4]2-

Explanation:
1. [Ni(CN)4]2- is square planar (dsp2) because CN- pairs the d8 electrons
2. The other three are tetrahedral (sp3)

Teacher's Note:
1. Strong field ligands with d8 give square planar geometry.

Question 3: Among Ni(CO)4, [Ni(CN)4]2- and [NiCl4]2-, which statement is correct?
(a) Ni(CO)4 and [NiCl4]2- are diamagnetic and [Ni(CN)4]2- is paramagnetic
(b) Ni(CO)4 and [Ni(CN)4]2- are diamagnetic and [NiCl4]2- is paramagnetic
(c) [NiCl4]2- and [Ni(CN)4]2- are diamagnetic and Ni(CO)4 is paramagnetic
(d) Ni(CO)4 is diamagnetic and the other two are paramagnetic
Show Answer & Explanation

Answer: (b) Ni(CO)4 and [Ni(CN)4]2- are diamagnetic and [NiCl4]2- is paramagnetic

Explanation:
1. Ni(CO)4 is Ni(0) d10 and diamagnetic
2. [Ni(CN)4]2- is square planar d8 with all electrons paired
3. [NiCl4]2- is tetrahedral Ni2+ with two unpaired electrons

Teacher's Note:
1. Square planar Ni(II) is diamagnetic but tetrahedral Ni(II) is paramagnetic.

Question 4: Which of the following compounds is not paramagnetic?
(a) NO
(b) NO2
(c) [Cu(NH3)4Cl2]
(d) [Ag(NH3)2]Cl
Show Answer & Explanation

Answer: (d) [Ag(NH3)2]Cl

Explanation:
1. Ag+ is d10, so [Ag(NH3)2]+ has no unpaired electron
2. Cu(II) has one, and NO and NO2 are odd-electron molecules

Teacher's Note:
1. A filled d10 shell gives a diamagnetic complex.

Question 5: Which of the following ions has the highest paramagnetism?
(a) [Zn(H2O)6]2+
(b) [Cu(H2O)6]2+
(c) [Fe(H2O)6]2+
(d) [Cr(H2O)6]3+
Show Answer & Explanation

Answer: (c) [Fe(H2O)6]2+

Explanation:
1. Unpaired electrons: Cr3+ d3 has 3, Fe2+ d6 (high spin) has 4, Cu2+ d9 has 1 and Zn2+ d10 has 0
2. Fe2+ has the most

Teacher's Note:
1. Water is a weak field ligand, so the electrons stay unpaired as far as possible.

Question 6: Which of the following factors tends to increase the stability of metal ion complexes?
(a) lower basicity of the ligand
(b) higher charge/size ratio of the metal ion
(c) lower ionisation potential of the metal ion
(d) higher ionic radius of the metal ion
Show Answer & Explanation

Answer: (b) higher charge/size ratio of the metal ion

Explanation:
1. A small, highly charged metal ion attracts the ligand lone pairs more strongly
2. Stronger basicity of the ligand also helps stability, so the last option is wrong

Teacher's Note:
1. Stability rises with charge density of the central ion.

Question 7: The complexes [Pt(NH3)4][PtCl6] and [Pt(NH3)4Cl2][PtCl4] are:
(a) optical isomers
(b) linkage isomers
(c) ionisation isomers
(d) coordination isomers
Show Answer & Explanation

Answer: (d) coordination isomers

Explanation:
1. The ligands are exchanged between the cationic and the anionic complex, and the metal oxidation states change
2. This is coordination isomerism

Teacher's Note:
1. Coordination isomerism needs both a complex cation and a complex anion.

Question 8: [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2 are related to each other as:
(a) coordination isomers
(b) ionisation isomers
(c) linkage isomers
(d) geometrical isomers
Show Answer & Explanation

Answer: (c) linkage isomers

Explanation:
1. NO2- is ambidentate and binds through N (nitro) in one and through O (nitrito) in the other
2. This is linkage isomerism

Teacher's Note:
1. Ambidentate ligands such as NO2-, SCN- and CN- cause linkage isomerism.

Question 9: The number of geometrical isomers of [Co(NH3)3(NO2)3] is:
(a) 2
(b) 3
(c) 4
(d) 0
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. The complex is of the Ma3b3 type
2. It has only the fac (facial) and mer (meridional) forms

Teacher's Note:
1. Enumerating the arrangements of three a and three b around an octahedron gives exactly two.

Question 10: Geometrical isomerism is found in coordination compounds having coordination number:
(a) 3
(b) 2
(c) 4 (tetrahedral)
(d) 6
Show Answer & Explanation

Answer: (d) 6

Explanation:
1. Linear, trigonal and tetrahedral arrangements have all positions equivalent for monodentate ligands
2. Octahedral (CN 6) and square planar (CN 4) complexes can show cis-trans forms

Teacher's Note:
1. Square planar and octahedral shapes have non-equivalent relative positions.

Question 11: Cis-trans isomerism is found in square planar complexes of the molecular formula (a and b are monodentate ligands):
(a) Ma3b
(b) Ma4
(c) Ma2b2
(d) Mab3
Show Answer & Explanation

Answer: (c) Ma2b2

Explanation:
1. Only Ma2b2 has two a ligands that can be adjacent (cis) or opposite (trans)
2. Ma4, Ma3b and Mab3 have a single arrangement

Teacher's Note:
1. Ma2b2 in square planar shape gives exactly two isomers.

Question 12: Geometrical isomerism can be shown by:
(a) [Ag(NH3)(CN)]
(b) [PtCl4I2]
(c) [Pt(NH3)3Cl][Au(CN)4]
(d) Na2[Cd(NO2)4]
Show Answer & Explanation

Answer: (b) [PtCl4I2]

Explanation:
1. [PtCl4I2] is an octahedral Ma4b2 complex with cis and trans forms
2. The other options are linear, tetrahedral or Ma3b-type, which have a single form

Teacher's Note:
1. Match the formula to the shape before looking for isomers.

Question 13: Which one of the following is not an organometallic compound?
(a) (CH3)4Sn
(b) RMgX
(c) C2H5ONa
(d) (C2H5)4Pb
Show Answer & Explanation

Answer: (c) C2H5ONa

Explanation:
1. Organometallic compounds have at least one metal-carbon bond
2. In sodium ethoxide the sodium is attached to oxygen

Teacher's Note:
1. The test is a direct M-C bond.

Question 14: The species having tetrahedral shape is:
(a) [PdCl4]2-
(b) [Pd(CN)4]2-
(c) [NiCl4]2-
(d) [Ni(CN)4]2-
Show Answer & Explanation

Answer: (c) [NiCl4]2-

Explanation:
1. Pd(II) is 4d8 and always square planar
2. [Ni(CN)4]2- is square planar since CN- is a strong field ligand
3. Cl- is weak, so [NiCl4]2- is tetrahedral (sp3)

Teacher's Note:
1. d8 of the 4d and 5d series is square planar regardless of ligand.

Question 15: The spin magnetic moment of cobalt in Hg[Co(SCN)4] is:
(a) sqrt(15)
(b) sqrt(24)
(c) sqrt(8)
(d) sqrt(3)
Show Answer & Explanation

Answer: (a) sqrt(15)

Explanation:
1. Co in this compound is Co2+, d7
2. SCN- (S-bonded) gives a tetrahedral high spin complex with 3 unpaired electrons
3. mu = sqrt(3 x 5) = sqrt(15) BM

Teacher's Note:
1. The tetrahedral d7 configuration is e4 t2^3 with three unpaired electrons.

Question 16: Which kind of isomerism is exhibited by the octahedral complex [Co(NH3)4Br2]Cl?
(a) geometrical and ionisation
(b) geometrical only
(c) optical and ionisation
(d) geometrical and optical
Show Answer & Explanation

Answer: (a) geometrical and ionisation

Explanation:
1. It is Ma4b2 type, so cis and trans forms exist
2. Exchange of Br- with Cl- gives [Co(NH3)4BrCl]Br, an ionisation isomer
3. Neither cis nor trans is chiral

Teacher's Note:
1. The trans and cis forms both have a plane of symmetry.

Question 17: The bond length of CO is 1.128 angstrom. The bond length of CO in Fe(CO)5 is expected to be:
(a) 1.118 angstrom
(b) 1.128 angstrom
(c) 1.178 angstrom
(d) 1.158 angstrom
Show Answer & Explanation

Answer: (d) 1.158 angstrom

Explanation:
1. Back-bonding from filled metal d orbitals into the pi* orbital of CO weakens and lengthens the C-O bond
2. So it is somewhat above 1.128, about 1.15 angstrom

Teacher's Note:
1. Synergic bonding reduces the C-O bond order below 3.

Question 18: Nickel(II) chloride reacts with KCN to give complex A, and with excess KCl (in the presence of HCl) to give complex B. Both A and B have coordination number 4. The IUPAC names of A and B are respectively:
(a) potassium tetracyanonickel(II) and potassium tetrachloronickel(II)
(b) potassium tetracyanonickelate(II) and potassium tetrachloronickelate(II)
(c) potassium cyanonickelate(II) and potassium chloronickelate(II)
(d) potassium cyanonickel(II) and potassium chloronickel(II)
Show Answer & Explanation

Answer: (b) potassium tetracyanonickelate(II) and potassium tetrachloronickelate(II)

Explanation:
1. A is K2[Ni(CN)4] and B is K2[NiCl4]
2. Both are anionic complexes, so the metal name ends in -ate
3. Prefix tetra is used for the four ligands

Teacher's Note:
1. Anionic complexes use the -ate form of the metal name.

Question 19: Nickel(II) chloride reacts with KCN to give complex A, and with excess KCl (in the presence of HCl) to give complex B. Both A and B have coordination number 4. The hybridisations of A and B are respectively:
(a) dsp2 for both
(b) sp2 and dsp2
(c) dsp2 and sp3
(d) sp3 for both
Show Answer & Explanation

Answer: (c) dsp2 and sp3

Explanation:
1. CN- is strong field, so Ni2+ (d8) pairs electrons and uses dsp2 (square planar) in A
2. Cl- is weak, so B uses sp3 (tetrahedral)

Teacher's Note:
1. Strong field ligand gives dsp2, weak field gives sp3 for Ni(II).

Question 20: Nickel(II) chloride reacts with KCN to give complex A, and with excess KCl (in the presence of HCl) to give complex B. Both A and B have coordination number 4. The magnetic natures of A and B are respectively:
(a) both diamagnetic
(b) A diamagnetic and B paramagnetic with one unpaired electron
(c) A diamagnetic and B paramagnetic with two unpaired electrons
(d) both paramagnetic
Show Answer & Explanation

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

Explanation:
1. A is square planar d8 with all electrons paired
2. B is tetrahedral d8 with e4 t2^4, giving two unpaired electrons

Teacher's Note:
1. Tetrahedral Ni(II) always has two unpaired electrons.

Question 21: Among the metal carbonyls [Mn(CO)6]+, [V(CO)6]-, [Cr(CO)6] and [Fe(CO)5], the C-O bond order is lowest in:
(a) [Mn(CO)6]+
(b) [Fe(CO)5]
(c) [V(CO)6]-
(d) [Cr(CO)6]
Show Answer & Explanation

Answer: (c) [V(CO)6]-

Explanation:
1. More negative charge on the metal gives more back donation to the pi* orbital of CO
2. [V(CO)6]- therefore has the lowest C-O bond order

Teacher's Note:
1. Back bonding is strongest in anionic carbonyls.

Question 22: The effective atomic numbers of the metal in [Fe(CO)2(NO+)2] and Co2(CO)8 are respectively:
(a) 34 and 36
(b) 34 and 35
(c) 36 and 36
(d) 36 and 35
Show Answer & Explanation

Answer: (c) 36 and 36

Explanation:
1. With two NO+ ligands (each a 2-electron donor) the neutral complex needs Fe in the -2 state
2. EAN of Fe = 26 + 2 + 4 x 2 = 36
3. Co2(CO)8: 27 + 8 + 1 (Co-Co bond) = 36

Teacher's Note:
1. Use EAN = Z - oxidation state + 2 x ligands, with a metal-metal bond adding one electron.

Question 23: Which of the following is an inner orbital complex as well as diamagnetic?
(a) [Cr(NH3)6]3+
(b) [Ni(NH3)6]2+
(c) [Ir(H2O)6]3+
(d) [Co(NH3)6]2+
Show Answer & Explanation

Answer: (c) [Ir(H2O)6]3+

Explanation:
1. Ir3+ is 5d6 and the 5d series always gives low spin (d2sp3), so it is inner and diamagnetic
2. [Cr(NH3)6]3+ is inner but d3 and paramagnetic
3. Ni2+ and Co2+ ammines are outer orbital

Teacher's Note:
1. Look for low spin d6 or d3 d-electron counts in the 4d/5d series.

Question 24: Which of the following statements is correct?
(a) [Ag(NH3)2]+ is linear in shape
(b) the oxidation state of iron in sodium nitroprusside Na2[Fe(CN)5NO] is +3
(c) in [Fe(H2O)6]3+ iron is d2sp3 hybridised
(d) in Ni(CO)4 the oxidation state of Ni is +1
Show Answer & Explanation

Answer: (a) [Ag(NH3)2]+ is linear in shape

Explanation:
1. Ag+ (d10) forms a linear sp complex
2. In nitroprusside NO is NO+, so Fe is +2
3. [Fe(H2O)6]3+ is outer orbital (sp3d2)
4. Ni in Ni(CO)4 is zero

Teacher's Note:
1. NO+ counts as a +1 ligand in nitroprusside.

Question 25: Which of the following statements is not correct?
(a) TiCl4 is a colourless compound
(b) [Cr(NH3)6]Cl3 is a coloured compound
(c) [Cu(NCCH3)4][BF4] is a colourless compound
(d) K3[VF6] is a colourless compound
Show Answer & Explanation

Answer: (d) K3[VF6] is a colourless compound

Explanation:
1. V3+ is d2, so d-d transitions give colour to K3[VF6]
2. Ti4+ (d0) and Cu+ (d10) have no d-d transitions and are colourless
3. Cr3+ is d3 and coloured

Teacher's Note:
1. Colour from d-d transitions needs a partly filled d subshell.

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