Download JEE MCQs for JEE Chemistry: Coordination Compounds
Review structured MCQ sets for JEE Chemistry Coordination Compounds. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Chapter-wise Objective Questions: Coordination Compounds
Access the complete set of multiple-choice questions for Coordination Compounds below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.
Multiple Choice Questions
(a) \(\ce{NO2^-}\)
(b) Oxalate ion
(c) Glycinate ion
(d) Diethylenetriamine (dien)
Show Answer & Explanation
Answer: (d) Diethylenetriamine (dien)
Explanation:
1. Denticity is the number of donor atoms bonded to the metal
2. \(\ce{NO2^-}\) is monodentate (ambidentate), oxalate and glycinate are bidentate
3. Dien, \(\ce{H2N-CH2CH2-NH-CH2CH2-NH2}\), has three nitrogen donor atoms
Teacher's Note:
1. Count donor atoms that attach to the same metal ion. Dien, with three, is tridentate.
(a) \(\ce{Cl^-}\)
(b) CO
(c) \(\ce{OH^-}\)
(d) \(\ce{NH3}\)
Show Answer & Explanation
Answer: (b) CO
Explanation:
1. Classical ligands are sigma donors only
2. \(\ce{CO}\) donates a lone pair and also accepts electrons into its empty pi* orbitals (synergic bonding)
3. So CO is a non-classical ligand
Teacher's Note:
1. Non-classical (pi-acceptor) ligands include CO, NO+ and phosphines.
(a) \(\ce{[Mn(CO)6]^+}\) < \(\ce{[Cr(CO)6]}\) < \(\ce{[V(CO)6]^-}\)
(b) \(\ce{[V(CO)6]^-}\) < \(\ce{[Cr(CO)6]}\) < \(\ce{[Mn(CO)6]^+}\)
(c) \(\ce{[V(CO)6]^-}\) < \(\ce{[Mn(CO)6]^+}\) < \(\ce{[Cr(CO)6]}\)
(d) \(\ce{[Cr(CO)6]}\) < \(\ce{[Mn(CO)6]^+}\) < \(\ce{[V(CO)6]^-}\)
Show Answer & Explanation
Answer: (b) \(\ce{[V(CO)6]^-}\) < \(\ce{[Cr(CO)6]}\) < \(\ce{[Mn(CO)6]^+}\)
Explanation:
1. More negative charge on the metal gives stronger back-donation into the pi* orbital of CO
2. Stronger back-donation weakens the C-O bond
3. Back-donation: \(\ce{[V(CO)6]^-}\) > \(\ce{Cr(CO)6}\) > \(\ce{[Mn(CO)6]^+}\), so C-O strength is the reverse
Teacher's Note:
1. Back bonding adds electrons to the antibonding orbital of CO. The more back-bonding, the longer and weaker the C-O bond.
(a) \(\ce{[Cu(NH3)4]^2+}\)
(b) \(\ce{[Ni(CO)4]}\)
(c) \(\ce{[NiCl4]^2-}\)
(d) \(\ce{[Ni(CN)4]^2-}\)
Show Answer & Explanation
Answer: (d) \(\ce{[Ni(CN)4]^2-}\)
Explanation:
1. \(\ce{[Ni(CN)4]^2-}\): Ni(II) is \(d^8\) and the strong-field CN pairs electrons, giving dsp2 hybridisation, square planar and diamagnetic
2. \(\ce{[Cu(NH3)4]^2+}\) is square planar but Cu(II) has one unpaired electron
3. \(\ce{[Ni(CO)4]}\) and \(\ce{[NiCl4]^2-}\) are tetrahedral
Teacher's Note:
1. Strong-field ligands such as CN- pair up the d8 electrons of Ni(II) and give a square planar complex. Weak-field ligands such as Cl- give tetrahedral complexes.
(a) \(\ce{[Fe(CN)6]^3-}\) and \(\ce{[Fe(NH3)6]^3+}\)
(b) \(\ce{[Cr(NH3)6]^3+}\) and \(\ce{[Cr(CN)6]^3-}\)
(c) \(\ce{[FeF6]^3-}\) and \(\ce{[Fe(CN)6]^3-}\)
(d) \(\ce{Ni(CO)4}\) and \(\ce{[Ni(CN)4]^2-}\)
Show Answer & Explanation
Answer: (d) \(\ce{Ni(CO)4}\) and \(\ce{[Ni(CN)4]^2-}\)
Explanation:
1. EAN = electrons of the metal ion + 2 per ligand
2. Fe(III): \(23+12=35\) for all three Fe complexes, so pairs 1 and 3 match; Cr(III): \(21+12=33\) for both, so pair 2 matches
3. \(\ce{Ni(CO)4}\): \(28+8=36\); \(\ce{[Ni(CN)4]^2-}\): Ni(II) \(26+8=34\), which are different
Teacher's Note:
1. EAN counts the metal electrons plus electrons donated by ligands. It depends on the oxidation state of the metal and the number of ligands.
(a) Tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)
(b) Dichlorotetraammineplatinum(IV) tetrachloroplatinate(II)
(c) Tetrachloroplatinum(II) tetraammineplatinate(IV)
(d) None of these
Show Answer & Explanation
Answer: (a) Tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)
Explanation:
1. The cation is \(\ce{[Pt(NH3)4Cl2]^2+}\): ligands in alphabetical order (ammine before chloro), Pt is +4
2. The anion is \(\ce{[PtCl4]^2-}\) with Pt +2 and is named tetrachloroplatinate(II)
3. The cation is named first, so the answer is the first option
Teacher's Note:
1. Name ligands alphabetically, ignoring numerical prefixes: ammine comes before chloro. The cation is named before the anion.
Free study material for Chemistry
Download Chapter MCQs: JEE Chemistry Coordination Compounds
About Coordination Compounds MCQs for JEE Chemistry
Review structured objective questions for JEE Chemistry Coordination Compounds. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.
How to Verify Your MCQ Answers
Built using the official NCERT book for JEE, these Chemistry objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.
Enhance Speed with Online MCQ Tests
Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.
FAQs
You can get most exhaustive JEE Chemistry Coordination Compounds MCQs Set 13 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.
Yes, our JEE Chemistry Coordination Compounds MCQs Set 13 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.
By solving our JEE Chemistry Coordination Compounds MCQs Set 13, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.
Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.
Yes, you can also access online interactive tests for JEE Chemistry Coordination Compounds MCQs Set 13 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.