CBSE Class 12 Chemistry Coordination Chemistry Assignment

LEVEL-1 QUESTIONS

Question 1. Define Ligand. Give one example also.
Answer: The ions or molecules which are coordinately bonded to the central metal atom in the coordination entity are called ligands.
Example: Chloride ion (\(\text{Cl}^-\)) or Ammonia (\(\text{NH}_3\)).
In simple words: Ligands are donor molecules or ions attached to the central metal in a complex.

Exam Tip: Always specify that ligands possess at least one unshared pair of valence electrons (lone pair).

Question 2. Why Coordination compounds are coloured?
Answer: Coordination compounds are coloured due to the d-d transition of electrons. When white light falls on the coordination compound, a part of it is absorbed, promoting an electron from one d-orbital to another (\(\text{t}_{2\text{g}}\) to \(\text{e}_{\text{g}}\)), and the remaining complementary colour is transmitted.
In simple words: Absorption of light energy causes electrons to jump between d-orbitals, producing visible colour.

Exam Tip: Mention d-d transition and presence of unpaired d-electrons as the primary reason for colour.

Question 3. What is Homoleptic complexes?
Answer: The complexes in which a metal is bound to only one kind of donor groups are called homoleptic complexes.
Example: \([\text{Co}(\text{NH}_3)_6]^{3+}\).
In simple words: Complexes containing only a single type of ligand are homoleptic.

Exam Tip: Contrast this with heteroleptic complexes where multiple types of ligands are present.

Question 4. What is the Oxidation no. of cobalt in \(\text{K}[\text{Co}(\text{CO})_4]\)?
Answer: Let the oxidation number of cobalt be \(x\).
Potassium (\(\text{K}\)) has an oxidation state of \(+1\).
Carbon monoxide (\(\text{CO}\)) is a neutral ligand with an oxidation state of \(0\).
\(1(+1) + x + 4(0) = 0\)
\(1 + x = 0\)
\(x = -1\)
In simple words: Setting up the algebraic sum of oxidation states to zero gives \(-1\) for cobalt.

Exam Tip: Metal carbonyl complexes can exhibit zero or negative oxidation states for the central metal.

Question 5. Amongst the following which is most stable complex is A) \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}\) B) \([\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-}\)
Answer: (B) \([\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-}\)
\(\text{C}_2\text{O}_4^{2-}\) (oxalate) is a bidentate ligand that forms stable five-membered chelate rings (chelate effect), making \([\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-}\) much more stable than the aqua complex.
In simple words: Complexes with chelating ligands are more stable due to the chelate effect.

Exam Tip: Chelation increases thermodynamic stability significantly.

Question 6. Write IUPAC name of the following: \([\text{Co}(\text{NH}_3)_6]\text{Cl}_3\)
Answer: Hexaamminecobalt(III) chloride
In simple words: Name the ligands first in alphabetical order, followed by the metal with its oxidation number, and lastly the counter anion.

Exam Tip: Note the spelling of 'ammine' with double 'm'.

Question 7. Which isomerisms are shown by following:
a) \([\text{Co}(\text{NH}_3)_5(\text{NO}_2)]\text{Cl}_2\) and \([\text{Co}(\text{NH}_3)_5(\text{ONO})]\text{Cl}_2\)
b) \([\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}\) and \([\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4\)
Answer:
(a) Linkage isomerism (due to ambidentate ligand \(\text{NO}_2^-\) coordinating via nitrogen or oxygen)
(b) Ionisation isomerism (due to exchange of groups between coordination sphere and ionization sphere)
In simple words: Linkage isomerism involves ambidentate ligands, while ionisation isomerism swaps ions inside and outside the bracket.

Exam Tip: Carefully identify the ambidentate ligand for linkage isomerism and counter ions for ionisation isomerism.

Question 8. Explain each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer:
(i) Coordination entity: A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules (ligands).
(ii) Ligands: The neutral molecules or negatively charged ions that surround the metal atom in a coordination entity possessing unshared pairs of valence electrons.
(iii) Coordination number: The total number of ligand donor atoms attached directly to the central metal atom.
(iv) Coordination polyhedron: The spatial arrangement of the ligand atoms directly attached to the central metal ion defining the geometry.
(v) Homoleptic complexes: Complexes in which a metal is bound to only one kind of donor groups.
(vi) Heteroleptic complexes: Complexes in which a metal is bound to more than one type of donor groups.
In simple words: These terms define the structural components and classifications of coordination compounds.

Exam Tip: Learn exact definitions as these frequently appear in 1-mark or 2-mark questions.

Question 9. What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Answer:
(a) Unidentate ligands: Ligands with only one donor site. Examples: \(\text{NH}_3\), \(\text{Cl}^-\).
(b) Didentate ligands: Ligands that have two donor sites. Examples: Ethane-1,2-diamine (\(\text{en}\)), Oxalate ion (\(\text{C}_2\text{O}_4^{2-}\)).
(c) Ambidentate ligands: Unidentate ligands that can ligate through two different atoms. Examples: \(-\text{NO}_2\) and \(-\text{SCN}^-\).
In simple words: Classification is based on the number of donor atoms or attachment points a ligand possesses.

Exam Tip: Memorize classic examples of ambidentate ligands like \(\text{NO}_2^-\) and \(\text{SCN}^-\).

Question 10. Write the name of each of the following complex compound:
(a) \([\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_3\)
(b) \([\text{Pt}(\text{NH}_3)_4][\text{NiCl}_4]\)
Answer:
(a) Tetraamminediaquacobalt(III) chloride
(b) Tetraammineplatinum(II) tetrachloridonickelate(II)
In simple words: Alphabetize neutral ligands before naming the metal and its oxidation state.

Exam Tip: Water is named as 'aqua' and ammonia as 'ammine'.

Question 11. Write the IUPAC name of \([\text{CrCl}_2(\text{en})(\text{NH}_3)_2]^+\).
Answer: Diamminedichlorido(ethane-1,2-diamine)chromium(III) ion
In simple words: List ammine, chlorido, and ethylenediamine in alphabetical order followed by chromium(III) ion.

Exam Tip: Multi-word ligand names like ethane-1,2-diamine use bis-, tris- if multiple, but here it is single so it is written directly in parentheses or alphabetical order rules apply.

Question 12. Why only transition metals are known to form \(\pi\) complexes?
Answer: Transition metals have empty d-orbitals into which electron pairs can be donated by ligands containing \(\pi\)-electrons (electrons in their \(\pi\)-molecular orbitals), forming synergic \(\pi\)-bonds.
In simple words: Vacant d-orbitals allow transition metals to accept back-donated electrons from ligands.

Exam Tip: Emphasize the role of vacant d-orbitals and back-bonding.

Question 13. CO is a stronger ligand than \(\text{NH}_3\). Why?
Answer: Due to back bonding. CO has vacant \(\pi^*\) molecular orbitals that can accept d-electrons from the metal atom, forming a synergic \(\text{M}\leftarrow\text{CO}\) \(\pi\)-bond which strengthens the metal-carbon bond.
In simple words: Carbon monoxide undergoes back bonding, making its bond with metals exceptionally strong.

Question 14. What is crystal field splitting energy?
Answer: The energy difference between the two sets of d-orbitals (\(\text{e}_{\text{g}}\) and \(\text{t}_{2\text{g}}\)) in a metal complex resulting from the ligand field is called crystal field splitting energy (\(\Delta_0\)).
In simple words: It is the energy gap created when d-orbitals split in the presence of ligands.

Exam Tip: Represented as \(\Delta_0\) for octahedral and \(\Delta_t\) for tetrahedral complexes.

Question 15. Differentiate between:
a) Didentate and Polydentate ligand
b) Chelate ligand and ambidentate ligand
Answer:
(a) Didentate ligand: Has exactly two donor binding sites.
Polydentate ligand: Has more than two donor binding sites.
(b) Chelate ligand: Binds through two or more donor sites to form a cyclic ring structure.
Ambidentate ligand: Binds through either of two different atoms, but only one at a time.
In simple words: Didentate/polydentate refer to donor site count, while chelate and ambidentate describe ring formation and dual-site attachment respectively.

Question 1. Write the possible isomers of \(\text{K}[\text{Cr}(\text{H}_2\text{O})_2(\text{C}_2\text{O}_4)_2]\).
Answer: Both geometrical (cis-, trans-) isomers can exist. Additionally, the cis-isomer is optically active and exists in two optical isomers (dextro and laevo), whereas the trans-isomer is optically inactive (symmetrical).
In simple words: It shows cis-trans geometrical isomerism, and the cis form exhibits optical activity.

Exam Tip: Trans-isomers with a plane of symmetry are generally optically inactive.

Question 2. Write the IUPAC name of:
i) \( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2\
ii) \([\text{PtCl}_2(\text{en})_2](\text{NO}_3)_2\)
Answer:
(i) Pentaamminechloridocobalt(III) chloride
(ii) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate
In simple words: Follow alphabetical rules for ligands, specify oxidation states in roman numerals.

Question 3. On the basis of VB theory that \([\text{Ni}(\text{CN})_4]^{2-}\) ion with square planar structure is diamagnetic and \([\text{NiCl}_4]^{2-}\) is tetrahedral is paramagnetic.
Answer:
In \([\text{Ni}(\text{CN})_4]^{2-}\), Ni is in \(+2\) oxidation state with \(3d^8\) configuration. \(\text{CN}^-\) is a strong field ligand, causing pairing of unpaired 3d electrons, leaving no unpaired electrons and resulting in \(\text{dsp}^2\) hybridization (square planar, diamagnetic).
In \([\text{NiCl}_4]^{2-}\), Ni is in \(+2\) state (\(3d^8\)). \(\text{Cl}^-\) is a weak field ligand, so pairing does not occur, leading to \(\text{sp}^3\) hybridization (tetrahedral, paramagnetic with 2 unpaired electrons).
In simple words: Strong ligands cause electron pairing (diamagnetic), while weak ligands do not (paramagnetic).

Question 4. Draw the crystal field splitting of d orbital in octahedral crystal field. Write the electronic configuration of \(\text{d}^5\).
Answer:
In an octahedral crystal field, the d-orbitals split into two sets: lower energy \(\text{t}_{2\text{g}}\) set and higher energy \(\text{e}_{\text{g}}\) set.
For \(\text{d}^5\):
- When \(\Delta_0 > P\) (strong field / low spin): \(\text{t}_{2\text{g}}^5\text{e}_{\text{g}}^0\)
- When \(\Delta_0 < P\) (weak field / high spin): \(\text{t}_{2\text{g}}^3\text{e}_{\text{g}}^2\)
In simple words: D-orbitals split into \(\text{t}_{2\text{g}}\) and \(\text{e}_{\text{g}}\); filling depends on pairing energy versus crystal field splitting.

Question 5. Write the optical isomer of \([\text{PtCl}_2(\text{en})_2]^{2+}\).
Answer: The cis-isomer of \([\text{PtCl}_2(\text{en})_2]^{2+}\) is optically active and exists as non-superimposable mirror images (dextro and laevo isomers).
In simple words: Cis-form has optical isomers (d and l forms) due to the absence of a plane of symmetry.

Question 6. For the complex \([\text{Fe}(\text{en})_2\text{Cl}_2]\text{Cl}\) (\(\text{en} = \text{ethylene diamine}\)), identify:
(i) The oxidation number of iron
(ii) The hybrid orbitals and the shape of the complex
(iii) The magnetic behavior of the complex
(iv) The number of geometrical isomers
(v) Whether there is an optical isomer also
(vi) Name of the complex. (At. No. of Fe = 26)
Answer:
(i) Oxidation number: \(x + 2(0) + 2(-1) + (-1) = 0 \implies x = +3\)
(ii) Hybridization: \(\text{sp}^3\text{d}^2\) (or \(\text{d}^2\text{sp}^3\)), Octahedral shape.
(iii) Paramagnetic (contains unpaired electrons).
(iv) 2 geometrical isomers (cis and trans).
(v) Yes, the cis-isomer shows optical isomerism.
(vi) Dichloridobis(ethane-1,2-diamine)iron(III) chloride.
In simple words: Iron is in \(+3\) state, forming an octahedral, paramagnetic complex with cis/trans and optical isomers.

Question 7. Explain structure of \([\text{Co}(\text{NH}_3)_6]^{3+}\) on the basis of valence bond theory.
Answer: In \([\text{Co}(\text{NH}_3)_6]^{3+}\), cobalt is in \(+3\) oxidation state with a \(3d^6\) electronic configuration. \(\text{NH}_3\) is a strong ligand causing pairing of electrons. It undergoes \(\text{d}^2\text{sp}^3\) hybridization using inner d-orbitals. The complex is octahedral and diamagnetic (no unpaired electrons), known as an inner orbital or low spin complex.
In simple words: Cobalt uses inner d-orbitals for hybridization, resulting in an inner-orbital octahedral diamagnetic complex.

Question 8. Write the formulas for the following coordination compounds:
i. Tetraamminediaquacobalt(III) chloride
ii. Potassium tetracyanonickelate(II)
iii. Tris(ethane-1,2-diamine)chromium(III) chloride
Answer:
i. \([\text{Co}(\text{H}_2\text{O})_2(\text{NH}_3)_4]\text{Cl}_3\)
ii. \(\text{K}_2[\text{Ni}(\text{CN})_4]\)
iii. \([\text{Cr}(\text{en})_3]\text{Cl}_3\)
In simple words: Translate the chemical names into formula units balancing charges properly.

Question 9. Write the IUPAC names of the following coordination compounds:
i. \([\text{Co}(\text{NH}_3)_6]\text{Cl}_3\)
ii. \([\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2\)
iii. \(\text{K}_3[\text{Fe}(\text{CN})_6]\)
Answer:
i. Hexaamminecobalt(III) chloride
ii. Pentaamminechloridocobalt(III) chloride
iii. Potassium hexacyanoferrate(III)
In simple words: Name ligands alphabetically followed by metal and oxidation number.

Question 10. Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:
i. \(\text{K}[\text{Cr}(\text{H}_2\text{O})_2(\text{C}_2\text{O}_4)_2]\)
ii. \([\text{Co}(\text{en})_3]\text{Cl}_3\)
iii. \([\text{Co}(\text{NH}_3)_5(\text{NO}_2)](\text{NO}_3)_2\)
Answer:
i. Geometrical (cis-trans) and optical isomerism.
ii. Optical isomerism (two optical isomers: d and l).
iii. Linkage isomerism, ionization isomerism, and optical isomerism.
In simple words: Complexes can exhibit multiple types of isomerism simultaneously.

Question 11. \([\text{NiCl}_4]^{2-}\) is paramagnetic while \([\text{Ni}(\text{CO})_4]\) is diamagnetic though both are tetrahedral. Why?
Answer: In \([\text{NiCl}_4]^{2-}\), \(\text{Cl}^-\) is a weak field ligand and does not cause pairing of 3d electrons, leaving 2 unpaired electrons (paramagnetic, \(\text{sp}^3\)). In \([\text{Ni}(\text{CO})_4]\), Ni is in zero oxidation state and CO is a strong field ligand which causes pairing of 3d electrons and shift of 4s electrons to 3d, leaving no unpaired electrons (diamagnetic, \(\text{sp}^3\)).
In simple words: Ligand strength determines electron pairing and magnetic behavior.

Question 12. Predict the number of unpaired electrons in the square planar \([\text{Pt}(\text{CN})_4]^{2-}\) ion.
Answer: In \([\text{Pt}(\text{CN})_4]^{2-}\), Pt is in \(+2\) state with \(5d^8\) configuration. Being a square planar complex involving \(\text{dsp}^2\) hybridization, and with \(\text{CN}^-\) being a strong field ligand, all electrons are paired. Thus, there are 0 unpaired electrons.
In simple words: Platinum(II) complexes with strong ligands are always diamagnetic with zero unpaired electrons.

Question 1. Explain the bonding in coordination compounds in terms of Werner's postulates.
Answer:
(i) Metals exhibit two types of valencies: primary valencies (ionizable, satisfied by negative ions, corresponding to oxidation number) and secondary valencies (non-ionizable, fixed, satisfied by neutral or negative ions, corresponding to coordination number).
(ii) Secondary valencies project in specific spatial directions, determining the geometry of the complex.
In simple words: Werner proposed primary (ionizable) and secondary (non-ionizable) valencies to explain coordination structures.

Question 2. Draw figure to show the splitting of d orbitals in an octahedral crystal field.
Answer: In an octahedral crystal field, the approach of ligands splits the degenerate d-orbitals into lower energy \(\text{t}_{2\text{g}}\) orbitals (\(\text{d}_{xy}, \text{d}_{yz}, \text{d}_{zx}\)) and higher energy \(\text{e}_{\text{g}}\) orbitals (\(\text{d}_{x^2-y^2}, \text{d}_{z^2}\)).
In simple words: D-orbitals split into two sets: lower \(\text{t}_{2\text{g}}\) and higher \(\text{e}_{\text{g}}\) in an octahedral field.

Question 3. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer: A spectrochemical series is the arrangement of common ligands in order of increasing crystal field splitting energy (CFSE) values. Weak field ligands cause small splitting and form high spin complexes, while strong field ligands cause large splitting and form low spin complexes.
In simple words: Spectrochemical series lists ligands by their splitting ability; strong ligands split d-orbitals more than weak ligands.

Question 4. Discuss the nature of bonding in metal carbonyls.
Answer: Metal-carbon bonds in metal carbonyls possess both \(\sigma\) and \(\pi\) character. A \(\sigma\)-bond is formed by donation of lone pair from carbonyl carbon to vacant metal orbital, and a \(\pi\)-bond is formed by back-donation of electrons from filled metal d-orbital into vacant anti-bonding \(\pi^*\) orbital of CO (synergic effect).
In simple words: Synergic bonding involving \(\sigma\)-donation and \(\pi\)-back-donation strengthens metal-carbonyl bonds.

Question 5. What is meant by the chelate effect?
Answer: When a multidentate ligand attaches to a metal ion forming a ring structure, the resulting complex is unusually stable. This enhanced stability of cyclic coordination complexes is known as the chelate effect.
In simple words: Ring formation by polydentate ligands makes coordination complexes extra stable.

Question 6. Discuss briefly giving an example in each case the role of coordination compounds in: (i) biological system (ii) medicinal chemistry (iii) analytical chemistry (iv) extraction/metallurgy of metals
Answer:
(i) Biological systems: Chlorophyll is a magnesium coordination compound essential for photosynthesis; hemoglobin is an iron coordination complex responsible for oxygen transport.
(ii) Medicinal chemistry: Cis-platin is used as an antitumor drug.
(iii) Analytical chemistry: EDTA is used in volumetric estimations of metal ions; specific colour changes identify basic radicals.
(iv) Extraction of metals: Gold and silver are extracted via cyanide complexation processes (\([\text{Au}(\text{CN})_2]^-\)).
In simple words: Coordination compounds play vital roles in biology, medicine, analysis, and metallurgy.

Question 7. What will be the correct order for the wavelengths of absorption in the visible region for the following: \([\text{Ni}(\text{NO}_2)_6]^{4-}, [\text{Ni}(\text{NH}_3)_6]^{2+}, [\text{Ni}(\text{H}_2\text{O})_6]^{2+}\)?
Answer: The CFSE values for the ligands increase in the order: \(\text{H}_2\text{O} < \text{NH}_3 < \text{NO}_2^-\).
Since energy is inversely proportional to wavelength (\(E = \frac{hc}{\lambda}\)), the wavelength of absorption will follow the reverse order:
\([\text{Ni}(\text{H}_2\text{O})_6]^{2+} > [\text{Ni}(\text{NH}_3)_6]^{2+} > [\text{Ni}(\text{NO}_2)_6]^{4-}\)
In simple words: Stronger ligands absorb higher energy (shorter wavelength), so the wavelength order is the reverse of ligand field strength.