Worksheet with Solutions: Coordination Compounds | Chemistry Class 12 - NEET PDF Download

Answer: square planar
Solution: In dsp² hybridisation, one d-orbital, one s-orbital, and two p-orbitals combine to form four hybrid orbitals arranged in a square planar geometry. This is commonly seen in low-spin d⁸ metal ions like Ni²⁺.

Answer: tetrahedral
Solution: In a tetrahedral structure, all four positions are equivalent, so geometric isomerism like cis-trans is not possible. It requires different spatial arrangements of ligands, which is only possible in square planar or octahedral geometries.

Answer: zero
Solution: Carbonyl (CO) is a neutral ligand. Since metal carbonyls like Ni(CO)₄ or Fe(CO)₅ are overall neutral compounds, the oxidation state of the metal must be zero.

Answer: sp³
Solution: The complex has a coordination number of 4, and Cl⁻ is a weak field ligand. This leads to sp³ hybridisation using outer orbitals, resulting in a tetrahedral shape.

Answer: greater
Solution: High-spin complexes have more unpaired electrons than low-spin complexes. Since magnetic moment increases with the number of unpaired electrons, it is greater in high-spin complexes.

Answer: –0.6 Δ₀
Solution: For a high-spin d⁴ configuration:

• 3 electrons occupy t₂g orbitals (–0.4Δ₀ each)

• 1 electron occupies e_g orbital (+0.6Δ₀)
  CFSE = (3 × –0.4) + (1 × 0.6) = –1.2 + 0.6 = –0.6 Δ₀

Answer: c

Explanation: The ligands are named in alphabetical order, regardless of charge:

• ammine (NH₃)

• bromido (Br⁻)

• chlorido (Cl⁻)

• nitro (NO₂⁻)

Central metal: platinum (Pt)

Oxidation state of Pt:

• Overall complex: [Pt(NH₃)₃Br(NO₂)Cl]⁺

• Charge of counter ion (Cl⁻) = –1

• So, the cation part has +1 charge

• Ligands: NH₃ (neutral), Br⁻ (–1), NO₂⁻ (–1), Cl⁻ (–1) ⇒ Total = –3

• So, Pt = +4

Final name: triamminebromidochloridonitroplatinum(IV) chloride

Answer: b

Explanation: NO₂⁻ is an ambidentate ligand: It can coordinate via N atom (nitro form) or O atom (nitrito form). When the same ligand binds through different atoms, the isomers formed are linkage isomers.

Answer: c

Explanation:

• [Ni(CN)₄]²⁻: Ni²⁺ is 3d⁸ configuration. CN⁻ is a strong field ligand, causing pairing.

  1. Hybridisation: dsp²

  2. Shape: square planar

• [NiCl₄]²⁻ and [Ni(CO)₄] are tetrahedral because Cl⁻ and CO are weak/moderate field ligands and do not cause pairing (or involve sp³ hybridization).

Answer: c

Explanation:

• In d⁶ (low spin) complex , two electrons get paired up to make two d orbitals empty. The hybridisation is d²sp³ (octahedral and the complex is low spin complex in the absence of half filled electrons.

Answer: c

Explanation:

• In aqueous solution, [Cu(NH₃)₄]²⁺ is the complex ion and SO₄²⁻ is the counter ion.

• Only the counter ion (SO₄²⁻) is free in solution and can be tested using BaCl₂ (white ppt of BaSO₄).

• Cu²⁺ is not free, it is complexed with NH₃ — so it doesn't give Cu²⁺ tests.

• Similarly, NH₃ is bound — no free NH₃ to detect with Nessler’s reagent.

Answer: c

Explanation:

• NF₃ is a weaker ligand than N(CH₃)₃ because in NF₃, the nitrogen’s lone pair is strongly pulled toward the highly electronegative fluorine atoms, reducing its ability to donate electrons.

• In N(CH₃)₃, the methyl groups are electron-donating and push electron density toward nitrogen, making it a better electron donor (stronger ligand).

• However, NF₃ does NOT ionize to give F⁻ ions in water. It's a covalent compound and remains mostly undissociated.

Answer: b

Explanation:

• [Fe(CN)₆]³⁻: Fe³⁺ has 3d⁵ configuration. CN⁻ is a strong field ligand, so pairing occurs, but still one unpaired electron remains ⇒ weakly paramagnetic.

• [Fe(CN)₆]⁴⁻: Fe²⁺ has 3d⁶ configuration. CN⁻ causes full pairing ⇒ diamagnetic.

• The oxidation states mentioned are correct, but they do not explain the magnetic behavior directly — the nature of the ligand (CN⁻) and crystal field splitting do.

Answer: a

Explanation:

• [Ti(H₂O)₆]³⁺ has Ti³⁺ (3d¹) ⇒ d–d transition possible ⇒ coloured.

• [Sc(H₂O)₆]³⁺ has Sc³⁺ (3d⁰) ⇒ no d-electrons ⇒ no d–d transitions ⇒ colourless.

• So, both statements are correct, and the reason correctly explains the assertion.

Answer: a

Explanation:

• Linkage isomerism occurs when a ligand can coordinate to the metal through two different atoms, e.g., NO₂⁻ can bind via N or O.

• These are ambidentate ligands, and their nature explains linkage isomerism.

Answer: In both the cases, Fe is in oxidation state +3. Outer electronic configuration of Fe+3 is :

In the presence of CN^–, the 3d electrons pair up leaving only one unpaired electron. The hybridisation involved is d²sp³ forming inner orbital complex which is weakly paramagnetic. In the presence of H₂O (a weak ligand), 3d electrons do not pair up. The hybridisation involved is sp³d² forming an outer orbital complex. As it contains five unpaired electrons so it is strongly paramagnetic.

Answer: In [Co(NH₃)₆]³⁺, the d-electrons of Co³⁺ ([Ar]3d⁶ 45°) get paired leaving behind two empty d-orbital and undergo d²sp³ hybridization and hence inner orbital complex, while in [Ni(NH₃)₆]²⁺ the d-electrons of Ni²⁺ ([Ar]3d⁸ 45°) do not pair up and use outer 4d subshell hence outer orbital complex.

Answer: (i) [Cr(NH₃)₂Cl₃(en)]Cl
IUPAC name : Diammine dichlorido ethylenediamine chromium (III) chloride.
(ii) [Co(NH₃)₅(ONO)]²⁺

Answer:
(i) [Fe(en)₂Cl₂] Cl or x + 0 + 2 (-1) + (-1) = 0
x + (- 3) = 0 or x = + 3
∴ Oxidation number of iron, x = + 3
(ii) The complex has two bidentate ligands and two monodentate ligands. Therefore, the coordination number is 6 and hybridization will be d²sp³ and shape will be octahedral.
(iii) In the complex ₂₆Fe³⁺ = 3d⁵ 4s⁰ 4p⁰

Due to presence of one unpaired electrons in d orbitals the complex is paramagnetic.
The number of geometrical isomers are two.
(v) In coordination complex of [Fe(en)₂Cl₂] Cl, only cis-isomer shows optical isomerism.

Answer:
(i) [COF4]2_ : Tetrafluorido cobalt (III) ion
Coordination number = 4 Shape = Tetrahedral Hybridisation = sp³

Answer:
(i) Crystal field splitting: It is the splitting of the degenerate energy levels due to the presence of ligands. When ligand approaches a transition metal ion, the degenerate d-orbitals split into two sets, one with lower energy and the other with higher energy. This is known as crystal field splitting and the difference between the lower energy set and higher energy set is known as crystal field splitting energy (CFSE)
Example : 3d⁵ of Mn²⁺
(ii) Linkage isomerism: When more than one atom in an ambidentate ligand is linked with central metal ion to form two types of complexes, then the formed isomers are called linkage isomers and the phenomenon is called linkage isomerism.
[Cr(H₂O)₅(NCS)]²⁺ Pentaaquathiocyanate chromium (III) ion
[Cr(H₂O)₅(NCS)]²⁺
Pentaaquaisothiocyanate chromium (III) ion
(iii) Ambidentate ligand: The monodentate ligands with more than one coordinating atoms is known as ambidentate ligand. Monodentate ligands have only one atom capable of binding to a central metal atom or ion. For example, the nitrate ion NO₂^– can bind to the central metal atom/ion at either the nitrogen atom or one of the oxygen atoms.
Example : — SCN thiocyanate, — NCS isothiocyanate

Answer: (i) [Co(NH₃)₆]⁺³ → Octahedral shape, d²sp³ hybridisation, diamagnetic
Formation of [Co(NH₂)₆]⁺³ → oxidation state of Co is +3.

Answer:
(i) The electronic configuration of Ni is [Ar] 3d⁸ 4s² which shows that it can only form two types of complexes i.e. square planar (dsp²) in presence of strong ligand and tetrahedral (sp³) in presence of weak ligand. There are four empty orbitals in Ni while octahedral complexes require six empty orbitals.
(ii) Due to presence of empty d-orbitals in transition metals, they can accept electron pairs from ligands containing π electrons and hence can form ic-bonding complexes.
Example : ligands like C₅H₅, C₆H₆ etc.
(iii) Due to greater magnitude of Δ₀, CO produces strong fields which cause more splitting of d-orbitals and moreover it is also able to form π bond due to back bonding.

Answer:
(i) [Cr(NH₃)_s4Cl₂]Cl :
Hybridization : d²sp³
Shape : Octahedral
Magnetic behaviour: Paramagnetic
(ii) [Co(en)₃] Cl₃ :
Hybridization : d2sp³
Shape : Octahedral
Magnetic behaviour : Diamagnetic
(iii) K₃[Ni(CN)₄] :
Hybridization ; dsp²
Shape : Square planar
Magnetic behaviour: Diamagnetic

Answer: (i) [Pt(NH₃)₂Cl(NO₂)]
Name : Diamine chloridonitroplatinum II
Structure :
Magnetic behaviour: paramagnetic
(ii) [Co(NH₃)₄Cl₂]Cl
Name : Tetraamminedichloridocobalt (III) chloride
Structure : octahedral
Magnetic behaviour : diamagnetic
(iii) Ni(CO)₄
Name : Tetracarbonylnickel (O)
Structure : tetrahedral
Magnetic behaviour : diamagnetic.