Chapter 9: Coordination Compounds
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: The coordination number of a central metal atom or ion in a complex is defined as the total number of coordinate bonds formed between the ligands and the central metal atom/ion.
Answer: Potassium hexacyanoferrate(III).
(Explanation: Potassium is the cation. The complex anion has 6 cyano ligands and iron. Let oxidation state of Fe be $x$. Then $3(+1) + x + 6(-1) = 0 \implies x = +3$).
Answer: Ethylenediamine ($H_2N-CH_2-CH_2-NH_2$) or Oxalate ion ($C_2O_4^{2-}$).
Answer: A complex in which the central metal atom or ion is bound to only one kind of donor group (ligand) is called a homoleptic complex. (Example: $[Co(NH_3)_6]^{3+}$).
2 Mark Questions (Short Answer-I)
Answer:
| Double Salt | Coordination Compound |
|---|---|
| 1. They dissociate completely into their simple constituent ions when dissolved in water. | 1. They do not dissociate completely in water; the complex ion remains intact. |
| 2. They lose their identity in aqueous solution. | 2. They retain their identity in the solid state as well as in aqueous solution. |
| 3. Give positive tests for all constituent ions in solution. | 3. Do not give tests for the ions present inside the coordination sphere. |
| 4. Example: Mohr's salt, Potash alum. | 4. Example: $K_4[Fe(CN)_6]$, $[Cu(NH_3)_4]SO_4$. |
Answer:
- Most elements exhibit two types of valency in coordination compounds: Primary valency (which is ionizable and corresponds to the oxidation state) and Secondary valency (which is non-ionizable and corresponds to the coordination number).
- The secondary valencies are directed towards fixed positions in space, which gives a definite geometry to the coordination complex (e.g., octahedral for 6 secondary valencies).
Answer:
Linkage Isomerism arises in coordination compounds containing an ambidentate ligand (a ligand that can coordinate through two different atoms). The isomers differ in the point of attachment of the ligand to the central metal atom.
Example:
- $[Co(NH_3)_5(NO_2)]Cl_2$: Here, the nitrite ligand is bonded through the Nitrogen atom (yellow colored).
- $[Co(NH_3)_5(ONO)]Cl_2$: Here, the nitrite ligand is bonded through the Oxygen atom (red colored).
3 Mark Questions (Short Answer-II / VBT Applications)
Answer:
Step 1: Oxidation State and Configuration
Let the oxidation state of Ni be $x$. Then $x + 4(-1) = -2 \implies x = +2$.
Electronic configuration of Ni ($Z=28$): $[Ar] 3d^8 4s^2$.
Electronic configuration of $Ni^{2+}$: $[Ar] 3d^8 4s^0 4p^0$.
Step 2: Nature of Ligand
The $CN^-$ ion is a strong field ligand. According to VBT, it forces the pairing of the two unpaired electrons in the 3d subshell against Hund's rule. This vacates one 3d orbital.
Step 3: Hybridization
To accommodate 4 pairs of electrons from 4 $CN^-$ ligands, the metal ion undergoes hybridization using the one empty inner 3d orbital, one 4s orbital, and two 4p orbitals. This results in $dsp^2$ hybridization.
Step 4: Geometry
The $dsp^2$ hybridization corresponds to a Square Planar geometry.
Step 5: Magnetic Behavior
Because the strong field ligand forced electron pairing, there are no unpaired electrons left in the complex. Therefore, $[Ni(CN)_4]^{2-}$ is diamagnetic ($\mu = 0$).
Answer:
Statement of EAN Rule:
The central metal atom or ion in a coordination complex accepts electron pairs from the ligands until the total number of electrons surrounding it (Effective Atomic Number) becomes equal to the atomic number of the nearest noble gas in the periodic table.
Calculation for $[Co(NH_3)_6]^{3+}$:
1. Atomic number of Co ($Z$) = 27.
2. Oxidation state of Co ($X$): Let it be $x$. Since $NH_3$ is neutral, $x + 6(0) = +3 \implies x = +3$. So, electrons lost = 3.
3. Number of electrons donated by ligands ($Y$): There are 6 $NH_3$ ligands, each donating 2 electrons. $Y = 6 \times 2 = 12$.
4. Formula: $$\text{EAN} = Z - X + Y$$
$$\text{EAN} = 27 - 3 + 12$$
$$\text{EAN} = 36$$
The calculated EAN is 36, which is the atomic number of the noble gas Krypton (Kr). Hence, the complex obeys the EAN rule.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 3-mark VBT question combined with a 1-mark IUPAC name question).
(b) Write the IUPAC name of $[Pt(NH_3)_2Cl_2]$. [1 Mark] March 2017, March 2021, March 2023
Answer (a): Application of VBT on $[CoF_6]^{3-}$
Step 1: Oxidation State and Configuration
Oxidation state of Co in $[CoF_6]^{3-}$: $x + 6(-1) = -3 \implies x = +3$.
Configuration of Co ($Z=27$): $[Ar] 3d^7 4s^2$.
Configuration of $Co^{3+}$: $[Ar] 3d^6 4s^0 4p^0 4d^0$.
Step 2: Nature of Ligand
The fluoride ion ($F^-$) is a weak field ligand. It cannot overcome the pairing energy, so it does not force the 3d electrons to pair up against Hund's rule. There will be 4 unpaired electrons in the 3d subshell.
Step 3: Hybridization
Since inner 3d orbitals are occupied and not available, the metal ion uses its outer orbitals. One 4s, three 4p, and two 4d orbitals mix to undergo $sp^3d^2$ hybridization to accommodate 6 pairs of electrons from 6 $F^-$ ligands.
Step 4: Geometry
The $sp^3d^2$ hybridization corresponds to an Octahedral geometry.
Step 5: Magnetic Property
Because there are 4 unpaired electrons present in the inner 3d subshell, the complex is highly paramagnetic. Since it uses outer d-orbitals, it is called an outer orbital complex or high-spin complex.
Answer (b): IUPAC Name
Ligands: Two ammine ($NH_3$), two chloro ($Cl^-$). Metal: Platinum (Pt). The complex is neutral, so oxidation state of Pt is +2.
IUPAC Name: Diamminedichloroplatinum(II).
(b) What is meant by a polydentate ligand? Give one example. [1 Mark] Oct 2013, March 2020
Answer (a): Geometric Isomerism
Definition: Geometric isomerism is a type of stereoisomerism that arises due to the different possible relative spatial arrangements of ligands around the central metal atom. It is common in heteroleptic square planar and octahedral complexes.
- Cis isomer: Identical ligands occupy adjacent positions (bond angle 90°).
- Trans isomer: Identical ligands occupy opposite positions (bond angle 180°).
Structures for $[Pt(NH_3)_2Cl_2]$ (Square Planar):
Cis-isomer
(Identical ligands are adjacent)
Trans-isomer
(Identical ligands are opposite)
Answer (b): Polydentate Ligand
A ligand that has two or more donor atoms and can bind to the central metal ion through multiple coordination sites simultaneously is called a polydentate ligand.
Example: EDTA (Ethylenediaminetetraacetate ion) is a hexadentate ligand (has 6 donor atoms: 2 Nitrogen and 4 Oxygen).
📚 Also Read
Lecture Notes🏆 Complete Maharashtra HSC Class 12 Chemistry Preparation
Prepare for the Maharashtra HSC Class 12 Chemistry Board Exam with chapter-wise revision notes, important questions, PYQs, formula sheets, mock tests, quick revision resources and exam-oriented study material. Everything you need to score high in one comprehensive learning hub.
🚀 Explore the Complete Maharashtra HSC Chemistry Hub
No comments:
Post a Comment