CBSE Class 12 Chemistry Chapter 9: Coordination Compounds NCERT Solutions

Solution: The stability of a coordination complex is directly related to its formation constant, K. A higher value of K indicates a more stable complex. The logarithm of the formation constant, log K, is provided for each reaction. We need to find the reaction with the highest log K value.

Comparing the given log K values:

• (a) $\backslash log\; K\; =\; 11.6$
• (b) $\backslash log\; K\; =\; 27.3$
• (c) $\backslash log\; K\; =\; 15.4$
• (d) $\backslash log\; K\; =\; 8.9$

The highest log K value is 27.3, which corresponds to the formation of the complex $[Cu(CN)\_\{4\}]^\{2-\}$. Therefore, this complex is the most stable among the given options.

Answer: (b) $Cu^\{2+\}\; +\; 4CN^\{-\}\; \to \; [Cu(CN)\_\{4\}]^\{2-\},\; \backslash log\; K\; =\; 27.3$

Solution: The color of coordination compounds is related to the energy of light absorbed, which causes electronic transitions (d-d transitions) within the metal ion. This energy is directly related to the crystal field splitting energy ($\backslash Delta\; E$). The relationship between energy ($\backslash Delta\; E$) and wavelength ($\backslash lambda$) is given by $\backslash Delta\; E\; =\; \backslash frac\{hc\}\{\backslash lambda\}$, where h is Planck's constant and c is the speed of light. This implies that $\backslash Delta\; E\; \backslash propto\; \backslash frac\{1\}\{\backslash lambda\}$, meaning higher splitting energy corresponds to shorter absorbed wavelength.

The strength of ligands determines the magnitude of crystal field splitting ($\backslash Delta\; E$). According to the spectrochemical series, the order of ligand strength is $CN^\{-\}\; >\; NH\_3\; >\; H\_2O$. Therefore, the order of crystal field splitting energy for the given complexes is:

$[Co(CN)\_6]^\{3-\}\; >\; [Co(NH\_3)\_6]^\{3+\}\; >\; [Co(H\_2O)\_6]^\{3+\}$

Since wavelength absorbed is inversely proportional to the splitting energy, the order of absorbed wavelength will be the reverse:

$[Co(H\_2O)\_6]^\{3+\}\; >\; [Co(NH\_3)\_6]^\{3+\}\; >\; [Co(CN)\_6]^\{3-\}$

Thus, $[Co(H\_2O)\_6]^\{3+\}$ absorbs the longest wavelength, and $[Co(CN)\_6]^\{3-\}$ absorbs the shortest wavelength.

Answer: (c) $[Co(H\_2O)\_6]^\{3+\}\; >\; [Co(NH\_3)\_6]^\{3+\}\; >\; [Co(CN)\_6]^\{3-\}$

Solution: Silver nitrate ($AgNO\_3$) reacts with free chloride ions ($Cl^-$) in solution to form a precipitate of silver chloride ($AgCl$). The number of moles of $AgCl$ precipitated is equal to the number of moles of free chloride ions present in the solution.

In this case, 0.1 mol of the complex $CoCl\_3(NH\_3)\_5$ yields 0.2 mol of $AgCl$ precipitate upon treatment with excess $AgNO\_3$. This means that there are 2 moles of free chloride ions for every 1 mole of the complex.

The chloride ions that precipitate are those located outside the coordination sphere. Therefore, the complex must contain 2 free chloride ions and 1 chloride ion coordinated to the cobalt ion. The general formula for such a complex would be $[Co(NH\_3)\_5Cl]Cl\_2$.

When this complex dissolves in water, it dissociates to form ions. The dissociation equation is:

$[Co(NH\_3)\_5Cl]Cl\_2(aq)\; \backslash longrightarrow\; [Co(NH\_3)\_5Cl]^\{2+\}(aq)\; +\; 2Cl^\{-\}(aq)$

This dissociation produces one complex cation ($[Co(NH\_3)\_5Cl]^\{2+\}$) and two chloride anions ($Cl^\{-\}$). Thus, the complex acts as a 1:2 electrolyte (1 cation : 2 anions).

Answer: (b) 1:2 electrolyte

Solution: The reaction with silver nitrate ($AgNO\_3$) precipitates chloride ions ($Cl^-$) that are present as counter-ions outside the coordination sphere. The number of moles of $AgCl$ precipitated corresponds to the number of free chloride ions.

Given that 1 mole of the hydrated chromium(III) chloride complex yields 3 moles of $AgCl$ precipitate upon reaction with excess $AgNO\_3$, it indicates that there are 3 free chloride ions per formula unit of the complex.

The general formula for a coordination compound is $[M(L)\_x]Y\_n$, where M is the metal, L are ligands coordinated to the metal, and Y are counter-ions. In this case, the metal is Chromium (Cr), the counter-ion is Chloride ($Cl^-$), and the ligands are water molecules ($H\_2O$). Since all three chloride ions are precipitated, they must be outside the coordination sphere. This means the coordination sphere contains only water molecules as ligands coordinated to the chromium ion.

The coordination number of chromium in such complexes is typically 6. Therefore, all six coordination sites are occupied by water molecules. The formula of the complex is thus $[Cr(H\_2O)\_6]Cl\_3$.

Answer: (d) $[Cr(H\_2O)\_6]Cl\_3$

Solution: To determine the IUPAC name of the coordination compound $[Pt(NH\_3)\_2Cl\_2]$, we follow the standard nomenclature rules:

1. Identify ligands and metal: The ligands are ammonia ($NH\_3$) and chloride ($Cl$). The central metal atom is Platinum (Pt).
2. Alphabetical order of ligands: Ligands are named before the metal. Ammonia is 'ammine' and chloride is 'chloro'. In alphabetical order, 'ammine' comes before 'chloro'. There are two ammonia ligands and two chloride ligands.
3. Prefixes for number of ligands: Since there are two of each ligand, we use the prefix 'di-'. So, we have 'diammine' and 'dichlorido'.
4. Name the metal: The metal is Platinum. Since the complex is neutral (no overall charge indicated), the metal name is used as is.
5. Determine the oxidation state of the metal: Let the oxidation state of Platinum be x. The oxidation state of $NH\_3$ is 0, and the oxidation state of $Cl$ is -1. The sum of oxidation states equals the overall charge of the complex, which is 0.

   $x\; +\; 2(0)\; +\; 2(-1)\; =\; 0$

   $x\; -\; 2\; =\; 0$

   $x\; =\; +2$

   The oxidation state of Platinum is +2, which is written in Roman numerals in parentheses: (II).
6. Assemble the name: Combine the parts in the correct order: Diammine (ligands) + dichlorido (ligands) + Platinum (metal) + (II) (oxidation state).

The complete IUPAC name is Diamminedichloridoplatinum(II).

Answer: (a) Diamminedichloridoplatinum (II)