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Calculate the overall complex dissociation equilibrium constant for the ${[Cu{(N{H_3})_4}]^{2 + }}$, given that stability constant $({\beta _4})$ for this complex is $2.1 \times {10^{13}}$:
A.$8.27 \times {10^{ - 13}}$
B.$4.76 \times {10^{ - 14}}$
C.$2.39 \times {10^{ - 7}}$
D.$1.83 \times {10^{14}}$

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Last updated date: 16th Jun 2024
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Answer
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Hint: The concept of stability constant in coordination chemistry is very important to study about the coordination compounds. . Stability constant is the equilibrium constant for the formation of a complex compound in a solution. The stability constant is also known as formation constant or binding constant.

Complete step by step answer:
As we know that the overall complex dissociation equilibrium constant of a coordination compound is the reciprocal of the stability constant of the coordination compound. In the question the coordination compound is ${[Cu{(N{H_3})_4}]^{2 + }}$. The value of the stability constant $({\beta _4})$ of the coordination compound ${[Cu{(N{H_3})_4}]^{2 + }}$is given to be $ \Rightarrow {\beta _4} = 2.1 \times {10^{13}}$. So we can directly find out the value of the complex dissociation equilibrium constant by using the formula
$ \Rightarrow {K_d} = \dfrac{1}{{{\beta _4}}}$. Here ${K_d}$ is the overall complex dissociation constant and $({\beta _4})$ is the stability constant. So:
$
   \Rightarrow {K_d} = \dfrac{1}{{{\beta _4}}} \\
   \Rightarrow {K_d} = \dfrac{1}{{2.1 \times {{10}^{13}}}} \\
   \Rightarrow {K_d} = 4.76 \times {10^{ - 14}} \\
 $
From the above explanation and calculation it is clear to us that

The correct answer of the given question is : B. $4.76 \times {10^{ - 14}}$

Additional information:
${[Cu{(N{H_3})_4}]^{2 + }}$ has coordination number of four. The oxidation state of $Cu$ is $ + 2$. The geometry of the coordination compound ${[Cu{(N{H_3})_4}]^{2 + }}$is square planar. It has one unpaired electron. The hybridisation of ${[Cu{(N{H_3})_4}]^{2 + }}$is $ds{p^2}$.The stability of a coordination compound in solution is the degree of association between the metal ion and the ligands involved in the state of equilibrium. The stability is expressed in terms of the equilibrium constant for the association.

Note:
Always remember that the overall complex dissociation equilibrium constant of a coordination compound is the reciprocal of the stability constant of the coordination compound. The formula we can use is : $ \Rightarrow {K_d} = \dfrac{1}{{{\beta _4}}}$ for ${[Cu{(N{H_3})_4}]^{2 + }}$. Always avoid silly mistakes and calculation errors while solving the numerical and using the formula.