Based on the given stability constants (hypot | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The stability constant $(K)$ of a coordination complex is a measure of the strength of the ligand-metal bond. $A$ higher value of the stability co

Vedclass · Accepted Answer

$Cu^{2+} + 4CN^- ightleftharpoons [Cu(CN)_4]^{2-} \quad (K = 2.0 	imes 10^{27})$

Vedclass · Answer

(B) The stability constant $(K)$ of a coordination complex is a measure of the strength of the ligand-metal bond. $A$ higher value of the stability constant indicates a more stable complex,which implies that the ligand is a stronger ligand. Comparing the given values: $(a)$ $K = 4.5 	imes 10^{11}$ $(b)$ $K = 2.0 	imes 10^{27}$ $(c)$ $K = 3.0 	imes 10^{15}$ $(d)$ $K = 9.5 	imes 10^8$ Since $2.0 	imes 10^{27}$ is the largest value,the complex $[Cu(CN)_4]^{2-}$ is the most stable,making $CN^-$ the strongest ligand among the given options.

Based on the given stability constants (hypothetical values),predict which is the strongest ligand?

Similar Questions

Compound $PdCl_4 \cdot 6H_2O$ is a hydrated complex; $1 \ m$ aqueous solution of it has a freezing point of $269.28 \ K$. Assuming $100\%$ ionization of the complex,calculate the molecular formula of the complex ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$,freezing point of pure water $= 273 \ K$).

How does ammonia react with a solution of $Cu^{2+}$?

$CuSO_4$ solution reacts with excess of $KCN$ solution to form:

The correct increasing order of stability of the complexes based on $\Delta_o$ value is $:$
$I. \ [Mn(CN)_6]^{3-}$
$II. \ [Co(CN)_6]^{4-}$
$III. \ [Fe(CN)_6]^{4-}$
$IV. \ [Fe(CN)_6]^{3-}$

Which complex cannot ionise in solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module