To determine the number of ambidentate ligands among the given species, we first need to understand what constitutes an ambidentate ligand. An ambidentate ligand can coordinate to a metal ion through two different atoms. We will analyze each ligand:
Thus, the ambidentate ligands among the listed are \(\text{NO}_2^{-}\), \(\text{SCN}^{-}\), and \(\text{CN}^{-}\). This gives us a total of 3 ambidentate ligands, which is within the specified range (3,3).
Step 1: Define ambidentate ligands
Ambidentate ligands are ligands that can coordinate to a metal center through two different atoms.
Step 2: Analyze the ligands
Step 3: Count ambidentate ligands
Ambidentate ligands are:
\[ \text{NO}_2^-, \, \text{SCN}^-, \, \text{CN}^-. \] Total number = 3.
Thus, the correct answer: 3.
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
