| List I (Complex) | List II (Hybridization) | ||
| A. | Ni(CO)4 | I. | sp3 |
| B. | [Cu(NH3)4]2+ | II. | dsp2 |
| C. | [Fe(NH3)6]2+ | III. | sp3d2 |
| D. | [Fe(H2O)6]2+ | IV. | d2sp3 |
For \([Fe(NH_3)_6]^{2+}\), \(\Delta_0 < P\), hence the pairing of electrons does not occur in \(t_{2g}\). Therefore, the complex is outer orbital and its hybridisation is \(sp^3d^2\).
(A) \([Ni(CO)_4]\) - \(sp^3\)
(B) \([Cu(NH_3)_4]^{2+}\) - \(dsp^2\)
(C) \([Fe(NH_3)_6]^{2+}\) - \(sp^3d^2\)
(D) \([Fe(H_2O)_6]^{2+}\) - \(sp^3d^2\)
Thus, the correct match is: \[ \text{A - I, B - II, C - IV, D - III.} \]
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 :

A coordination compound holds a central metal atom or ion surrounded by various oppositely charged ions or neutral molecules. These molecules or ions are re-bonded to the metal atom or ion by a coordinate bond.
A coordination entity composes of a central metal atom or ion bonded to a fixed number of ions or molecules.
A molecule, ion, or group which is bonded to the metal atom or ion in a complex or coordination compound by a coordinate bond is commonly called a ligand. It may be either neutral, positively, or negatively charged.