| 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: \[ \text{A - I, B - II, C - IV, D - III.} \]
The steam volatile compounds among the following are:

If A and B are two events such that \( P(A \cap B) = 0.1 \), and \( P(A|B) \) and \( P(B|A) \) are the roots of the equation \( 12x^2 - 7x + 1 = 0 \), then the value of \(\frac{P(A \cup B)}{P(A \cap B)}\)
A wire of resistance $ R $ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $ A $ and $ B $ is $ \frac{R}{n} $. The value of $ n $ is:
Let the matrix $ A = \begin{pmatrix} 1 & 0 & 0 \\1 & 0 & 1 \\0 & 1 & 0 \end{pmatrix} $ satisfy $ A^n = A^{n-2} + A^2 - I $ for $ n \geq 3 $. Then the sum of all the elements of $ A^{50} $ 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.