The given complex is
\[[\text{Fe}(\text{NH}_3)_2(\text{CN})_4]^{3+}\]
Here, the oxidation state of iron is \(+3\), which corresponds to a \(d^5\) electronic configuration in the high-spin state. The ligands cyanide (\(\text{CN}^-\)) and ammonia (\(\text{NH}_3\)) are arranged such that the \(e_g\) orbitals remain unoccupied.
Given values:
\(x = 2 \, (number\ of \text{NH}_3 \, \text{ligands})\),
\(y = 4 \, (\text{number of } \text{CN}^- \, \text{ligands})\)
Thus, \(x + y = 2 + 4 = 6\).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: