[Co(NH$_3$)$_6$]$^{3+}$:} In this complex, cobalt is in the $+3$ oxidation state with an electronic configuration of [Ar]3$d^6$. Since ammonia (NH$_3$) is a weak field ligand, it causes a small splitting of d-orbitals in the octahedral field, leading to a spin paired (low-spin) configuration. Therefore, [Co(NH$_3$)$_6$]$^{3+}$ is a spin paired complex.
[CoF$_6$]$^{3-}$: In this complex, cobalt is also in the $+3$ oxidation state with an electronic configuration of [Ar]3$d^6$. Fluoride (F$^-$) is a weak field ligand, causing less splitting of d-orbitals in the octahedral field, resulting in a spin free (high-spin) complex configuration. Therefore, [CoF$_6$]$^{3-}$ is a spin free complex.
Conclusion: [Co(NH$_3$)$_6$]$^{3+}$ is a spin paired complex and [CoF$_6$]$^{3-}$ is a spin free complex.
Given below are two statements:
Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism
Statement II: cis- and trans-platin are heteroleptic complexes of Pd.
In the light of the above statements, choose the correct answer from the options given below
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: