[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.
Match the following List-I with List-II and choose the correct option: List-I (Compounds) | List-II (Shape and Hybridisation) (A) PF\(_{3}\) (I) Tetrahedral and sp\(^3\) (B) SF\(_{6}\) (III) Octahedral and sp\(^3\)d\(^2\) (C) Ni(CO)\(_{4}\) (I) Tetrahedral and sp\(^3\) (D) [PtCl\(_{4}\)]\(^{2-}\) (II) Square planar and dsp\(^2\)
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32