To analyze the properties of K$_3$[Co(CN)$_6$], let's break it down:
Oxidation state of cobalt: The overall charge on the complex is 0. Let $x$ represent the oxidation state of cobalt: \[ 3 + x - 6 = 0 \implies x = +3 \] So, Co is in the +3 oxidation state.
Electronic configuration of \(Co^{3+}\): Cobalt's ground state is [Ar] \(3d^7 4s^2\).
After losing 3 electrons, the configuration becomes \(3d^6\).
Ligand strength: CN$^-$ is a strong field ligand (SFL) as per the spectrochemical series. Strong field ligands cause significant splitting of the d-orbitals, leading to pairing of electrons in the lower energy orbitals.
Electron pairing: In the presence of CN$^-$, the 3d electrons pair as follows: \[ \text{Before pairing: } \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \] \[ \text{After pairing: } \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \] As all electrons are paired, the complex is diamagnetic.
Geometry: The coordination number is 6, and with CN$^-$ being a strong ligand, the geometry is octahedral.
Stability: CN$^-$ forms strong bonds with the metal center due to its strong field nature, making K$_3$[Co(CN)$_6$] the most stable complex among the options.
Thus, K$_3$[Co(CN)$_6$] is octahedral, diamagnetic, and the most stable.
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).

Given below are two statements:
Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with \( \text{CH}_3\text{COCl} / \text{AlCl}_3 \) will give the product shown. 
Statement II: \( -\text{NO}_2 \) group is a meta-directing and deactivating group.
In the light of the above statements, choose the most appropriate answer from the options given below.