The compound $\mathrm{CoCl}{}_3\cdot 5\mathrm{NH}{}_3$ shows molar conductivity of a 1:2 electrolyte. Excess $\mathrm{AgNO}{}_3$ is added to one mole of it. Treating $\mathrm{NH}{}_3$ as strong-field towards $\mathrm{Co}{}^{3+}$ (low spin), which option correctly gives the formula, moles of AgCl precipitated and the electronic details?

1. A$[\mathrm{Co}(\mathrm{NH}{}_3){}_5\mathrm{Cl}]\mathrm{Cl}{}_2$; 2 mol AgCl; 4 unpaired electrons, $s{p}^3{d}^2$
2. B$[\mathrm{Co}(\mathrm{NH}{}_3){}_4\mathrm{Cl}{}_2]\mathrm{Cl}$ with one free $\mathrm{NH}{}_3$; 1 mol AgCl; 0 unpaired electrons, $d^{2}s{p}^3$
3. C$[\mathrm{Co}(\mathrm{NH}{}_3){}_5\mathrm{Cl}]\mathrm{Cl}{}_2$; 3 mol AgCl; 0 unpaired electrons, $d^{2}s{p}^3$
4. D$[\mathrm{Co}(\mathrm{NH}{}_3){}_5\mathrm{Cl}]\mathrm{Cl}{}_2$; 2 mol AgCl; 0 unpaired electrons, $d^{2}s{p}^3$