Considering that $∆$₀  >P , the magnetic moment of [Ru(H₂O)₆]^{2 +}would be _____ BM.

• In 4d and 5d  series in periodic table  all the metals form low spin complex with any ligands. Here ${\mathrm{H}}_2\mathrm{O}$ is act as strong field ligand Ru ²⁺ = 3d⁶(∆⁰ > P) hence electronic configuration is ${\mathrm{t}}_{2\mathrm{g}}^6{\mathrm{e}}_{\mathrm{g}}^0$, so there is no unpaired electron in this complex. Therefore spin only magnetic moment, $\mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)}$  (n= number of unpaired electron)here n=0

$\mathrm{\mu }=\sqrt{0(0+2)}$; $\mathrm{\mu }=0$

Hence, correct option is: (A) 0