Consider the following complex ions, P, Q and R. The correct order...

$F^{-}$ and ${H}_{2} \text{O}$ are weak field ligands so pairing of electrons in metal cations 3d subshell will not take place.

now we have to cosider hybridization that is given in following diagram.


${μ}_{P} = \sqrt{n\left(n + 2\right)}n = 5 = \sqrt{5 \times 7} = \sqrt{35}\text{BM}$

${μ}_{Q} = \sqrt{n ( n + 2 )}n = 3 = \sqrt{3 \times 5} = \sqrt{15}\text{BM}$

${μ}_{R} = \sqrt{n ( n + 2 )}n = 4 = \sqrt{4 \times 6} = \sqrt{24}\text{BM}$
${μ}_{Q} < {μ}_{R} < {μ}_{P}$
$\text{Q < R < P}$