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**Hint:**If any complex has more unpaired electrons means it has a high spin, while less unpaired electron results in low spin. Inner orbital complexes contain the central atom using inner d orbital, while outer orbital complexes use outer d orbitals in the central metal.

**Complete step-by-step answer:**We have been given a complex ${{[Mn{{(CN)}_{6}}]}^{4-}}$, Mn has a configuration $M{{n}^{2+}}[Ar]3{{d}^{5}}$ , with CN being a ligand, this complex consist of unpaired electrons, hence it is paramagnetic, while it is an inner orbital complex with low spin due to the nature of the ligand CN. So, now we will take out its magnetic moment.

Magnetic moment of any complex is its behavior due to an unpaired electron under the influence of a magnetic field. It is known as a spin only magnetic moment as it is based on the spin of the unpaired electron. The formula for calculation of magnetic moment is, ${{\mu }_{s}}=\sqrt{n(n+2)BM}$,where ${{\mu }_{s}}$, is the spin magnetic moment, n is number of unpaired electron, BM is known as Bohr magneton.

Now, taking out the magnetic moment, we have n = 1 for this complex, so,

${{\mu }_{s}}=\sqrt{1(1+2)BM}$

${{\mu }_{s}}=\sqrt{3BM}$

${{\mu }_{s}}$= 1.73 BM

**So, the magnetic moment is 1.73 BM, hence, option C is correct for this complex.**

**Note:**

Outer orbital complexes have the hybridization of dsp order, while inner orbital complexes have the hybridization of spd order. The nature of ligands lead to paramagnetic or diamagnetic behavior in coordination complexes.

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