A transition metal '$$M$$' among $$Sc, Ti, V, Cr, Mn$$ and $$Fe$$ has the highest second ionisation enthalpy. The spin-only magnetic moment value of $$M^+$$ ion is ______ BM (Near integer) (Given atomic number $$Sc : 21, Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26$$)

Among Sc, Ti, V, Cr, Mn, Fe, the one with highest second ionization enthalpy is Cr.

Cr: [Ar]3d⁵4s¹. After removing first electron (4s¹): Cr⁺ = [Ar]3d⁵ (half-filled, very stable).

Second IE involves removing from the stable half-filled 3d⁵ configuration, which requires very high energy.

Cr⁺ = [Ar]3d⁵: number of unpaired electrons = 5.

Spin-only magnetic moment = $$\sqrt{n(n+2)} = \sqrt{5 \times 7} = \sqrt{35} = 5.92 \approx 6$$ BM.

The answer is $$\boxed{6}$$ BM.