The Spin only magnetic moment value of square planar complex $$[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$$ is _____ B.M.
(Nearest integer)
(Given atomic number for Pt = 78)

$$Pt$$ has atomic number 78. Its electronic configuration is $$[Xe]4f^{14}5d^96s^1$$.

In the complex $$[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$$, the inner complex is $$[Pt(NH_3)_2Cl(NH_2CH_3)]^+$$, so Pt is in +2 oxidation state.

$$Pt^{2+}$$: $$[Xe]4f^{14}5d^8$$

In a square planar complex (which Pt(II) typically forms), the crystal field splits the d-orbitals such that all 8 electrons pair up in the 4 lower energy orbitals.

Number of unpaired electrons = 0.

Spin-only magnetic moment = $$\sqrt{0(0+2)} = 0$$ B.M.

The answer is $$\boxed{0}$$ B.M.