= Solution
The <chloride> complex has Pt(II), $d^8$, whereas the neutral bis(phosphine) complex has Pt(0), $d^{10}$ in the neutral-ethene assignment. \b[The lower-valent, phosphine-supported Pt(0) centre back-donates more strongly into <ethene> $\pi^*$], weakening and lengthening its internal C–C bond. This accounts for the longer $1.430\,\text{Å}$ distance compared with $1.375\,\text{Å}$ in the Pt(II) complex.
Both are formally 16-electron complexes: $8+3(2)+2=16$ for the <chloride> species and $10+2(2)+2=16$ for the phosphine species. Thus the difference is not simply a difference in total valence count. It reflects the energies and occupations of the <metal>–ligand <molecular orbitals>. The increased reduction of the <ethene> ligand gives the Pt(0) complex more <metallacyclopropane> character; the two descriptions are bonding limits, not distinct products inferred from the distances alone.
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