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A complex is synthesised by combining hydrated solutions of $\ce{NiCl2}$ $(\pu{0.6 g})$ with $\pu{1.35 g}$ triphenylphosphine $(\ce{PPh3}).$ After cooling the product precipitates as a solid with formula $\ce{NiCl2(PPh3)_x}.$ Determine $x.$

I first found the molar ratios

$$n(\ce{NiCl2}) = \frac{\pu{0.6 g}}{\pu{129.59 g mol^-1}} \approx \pu{4.6 mmol}\tag{1}$$

$$n(\ce{PPh3}) = \frac{\pu{1.35 g}}{\pu{262.29 g mol^-1}} \approx \pu{5.1 mmol}\tag{2}$$

The smallest ratio these go into gives $x \approx 1,$ but I don’t think this is right.

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    $\begingroup$ Have you taken into consideration the possibility of having nickel chloride as a hydrate? x=2 for the hexahydrate. $\endgroup$
    – TheRelentlessNucleophile
    Nov 8, 2020 at 11:28
  • $\begingroup$ Oh gosh what a silly mistake to make, I didn’t take that in thank you! $\endgroup$
    – user85551
    Nov 8, 2020 at 11:48
  • $\begingroup$ Nothing looks silly here. Only one issue: the question contained contradictory information: trimethylphosphine $(\ce{PPh3}).$ I assume it was supposed to be triphenylphosphine judging by the molar mass you are using. $x = 2$ looks the most correct answer to me as it both justifies a typical C.N. = 4 for nickel, but still leaves enough room for two bulky triphenylphosphine ligands. $\endgroup$
    – andselisk ♦
    Apr 1, 2021 at 11:06

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