# Determining the number of triphenylphosphine ligands in nickel(II) complex

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.

• Have you taken into consideration the possibility of having nickel chloride as a hydrate? x=2 for the hexahydrate. Nov 8 '20 at 11:28
• Oh gosh what a silly mistake to make, I didn’t take that in thank you! Nov 8 '20 at 11:48
• 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. Apr 1 '21 at 11:06