Question

Trimethylphosphine, \(P{\left( {C{H_3}} \right)_3}\) can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel \(\left( {II} \right)\) chloride in acetone, a blue compound that has a molecular mass of approximately \(270 g\) and contains \(21.5\% Ni,26.0\% Cl,\)and \(52.5\% P{\left( {C{H_3}} \right)_3}\) can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?

Short answer

• First let us calculate the number of moles for each compound using the simple formula: \(n = \frac{m}{M}\).

Step-by-step solution

of 2: Calculation

1) Number of moles for\(Ni\):\(\begin{aligned}{}n(Ni) & = \frac{{m(Ni)}}{{M(Ni)}}\\n(Ni) &= \frac{{21.5\;{\rm{g}}}}{{58.7{\rm{gmo}}{{\rm{l}}^{ - 1}}}}\\n(Ni) &= 0.366\;{\rm{mol}}\end{aligned}\)

2) Number of moles for\(Cl\):\(\begin{aligned}{}n(Cl) &= \frac{{m(Cl)}}{{M(Cl)}}\\n(Cl) &= \frac{{26.0\;{\rm{g}}}}{{35.45{\rm{gmo}}{{\rm{l}}^{ - 1}}}}\\n(Cl) &= 0.733\;{\rm{mol}}\end{aligned}\)

3) Number of moles for\({\rm{P}}{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}\):

\(\begin{aligned}{}n\left( {P{{\left( {C{H_3}} \right)}_3}} \right) = \frac{{m\left( {P{{\left( {C{H_3}} \right)}_3}} \right)}}{{M\left( {P{{\left( {C{H_3}} \right)}_3}} \right)}}\\n\left( {P{{\left( {C{H_3}} \right)}_3}} \right) = \frac{{52.5\;{\rm{g}}}}{{76.078g_{mol}^{ - 1}}}\\n\left( {P{{\left( {C{H_3}} \right)}_3}} \right) = 0.690\;{\rm{mol}}\end{aligned}\)

of 2:Explanation

• Now, let us calculate the ratio of number of moles to get the empirical formula.
• We need to divide each mole value by the smallest number of moles calculated and then round off to the nearest whole number.
• From all \({\bf{3}}\)numbers the smallest is \({\bf{0}}.{\bf{366}},\)so we get:

\(\begin{aligned}{}N(Ni) &= \frac{{0.366}}{{0.366}} &= 1\\N(Cl) &= \frac{{0.733}}{{0.366}} &= 2\\N\left( {P{{\left( {C{H_3}} \right)}_3}} \right) &= \frac{{0.690}}{{0.366}} &= 1.89 &= 2\end{aligned}\)

• It means the empirical formula for the complex is: \({\rm{NiCl}}{l_2}{\left( {{\rm{P}}{{\left( {{\rm{C}}{{\rm{H}}_3}} \right)}_3}} \right)_2}\)
• Now we need to determine the geometry of the molecule. We can see that we have a metal in this compound which is \({\bf{Ni}}\)and that means it is in the center of our molecule.
• As the metal are attached \(2{\rm{Cl}}\)ions and \(2{\rm{P}}{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}\)it means this complex has \({\bf{4}}\)as the coordination number.
• There are only \({\bf{2}}\)structures present if we have a coordination number of\({\bf{4}}\): square planar and tetrahedral
• The main difference between these two structures is that square planar structure has \({\bf{2}}\)isomers: cis and trans, and in tetrahedral geometry there are no isomers.
• Hence, this compound does not have any isomeric forms, which means its geometry is tetrahedral.

Result

The molecular formula of the blue compound is \({\rm{NiCl}}{l_2}{\left( {{\rm{P}}{{\left( {{\rm{C}}{{\rm{H}}_3}} \right)}_3}} \right)_2}\)and it has tetrahedral geometry.