Question

the Ni(en)." complexes can be determined using the equation: n = x/(1-x) Under Results tabulate the maximum x-values at each wavelength, together with the calculated values of n and the formulae of the Ni(ll) complexes present in solution. If you do not obtain values of n=1, 2 and 3 from the given data, you need to provide a reasonable explanation, in the Conclusion of your report, why the complex with nex was not formed.

Answer 1

The n values corresponding to each x values for each wavelength have been given and the maximum absorbance values corresponding to each wavelength have also marked in the table below.

ABSORBANCE 530 nm 545 nm 575 nm 620 nm 640 nm x/(1-x) Abs Abs Abs Abs Abs 0.069 0.080 0.122 0.295 0.315 0.11 0.091 0.095 0.172 0.345 0.342 0.25 0.102 0.120 0.210 0.400 0.405 0.150 0.270 0.432 0.443 0.43 0.66 0.110 0.120 0.190 0.300 0.465 0.470 1.00 0.165 0.236 0.365 0.480 0.475 1.50 0.234 0.306 0.420 0.422 0.375 2.33 0.365 0.440 0.444 0.288 0.230 4.00 0.346 0.324 0.305 0.130 0.052 9.00 0.188 0.195 0.148 0.059 0.040 0.019 0.015 0.012 0.005 0.009

The required equation for the determination of the formula of the complex is given as  [Ni(H2O)6-2n(en)n]2+

According to the table, the Xmax value for the wavelengths 530 nm, 545 nm, and 545 nm is 0.7, and thus the corresponding n value is n = 2.33

and the corresponding formula of the complex is = [Ni(H2O)6-2x2.33(en)2.33]2+

= [Ni(H2O)1.34(en)2.33]2+

According to the table, the Xmax value for the wavelengths 620 nm and 640 nm is 0.5, and thus the corresponding n value is n = 1

and the corresponding formula of the complex is = [Ni(H2O)6-2x1(en)1]2+

= [Ni(H2O)4(en)]2+

4. The value of n = 1 is obtained because X=0.5 will give n = 1, but n = 2 and 3 are not obtained because of the value of X is taken only in one decimal place, the intermediate values like 0.11, 0.12 and so on is not taken into account. For obtaining n = 2, X value of 0.66 and for obtaining n = 3, X value of 0.75 was needed.