Question

Questions 1 and 2 please

Post-lab Questions (Show logic and calculations for full credit) Sodium trioxalatocobaltate(II) trihydrate is prepared by the following reactions: What is the percent yield of (Assume excess amounts of the other reagents.) Na, [Co(C,O) H,O if7.6 g are obtained from 12.5 g of [Co(H,O),JCl,? 2. Why are K,[Cr(C,O,)] 3HsO, K,[Cu(C,O) 3H,O, and K,[Fe(C,o) 3H,O colored, whereas [AI(C,0)J 3H2O is colorless? 3. Predict the magnetic susceptibilities (para- or diamagnetic) and determine the number of unpaired elec- trons of each of the following complexes (consider C,0, and H,O as strong field ligands which cause a large splitting of the d-orbitais).

Answer 1

1. From chemical reaction it is clear that

1 mol of [Co(H2O)6]Cl2 gives 1 mol of Na3 [Co(C2O4)3].H2O

*Also, 1 unit [Co(H2O)6]Cl2 has 1 Co and 1 unit formula of Na3[Co(C2O4)3].H2O has 1 Co.

So, number of moles of Na3[Co(C2O4)3].H2O produced is equal to the number of moles of [Co(H2O)6]Cl2 reacted.

Moles = given mass / molar mass

Moles of [Co(H2O)6]Cl2

= 12.5 g / 237.93 g mol^-1= 0.0525 mol

So, theoretical possible number of moles of Na3 [Co(C2O4)3]H2O =0.0525 mol

Theoretical yield = moles × molar mass

= 0.0525 mol × 428 g mol^-1 = 22.47 g

Now,

Percent yield = (actual yield / theoretical yield ) ×100

=(7.6 g /22.47 g) ×100 = 33.82 %

2.

The colour of the complex is due to presence of unpaired electron present in the central metal atom.

The number of unpaired electron can be calculated by using oxidation number of different species involved in the complex.

For complex of Al

3 × (+1) + x + 3×(-2) = 0

x = +3

Al - 3s2 3p1

Al(+3) - 3s03p0

No valence electron so, colourless.