Based on the crystal-field strengths f- < ch2cn < nh3 < no2- < cn- which co(iii) complex is most likely high-spin?
a. [co(nh3)6]3+
b. [co(no2)6]3-
c. [co(cn)6]3-
d. [cof6]3-
e. [co(ch2cn)6]3+
f. [cof5(nh3)]
g. none of the above
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D. [CoF₆]³⁻

Explanation:

In the weak field octahedral complexes of 3d series, the value of the ∆₀ is very small as there will be no pairing of d- electrons. These type of complexes have the maximum number of the unpairing electrons and are called high spin or spin free complexes.

As stated above, The complex with weak field ligands corresponds to high spin complex. The trend for the crystal field strength is given in the question. So, F⁻ corresponds to the least weak field ligand and thus, D is the most likely high spin complex.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.