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.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.