'A' Level Chemistry Problem Analysis: Dipyridineiodine(I) Nitrate(V) And Its Reaction With Potassium Iodide

It is given that the pyridine molecule is the benzene ring, but with a nitrogen atom replacing a carbon atom, ie. C5N1H6.

1) Draw the displayed structural formula for dipyridineiodine(I) nitrate(V).

2) Explain the effect of adding KI to dipyridineiodine(I) nitrate(V).

Thought process:

Regarding the structure of dipyridineiodine(I) cation; first notice that this is a cation, so an electron has been lost from the species, and from the aromatic nature of pyridine, you would be able to correctly deduce that the electron has been lost from iodine.

Because nitrogen does not have empty d orbitals to expand its octet, (and) in order to preserve aromaticity, pyridine would act as a nucleophile attacking the I+ ion (because the lone pair on N is not delocalized into the benzene ring, it's a strong nucleophile and base). Hence the dipyridineiodine(I) species would see the iodine atom having two bond pairs, one with each pyridine.

So each N atom would have a postive formal charge (having donated its lone pair for dative bonding), and the iodine atom, having 3 lone pairs (recall that it has lost an electron) and 2 bond pairs, giving the iodine atom a negative formal charge.

Overall ionic charge = sum of formal charges = (+1) + (+1) + (-1) = +1 cationic charge.

Oxidation State (O.S.) of iodine = formal charge + electronegativity consideration = (-1) + (+1) + (+1) (because the iodine has 2 covalent bond pairs with more electronegative nitrogen atoms) = +1

Because the oxidation state of iodine in dipyridineiodine(I) cation is +1, hence the stock name dipyridineiodine(I).

Check your displayed structural formula for the dipyridineiodine(I) cation :

Each of the N atoms has

(i) a formal charge of +1 (4 valence electrons (close to nucleus) on a Grp V element; donated to form a dative bond in a nucleophilic attack)

(ii) an oxidation state of -3 (ie. (+1) + 4(-1); N is more electronegative compared to I and C)

(iii) a stable octet (4 bond pairs = 8 valence electrons (in total)).

The I atom has

(i) a formal charge of -1 (8 valence electrons (close to nucleus) on a Grp VII element; initially lost an electron but subsequently accepted 2 electrons in the form of 2 dative bonds)

(ii) an oxidation state of +1 (ie. (-1) + 2(+1) ; N is more electronegative compared to I)

(iii) expanded its octet (2 bond pairs + 3 lone pairs = 10 valence electrons (in total). Iodine is in period 5 and can expand its octet by accepting electrons into its vacant 5d orbitals / subshell).

To draw the nitrate(V) ion, NO3 - (do not confuse with the nitrate(III) ion, aka nitrite ion)

Central N atom is bonded to 3 O atoms. Since it possesses a uni-negative ionic charge, there would be at least 1 uni-negative formal charge, obviously on O.So we have 1 singly bonded -ve formal charged O atom (ie. 1 bond, 3 lone pairs). Assuming no other formal charges, the other two O atoms are doubly bonded (ie. 2 bond, 2 lone pairs). However, this would violate N's octet (since N doesn't have energetically accessible vacant 3d orbitals to expand its octet). Hence, one of the pi bonds is shifted to become a lone pair on one of the neutral O atoms.

So now we have 2 singly bonded O atoms, each with a uni-negative formal charge. And the N atom now has its stable octet and a uni-positive formal charge (ie. 4 bonds, 0 lone pairs).One of bonds from the N to O- atoms is a dative covalent bond, which explains the presence of two atoms with opposite formal charges (positive formal charge on the donor, negative formal charge on the recipient). The negative formal charge on the other singly bonded O atom is the result of the loss of a proton (from its conjugate acid, nitric(V) acid).

Explain the effect of adding KI to dipyridineiodine(I) nitrate (Hint: redox reaction)

The oxidation state of I in KI is -1, while the O.S. of I in dipyridineiodine(I) nitrate is +1. So the probable resulting redox reaction is quite easily deduced - molecular iodine aka diiodine (oxidation state = zero) is produced, along with pyridine (in its protonated conjugate acid form if an acidic protic solvent is used) and potassium nitrate(V) (as spectator ions).


The above content is contributed by Mr Heng, owner and 'A' Level Chemistry tutor at Bedok Funland JC. He also goes by the handle UltimaOnline on various online popular homework forums.


'A' Level Chemistry Problem Analysis: Similar Positive Iodoform Test Results For Alcohols And Carbonyl Compounds

'A' Level Chemistry Problem Analysis: Determining if Ink Smudges Are Ionic or Covalent In Nature

'A' Level Chemistry Problem Analysis: Solubility of Magnesium Chloride And Gibbs Free Energy