Dot and Cross diagrams

anpanman SG

Have problems drawing the dot and cross diagrams of HNO3,H2SO4 and HSO4-. Can someone please teach me how to do so? Anyway is the Dot and Cross diagram of I3- that of the left and right I having 8 electrons and the centre I having10 electrons?

UltimaOnline SG

About the drawing of structure formulae in Chemistry :

To be honest, the best (ie. fastest and most effective and clearest) way to learn the correct way of drawing structural formulae (eg. Kekule, Lewis, dot-&-cross, etc) is to attend my tuition (it's very difficult to teach certain aspects of Chemistry, such as drawing structures, over the internet). My tuition students can draw structures much better than their JC peers who do not have the benefit of my tuition, because this is one area (out of many) that JC teachers do not teach well at all. (As an ex-MOE teacher, I can empathize and sympathize, and don't blame the JC teachers, it's simply the fault of the system, with all its constraints and problems.)

Always draw Kekule structures first, then translate into dot-&-cross if the exam question requires one. Normal covalent bonds are a dot-cross, while dative bonds are a dot-dot or a cross-cross.

If you have difficulty drawing conjugate acids, simply draw the conjugate bases first, then add the protons one by one. For instance, draw the conjugate base sulfate(VI) ion, SO4 2-, then protonate once to obtain HSO4 -, and protonate again to obtain H2SO4.

Protonating an negatively (formal) charged O- atom, means to use one of its lone pairs to become a bond pair with the proton. In dot-cross-terminology, in the conjugate base one of the lone pairs is a dot-cross (while the other 2 lone pairs are dot-dot), and in the conjugate acid the bond pair is a dot-cross.

So to draw nitric(V) acid HNO3, first draw nitrate(V) ion NO3 -.

Begin by drawing an N aton in the center, singly bonded to 3 O atoms around it.

Since the ionic charge is uninegative, you expect a uninegative formal charge on one of the O atoms.

Which means that O atom with the uninegative formal charge has 3 lone pairs and 1 bond pair, while the other 2 O atoms have 2 lone pairs and 2 bond pairs.

But that violates N's octet (which is in Period 2 and cannot expand its octet).

Hence shift one pi bond to become a lone pair on the O atom.

This gives you 2 singly bonded O atoms with a uninegative formal charge, and an N atom with a unipositive formal charge (since it now has 4 bond pairs and no lone pairs, and it is in Group V).

Check that your uninegative ionic charge is the sum of formal charges (and it is also the sum of oxidation states, but we'll talk about that another time... during tuition).

How do you account for the separation of unipositive and uninegative formal charges? One of the N-O bonds must obviously be dative in nature.

But what about the other N-O bond where the O also has a uninegative formal charge? Is that also dative?

If it were also that, then the formal charge on the N atom would be dipositive, not unipositive. But since its formal charge is unipositive (a Group V atom having 4 bond pairs and 0 lone pairs), hence we deduce only one of the N-O bonds is dative.

Then how do you account for the remaining uninegative formal charged O- atom? Bear in mind that this entire structure is of a conjugate base, and therefore it follows that the uninegative formal charge on the remaining singly bonded O atom arose from the loss of a proton (ie. HNO3 ---> H+ + NO3 -).

The electron and molecular/ionic geometries of the NO3-/HNO3 species, are both trigonal planar (electron geometry = molecular/ionic geometry since there are no lone pairs on the central atom).

The sulfate(VI) ion, SO4 2- is simpler.

Begin by drawing an S atom in the center, followed by 4 O atoms around. Since the ionic charge is dinegative, you expect two of the O atoms to be uninegative and singly bonded, having 3 lone pairs and 1 bond pair. The remaining two doubly bonded O atoms thus have 2 lone pairs each, and no formal charge.

Notice that the S atom has expanded its octet (it's allowed to do so, being in Period 3, it has energetically accessible vacant 3d orbitals to use), and it has no formal charge (a Group VI atom having 6 bond pairs and 0 lone pairs), which is consistent with the fact that the ionic charge is dinegative.

As for the triiodide ion, notice that the I- nucleophile (having a uninegative formal or ionic charge, thus being electron rich) is capable of inducing (due to inter-electron repulsion) a temporary dipole in the non-polar I2 molecule, making it electrophilic. The nucleophile then attacks (ie. sends out a lone pair to become a dative bond pair) the partially positively charged (ie. delta +ve) I atom in the I2 molecule.

Hence in the triiodide ion, I3 -, the dative bond is from a terminal I atom to the central I atom, and the uninegative formal charge (for the uninegative ionic charge) rests on the central I atom which is the dative bond acceptor or recipient : a Group VII atom having 3 lone pairs and 2 bond pairs, has 8 valence electrons thus having a uninegative formal charge (note : in terms of an expanded octet it has 10 valence electrons).

The oxidation state (OS) of the uninegatively formal charged central I atom is (Oxidation State = formal charge + electronegativity considerations) -1 + 0 = -1. The other two (terminal) I atoms each has an OS of 0.

Ionic charge = sum of formal charges = 0 + -1 + 0 = 1-

Ionic charge = sum of OS = 0 + -1 + 0 = 1-

(Note :

In organic chemistry presentation, show a curved arrow from a lone pair on the I- nucleophile becoming a bond pair with the delta +ve I atom of the I2 molecule.

In physical/inorganic chemistry presentation, you can either show formal charges before, or after, dative bond formation. Some JC teachers and authors (including 'A' level study guide author CS Toh) favour the former, while other JC teachers and authors (including myself) favor the latter, which honestly makes more sense. Some JC teachers and authors even insist you show *the same* lone pair next to *the same* dative bond, which reveals a confusion between organic chemistry and physical/inorganic chemistry presentation : it is either a lone pair (ie. before) or a dative bond (ie. after), but you should *not* show it existing in two forms at the same time; unless you're intending to use a curved arrow, not a straight arrow, to represent the mechanism for the formation of its dative bond, which is generally used in organic chemistry, rather than the structure after the dative bond is formed, which is generally used in physical/inorganic chemistry).

Since the central I atom in the triiodide ion has 3 lone pairs and 2 bond pairs, hence it's electron geometry is trigonal bipyramidal, and consequently its ionic geometry is linear (unlike octahedral electron geometry, lone pairs in trigonal bipyramidal electron geometries preferentially occupy the equatorial, not axial, positions).

You'll probably want to google out these structures after you've drawn them, to double check. In addition to Google and Wikipedia (which provide lots more useful info on any chemical species other than just its structure), you can also check a species' structure using the Chemical Search Engine :

http://www.chemindustry.com/apps/chemicals?

I'll leave you with two awesomely exciting draw-the-structure challenge questions :

Without cheating and googling/wikipeding/searching out its structure on the internet, attempt to deduce and elucidate the structure (eg. Kekule, Lewis, dot-&-cross, etc) of :

i) the nitrogen dioxide molecule, and

ii) the dipyridineiodine(I) cation (eg. in dipyridineiodine(I) nitrate(V), of which I've earlier already shown you how to draw the nitrate(V) ion).

In case you didn't already know :

The pyridine molecule is a benzene ring in which one of the (C-H) of the ring has been replaced by an N: (with a lone pair). So the di in dipyridine means two such pyridines, connected to each other via the iodine (BenzeneN-I-NBenzene). Hence this is called the dipyridineiodine(I) ion.

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Solution for dipyridineiodine(I) cation :

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, 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 O.S. of iodine in dipyridineiodine(I) cation is +1, hence the stock name dipyridineiodine(I).

Check that in your structure of the dipyridine(I) ion, 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 empty 5d orbitals / subshell).

>>> 1) Explain the effect of adding KI to dipyridineiodine(I) nitrate. (Hint. redox) <<<

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

jayh272416 SG

i just got hit by a wall of text.