Bond angles

bonkysleuth SG

Hello. Have a question on bond angles of H2S and SO2 in which I was asked to determine whether bond angle of H2S < that of SO2.

I found that H2S has 2 bonding pair of e- and 2 lone pair of e- while SO2 has 2 bond pair of e- and 1 lone pair of 2-. Which gives H2S a bent shape of 105deg and SO2 also a bent shape of 118deg. Thus H2S has smaller bond angle than SO2.

Just verifying, the answer my tutor has provided didnt indicate that this statement is true. Thank you!

UltimaOnline SG

Hello. Have a question on bond angles of H2S and SO2 in which I was asked to determine whether bond angle of H2S < that of SO2.

Just verifying, the answer my tutor has provided didnt indicate that this statement is true. Thank you!

SO2 has electron geometry of trigonal planar, molecular geometry of bent / v-shape. The S atom has 3 electron charge clouds : 1 lone pair, 2 double bonds. Bond angle approx 120 deg, modified to slightly less (eg. 118 deg) due to greater repulsion between lone pair - bond pair, compared to bond pair - bond pair repulsion (this is based on complete sp2 hybridization as taught in H1/H2/H3 'A' level Chemistry; in reality extent of hybridization is incomplete for some compounds of period 3 elements, due to less repulsion from a larger atomic radius; in SO2 the bond angle is actually 119 deg, so indeed sigma bonds have significant sp2-sp2 character).

H2S has electron geometry tetrahedral, molecular geometry of bent / v-shape. The S atom has 4 electron charge clouds : 2 lone pairs, 2 single bonds. Bond angle approx 109.5 deg, modified to somewhat less (eg. 105 deg) due to greater repulsion between lone pairs, compared to lone pair - bond pair repulsion, compared to bond pair - bond pair repulsion (this is based on complete sp3 hybridization as taught in H1/H2/H3 'A' level Chemistry; in reality extent of hybridization is minimal for some compounds of period 3 elements, due to minimal repulsion from a larger atomic radius; in H2S the bond angle is 92.1 deg, sigma bonds have significant s-p character, rather than s-sp3).

At H1/H2/H3 'A' levels, the actual Cambridge paper will presumably (though Cambride isn't always so reliable) set the question such that whether you use your limited (and not completely correct) 'A' level knowledge, or you use higher level (more correct) knowledge, either way you should get the same correct answer (eg. if MCQ) or still score the marks (eg. if P2 or P3 question).

bonkysleuth SG

Thanks for explaining so clearly.

I have 2 more questions on this topic which I am unsure of.

1. The N2O5 molecule is present in the vapour state, but the structure is ionic in the solid state, existing as NO2+ NO3-.

Later on, I was asked to draw the displayed formula of NO2+ and NO3-, which was done easily.

Then the part I am unsure of is: Draw a possible displayed formula for the N2O5 molecule and suggest values for the bond angles in the molecule.

I realized I could not really make "complete" use of the 2 displayed formula I was asked to draw earlier on because the answer showed merely TWO of the NO3- oxoanion, repetitively. Why is NO2+ not included in helping me draw this N2O5 molecule? What are the steps I should take in deriving the displayed formula for N2O5?

2. Nitrogen and phosphurous are elements of Group V in the Periodic Table.

Nitrogen exists naturally as gaseous diatomic molecules, with triple bonds, whereas phosphorus is a solid and exists as P4 molecules comprising of P-P single bonds.

SUggest why phosphorus does not occur naturally as triple bonds.

My sch tutor said it is because P is a relatively big atom with diffused orbitals, so side-on overlap of p-orbitals to form pi bonds is much less effective than the head-on overlap to form sigma bond.

Well, dont really get why a big atom will be less effective in having side-on overlap...

UltimaOnline SG

Thanks for explaining so clearly.

I have 2 more questions on this topic which I am unsure of.

1. The N2O5 molecule is present in the vapour state, but the structure is ionic in the solid state, existing as NO2+ NO3-.

Later on, I was asked to draw the displayed formula of NO2+ and NO3-, which was done easily.

Then the part I am unsure of is: Draw a possible displayed formula for the N2O5 molecule and suggest values for the bond angles in the molecule.

2. Nitrogen and phosphurous are elements of Group V in the Periodic Table.

Nitrogen exists naturally as gaseous diatomic molecules, with triple bonds, whereas phosphorus is a solid and exists as P4 molecules comprising of P-P single bonds.

SUggest why phosphorus does not occur naturally as triple bonds.

Well, dont really get why a big atom will be less effective in having side-on overlap...

NO2+ has a unipositive formal charge on the central N atom.

NO3- has a unipositive formal charge on the central N atom, and 2 uninegative formal charges on the 2 singly bonded O atoms. There are a total of 3 formal charges, and one dative bond.

NO2 (nitrogen dioxide) has an N atom with a unipositive formal charge as well as an unpaired electron on the central N atom, doubly bonded to an O atom with no formal charge, and singly bonded to a O atom with a uninegative formal charge. Hence NO2 is a free radical, with 2 formal charges, and one dative bond.

N2O4 is simply a dimer of two NO2 molecules joined together using their unpaired electrons. Hence N2O4 is a non-radical, with 4 formal charges, and two dative bonds.

N2O5 is similar to N2O4, except that instead of directly joining two NO2 free radical molecules together using their unpaired electrons, you join the two NO2 free radical molecules together using an O atom in the center, using the unpaired electrons of the two NO2 molecules. N2O5 has a total of 4 formal charges, with two dative bonds.

The bond geometries in N2O5 are hence :

V-shaped about the O atom (ie. 105 deg).

trigonal planar about the 2 N atoms (ie. 120 deg).

It is difficult and unfeasible to attempt to give a comprehensive, adequate explanation of the process of elucidating structural formulae via this forum. Tuition is ideal for this purpose (not trying to hard sell my tuition services, just being frank and honest about the limitations of online forum discussions).

2. Yes, your JC teacher is pretty much right about the reason. But I'll elaborate a little here, to help you understand a little better.

Note that the inter-nuclei or inter-atomic distance is significantly larger between two P atoms, compared to two N atoms. Remember that single bonds are longer (sigma bonds are end-on or head-on overlap of any orbitals, usually hybridized), and double or triple bonds are much shorter (a sigma bond and one or two pi bonds, which are formed by the side-on overlap of unhybridized p orbitals).

Hence, P atoms find it easier (ie. feasibility under standard conditions) to form single bonds with each other, while N atoms find it easier (ie. feasibility under standard conditions) to form triple bonds with each other.

The specific reason for the greater ease for nitrogen, and a decreased ease for phosphorus, to achieve p electron sideways orbital overlap to form pi bonds, can be traced to their electron configuration :

Phosphorus, being in period 3, will have to utilize its significantly more diffuse 3p electron orbitals to form much weaker side-on overlaps, which means the consequent attempted triple bond will be so weak, that it is unable to stabilize the P atoms and molecule well.

Nitrogen, being in period 2, is able to utilize its significantly more concentrated 2p electron orbitals to form much stronger side-on overlaps, which means the consequent triple bond formed will be so strong, that it is able to stabilize the N atoms and molecule well.

And remember that in Chemistry, whichever possible path achieves greater stability, will be the more feasible one. (Under standard conditions) A triply bonded N2 molecule would be more stable than a singly bonded N4 molecule, and is hence favoured. (Under standard conditions) A singly bonded P4 molecule would be more stable than a triply bonded P2 molecule, and is hence favoured. (There are actually many allotropes of phosphorus, but P4 is the most common.)