Chemical Bonding

anpanman SG

1 Sulfur reacts with fluorine to form fluorides with formulae SF2, SF4 and even SF6 which are all gases at r.t.p. However Mg reacts with Fluorine to form MgF2 which is a solid at r.t.p Explain why Mg only forms MgF2 while sulphur can form a variety of fluorides.

I'll start off by explaining that sulfur is a period 3 element so it can actually contain more than 8 electrons in its outer shell. This is due to the vancat and easily accessible d-orbitals present so Sulphur can form a variety of fluorides. As for MgF2, do I merely explain about its ionic characteristics?

I dont think this will suffice, anyone will provide me tips on how to work on this?

2. Give the dot and cross diagram of the compounds that can be formed between the 2 given elements. Describe thebonding that is formed between these elements.

Sodium and hydrogen - Not sure on how to explain the bonding that is formed. Can I explain through the no.of electrons transferred, etc?

oxygen and fluorine - between these 2 elements, we can form a number of compounds, so which one am i supposed to draw? hmm, and how do i know what are the compounds formed?

UltimaOnline SG

I dont think this will suffice, anyone will provide me tips on how to work on this?

2. Give the dot and cross diagram of the compounds that can be formed between the 2 given elements. Describe thebonding that is formed between these elements.

Sodium and hydrogen - Not sure on how to explain the bonding that is formed. Can I explain through the no.of electrons transferred, etc?

oxygen and fluorine - between these 2 elements, we can form a number of compounds, so which one am i supposed to draw? hmm, and how do i know what are the compounds formed?

Q1. You're correct regarding sulfur. For Mg, explain that the energectic feasibility of ionic compound formation is a balance between ionization energies and lattice energies. Mg+ wouldn't generate sufficient lattice energy. Mg3+ would require too much ionization energy. Hence, only the formation of MgF2 is energetically feasible.

Q2. Na+ H- sodium hydride is an ionic compound.

The simplest and most straightforward oxygen fluoride compound would be, naturally, oxygen difluoride, or OF₂, in which the valencies of O and F are 2 and 1 respectively.

The only other common oxygen fluoride compound would be dioxygen difluoride, O₂F₂., which is even less stable than OF₂.

The interesting aspect is the rarity of positive oxidation states of oxygen, as seen in these compounds.

anpanman SG

Thank you UltimaOnline.

I have another question.

1. Nitrogen andphosphurs are elements of group V in the Periodic Table. Nitrogen exists naturally as gaseous diatomic N - - - N (supposed to be 3 lines stacked on top of each other in indicate triple bond) whereas P is a solid and exists as P4 molecules comprising P-P single bonds

Suggest why P does not occur naturally as triple P-P bonds

2. Explain why it is not possible for Nitrogen to form an oxoanion with formula of NO4 3- (different question)

Thanks!

UltimaOnline SG

Thank you UltimaOnline.

I have another question.

Suggest why P does not occur naturally as triple P-P bonds

2. Explain why it is not possible for Nitrogen to form an oxoanion with formula of NO4 3- (different question)

Thanks!

You're welcome, anpanman.

Q1. Unlike the N atom with 2 electron shells, the P atom has 3 electron shells and hence the internuclear distance between P atoms is too large for pi bonds to be formed (ie. due to the greater internuclear distance and minimum bond length (remember pi bonds are by their nature shorter than sigma bonds) between P atoms compared to N atoms, there is a decreased ease of overlap of the p orbitals of phosphorus to form pi-bonds, compared to nitrogen; hence hybridization of s and p orbitals in phosphorus allow for the formation of 3 sp3-sp3 sigma bonds instead, in the P4 molecule). Hence only sigma bonds are formed in the tetrahedral P4 molecule of the white phosphorus allotrope.

Q2. For such a trinegative oxoanion, the N atom needs to expand its octet (ie. the N atom would have 5 bond pairs : 3 singly bonded O atoms with a -ve charge, and a doubly bonded O atom with no charge). Since N is in period 2, it means the N atom has no energetically accessible vacant orbitals to utilize to expand its octet, and hence cannot form the NO4 3- ion.

SBS n SMRT SG

clarify: Is this H2 chem or uni chem???

UltimaOnline SG

clarify: Is this H2 chem or uni chem???

Both. Some aspects discussed in this thread may be somewhat beyond the conventional limits of the H2 syllabus, but they may be regarded as 'good to know' and still helpful for the H2 exam candidate, particularly candidates who aspire to score a distinction.

My tuition students get the benefit of an 'integrated program' or 'H3' or 'advance-reconnaissance' approach into University level Chemistry as part of their H2 Chemistry education, which is helpful for students aspiring towards a distinction in H2 Chemistry, and/or students who intend to read Chemistry-based courses in the University (which is > 50% of all Science courses in the University).

Edited :

Added more elaboration and elucidation to my earlier post in this thread on the bonding in P4 molecule compared to the N2 molecule :

Unlike the N atom with 2 electron shells, the P atom has 3 electron shells and hence the internuclear distance between P atoms is too large for pi bonds to be formed (ie. due to the greater internuclear distance and minimum bond length (remember pi bonds are by their nature shorter than sigma bonds) between P atoms compared to N atoms, there is a decreased ease of overlap of the p orbitals of phosphorus to form pi-bonds, compared to nitrogen; hence hybridization of s and p orbitals in phosphorus allow for the formation of 3 sp3-sp3 sigma bonds instead, in the P4 molecule). Hence only sigma bonds are formed in the tetrahedral P4 molecule of the white phosphorus allotrope.

SBS n SMRT SG

Edited :

Added more elaboration and elucidation to my earlier post in this thread on the bonding in P4 molecule compared to the N2 molecule :

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