Why can't nitrogen form NCl5?

As usual, pisgachemist is right. Textbooks have been pushing the "d orbitals on heavier main-group atoms" argument for way too long. Accurate quantum chemical computations have shown that 3d participation is modest for molecules like PCl₅, for example. (There is some contribution, and more when highly electronegative ligands are present, but it isn't very important. Hence, questions like these, in which a nitrogen/phosphorus comparison is implicit, lead students astray.) The crowding of chlorine atoms around the smaller nitrogen atom is a more satisfying explanation for the nonexistence of NCl₅.

Pisgachemist's only mistake is a typo: I'm sure he meant Mg₃N₂, not Mg₃N₃, in his second paragraph.

The only difference I have with pisgachemist is a very small quibble: It isn't technically correct to say the is NO 3d orbital participation in the bonding for PCl₅. The 3d orbitals have the right symmetry to mix with the chlorine donor orbitals and if the 3d orbitals are included in the basis of a quantum mechanical calculation, they will make "some" contribution to the wavefunction. It just isn't very significant. I'm in complete agreement that textbooks should eliminate sp³d and sp³d² hybridization stuff from qualitative bonding descriptions. There is enough to teach without it, and in higher level upper division courses and graduate courses, I end up "unteaching" that stuff anyway!

there's no longer sufficient room interior the nitrogen's digital configuration for the ten electrons - nitrogen in basic terms has 2s and 2p (the variety is the row on the periodic table, no longer what number there are of them) orbitals obtainable. the 1st can carry 2 electrons and the 2d can carry six. this is why nitrogen can variety a quaternary ammonium salt (i.e. it could make NCl4 +), for the reason that each bond involves 2 electrons. Phosphorus is different - no longer in basic terms does it have 3s (word the variety substitute 'cos we've moved down the table?) and 3p orbitals obtainable, it additionally has the three-D orbitals no longer too intense up in power, so those could be used to settle for electrons. you're able to no longer have heard of the d orbitals yet simply by fact they are many times taught as component of transition metallic lecture classes at uni. Regardless, those can settle for electrons from a miles better form of donors, meaning PCl5 is available. It additionally facilitates that phosphorus atoms are greater, so greater atoms can in fantastic condition around it, and that that is 'softer', so it may variety ions greater actual. this is crucial simply by fact it always exists as a mixture of (PCl4 +)(Cl -) ions - evaluate this with the (NCl4 +) difficulty that nitrogen can do. in actuality: nitrogen's too small and would not have the greater 'electron room' that phosphorus has. wish this is clean adequate :)

there is not any longer sufficient room in the nitrogen's digital configuration for the ten electrons - nitrogen in basic terms has 2s and 2p (the quantity is the row on the periodic table, no longer what share there are of them) orbitals accessible. the 1st can carry 2 electrons and the 2d can carry six. for this reason nitrogen can type a quaternary ammonium salt (i.e. it could make NCl4 +), considering the fact that each and each bond includes 2 electrons. Phosphorus is distinctive - no longer in basic terms does it have 3s (word the quantity exchange 'cos we've moved down the table?) and 3p orbitals accessible, it additionally has the three-D orbitals no longer too extreme up in power, so those would be utilized to settle for electrons. you will desire to no longer have heard of the d orbitals yet because of fact they're usually taught as area of transition steel lecture classes at uni. Regardless, those can settle for electrons from a greater robust style of donors, meaning PCl5 is a hazard. It additionally enables that phosphorus atoms are greater advantageous, so greater atoms can greater wholesome around it, and that it incredibly is 'softer', so it may type ions greater actual. it incredibly is significant because of fact it frequently exists as a mix of (PCl4 +)(Cl -) ions - evaluate this with the (NCl4 +) difficulty that nitrogen can do. in actuality: nitrogen's too small and would not have the greater advantageous 'electron room' that phosphorus has. wish it incredibly is obvious sufficient :)

NCl5 doesn't form because the chlorine atoms are too large to fit five of them around a tiny nitrogen atom. There is some speculation that NF5 should be possible, but it hasn't been synthesized yet. Fluorine atoms are much smaller than chlorine and could be possible to fit five of them around a nitrogen atom.

None of your choices, including #1 are particularly satisfying. Nitrogen certainly can have an oxidation state of +5. In N2O5 and the nitrate ion, NO3^-, the oxidation state of nitrogen is +5, and there are many more. Nitrogen is pretty nonreactive at room temperature, but under conditions of increased temperature and pressure, nitrogen becomes much more reactive. You have (or will do) the experiment where Mg is burned in air to make MgO. Some of the white powder that is produced is Mg3N3, magnesium nitride. At the elevated temperature of burning magnesium, nitrogen reacts. Nitrogen is also quite reactive at the temperatures and pressures inside a car engine. Several oxides of nitrogen are produced during gasoline combustion.

Consider PCl5 vs NCl5. PCl5 forms because phosphorous can be "hypervalent", that is, it can have more than 8 electrons around it. Elements in period three and up will often exhibit hypervalency, while elements in period 2 rarely do. I suggest to my students that given the choice of following the octet rule or being hypervalent, period 2 elements will generally follow the octet rule.

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Kenny says, "You can "easily" strip five electrons off of P to make a P+5." Nonsense! Phosphorous doesn't lose any electrons in the formation of PCl5. The bonds in PCl5 are quite covalent. There are NO P^5+ ions being formed. Besides, there is nothing "easy" about stripping 5 electrons from an atom. Look up the 5th ionization energy for phosphorous. You will see that it is quite high.

Ryan says, "For phosphorus to form PCl5 it takes on the hybridization to form sp3d hybrid orbitals resulting in the possibility to form 5 bonds." Nope! Textbooks are just catching up to what chemists have known for some time, that there is NO d-orbital involvement in hypervalent bonding. Instead, PCl5 can quite easily be explained by 3-center, 4-electron bonding. Read more about 3C4E bonding here: http://en.wikipedia.org/wiki/Three-center_four-ele... and here http://www.raleighcharterhs.org/faculty/egrunden/A...

And don't go with Ryan's or Kenny's suggestion that #2 is the best answer. Clearly, N2O5 and the ubiquitous nitrate ion, NO3^-, demonstrate that nitrogen can exist in the +5 oxidation state.

In the olden days it was said that PCl5 could form and NCl5 couldn't because nitrogen had no empty d-orbitals (no such thing as a 2d orbital) with which to form hybrid orbitals. That argument is no longer valid since we now know that NO non-metallic elements us the d-orbital for hypervalent bonding. Therefore, you should NEVER use sp3d or sp3d2 to "explain" the bonding for trigonal bipyramidal and octahedral geometries.

And while we are at it, nitrogen isn't the only group 15 element not to form XCl5. Arsenic forms AsCl3, but AsCl5 and antimony(V) chloride, SbCl5 is only stable below -50C, and BiCl5 has never been observed. Even PCl5, the one which so often comes up in discussion of trigonal bipyramidal geometry is unstable and reacts violently with water.

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Magnesium nitride is indeed Mg3N2.

It is much simple. The only point of difference between P and N regarding your question is that P contains vacant d-orbitals whereas N having only two energy levels doesn't have vacant d-orbitals(for principal quantum number n=2, azimuthal quantum number l = 0,1 , i.e.,only the s and p orbitals.

Under excitation the valence electrons of P can jump to the vacant d-orbitals to form the oxidation state +5 and then form PCl5 by sp3d hybridization.[PCl5 is a case of expanded octet]

However, no vacant d-orbitals in N indicates that there is no scope of expanded octet nor the formation of NCl5

in order to explain the answer to your question you must be familiar with the quantum mechanical model of the atom (where electrons can be found in orbitals: s,p,d and f). For phosphorus to form PCl5 it takes on the hybridization to form sp3d hybrid orbitals resulting in the possibility to form 5 bonds. however this hybridization can only happen as phosphorus has the third energy level which has the d sub shell, but nitrogen cannot form sp3d orbitals as it only has up to the second energy level which does not have the d sub shell; therefore the formation of sp3d hybrid orbitals is impossible.

p.s. the correct answer is therefore no. 2 as no. 1 is an overly simplistic answer and no. 3 is just plain wrong, although with difficulty (meaning requiring high pressures and temperatures) nitrogen does react.