Chemistry Alevel Electronegativity
Why is Sulfur more electronegative than Beryllium however Nitrogen is more electronegative than Sulfur, using the explanation of shielding/nuclear charge/distance to outer electrons?
Hello chemhelp!
Sulphur has a higher electronegativity than beryllium due to its stronger nucleus charge and larger atomic radius, two factors that draw the outer electrons inwards. Nitrogen has a higher electronegativity than sulphur because its inner flled shells shield the positive charge of the nucleus, allowing it to attract electrons more effectively.
Krgds,
Sandro
Sulphur has a higher electronegativity than beryllium due to its stronger nucleus charge and larger atomic radius, two factors that draw the outer electrons inwards. Nitrogen has a higher electronegativity than sulphur because its inner flled shells shield the positive charge of the nucleus, allowing it to attract electrons more effectively.
Krgds,
Sandro
(edited 1 month ago)
Electronegativity can be defined as a relative measure of how well something is able to attract a shared pair of electrons in a covalent bond. The more electronegative the element is, the more likely it is to pull the electron pair closer to itself when forming a covalent bond.
As such, it is worth considering two factors - the (effective) nuclear charge and the atomic radius.
Larger (effective) nuclear charges generally result in higher electronegativities - the greater the charge experienced by the electrons, the harder the nucleus is pulling them towards it - compare the electronegativities along a period and you will find that this is generally true (for example along period 2, if you look at the electronegativities on the Pauling scale, Li is 0.98, Be is 1.57, B is 2.04, C is 2.55, N is 3.04, O is 3.44 and F is 3.98). Whilst not a concept discussed at A level, effective nuclear charges are worth knowing about - they are themselves dependent on proton number and how much shielding the outer electrons experience. Along a period, the proton number increases much more quickly than the extent of the shielding and so the effective nuclear charge increases.
Larger atomic radii typically result in a decrease in electronegativity - compare the electronegativities down a group and you will find that this is generally true (for example down group 7/17, if you look at the electronegativities on the Pauling scale, F is 3.98, Cl is 3.16, Br is 2.96, I is 2.66 and At is 2.20). The greater the distance between the nucleus and the electrons, the less influence the nucleus has on them.
It's also worth noting that for atoms of very similar atomic radii, the most important factor is effective nuclear charge / proton count and if atoms are very differently sized, the most important factor is the atomic radii. This can be rationalised using Coulomb's law (an A level physics topic, but one that occasionally sneaks its way into A level chemistry textbooks very briefly).
With the above two factors outlined, let's consider the three cases.
Although sulphur is a period lower down than beryllium, it has considerably more protons (16 vs 4) and it is only one period lower down, so the shielding doesn't increase enough to outweigh the effect of there being more protons. Interestingly, sulphur is also actually a (slightly) smaller atom than beryllium according to atomic radii data (100 pm vs 112 pm - this is again in part because the shielding isn't that much greater and there are far more protons able to attract the outer electrons). All things considered, sulphur must be more electronegative than beryllium.
Comparing sulphur to nitrogen is more tricky. Sulphur has considerably more protons than nitrogen (16 vs 7) and is one period lower down, but nitrogen is a considerably smaller atom than sulphur (65 pm vs 100 pm) and this makes all the difference. As such, nitrogen is the more electronegative of the two elements.
As such, it is worth considering two factors - the (effective) nuclear charge and the atomic radius.
Larger (effective) nuclear charges generally result in higher electronegativities - the greater the charge experienced by the electrons, the harder the nucleus is pulling them towards it - compare the electronegativities along a period and you will find that this is generally true (for example along period 2, if you look at the electronegativities on the Pauling scale, Li is 0.98, Be is 1.57, B is 2.04, C is 2.55, N is 3.04, O is 3.44 and F is 3.98). Whilst not a concept discussed at A level, effective nuclear charges are worth knowing about - they are themselves dependent on proton number and how much shielding the outer electrons experience. Along a period, the proton number increases much more quickly than the extent of the shielding and so the effective nuclear charge increases.
Larger atomic radii typically result in a decrease in electronegativity - compare the electronegativities down a group and you will find that this is generally true (for example down group 7/17, if you look at the electronegativities on the Pauling scale, F is 3.98, Cl is 3.16, Br is 2.96, I is 2.66 and At is 2.20). The greater the distance between the nucleus and the electrons, the less influence the nucleus has on them.
It's also worth noting that for atoms of very similar atomic radii, the most important factor is effective nuclear charge / proton count and if atoms are very differently sized, the most important factor is the atomic radii. This can be rationalised using Coulomb's law (an A level physics topic, but one that occasionally sneaks its way into A level chemistry textbooks very briefly).
With the above two factors outlined, let's consider the three cases.
Although sulphur is a period lower down than beryllium, it has considerably more protons (16 vs 4) and it is only one period lower down, so the shielding doesn't increase enough to outweigh the effect of there being more protons. Interestingly, sulphur is also actually a (slightly) smaller atom than beryllium according to atomic radii data (100 pm vs 112 pm - this is again in part because the shielding isn't that much greater and there are far more protons able to attract the outer electrons). All things considered, sulphur must be more electronegative than beryllium.
Comparing sulphur to nitrogen is more tricky. Sulphur has considerably more protons than nitrogen (16 vs 7) and is one period lower down, but nitrogen is a considerably smaller atom than sulphur (65 pm vs 100 pm) and this makes all the difference. As such, nitrogen is the more electronegative of the two elements.
(edited 1 month ago)
Original post by chemhelp101
Why is Sulfur more electronegative than Beryllium however Nitrogen is more electronegative than Sulfur, using the explanation of shielding/nuclear charge/distance to outer electrons?
As it was written before, there are two main things having an influence on the electronegativity: the nuclear charges and the atomic radius. Explanation in short:
- the more protons are in a nucleus, the stronger the charge to attract the electrons, that is to say the electronegativity decreases.
- the greater the atomic radius, the weaker the charge of these nuclear particles is to attract electrons, that is to say the electronegativity increases.
(edited 1 month ago)
Original post by Kallisto
As it was written before, there are two main things having an influence on the electronegativity: the nuclear charges and the atomic radius. Explanation in short:
- the more protons are in a nucleus, the stronger the charge to attract the electrons, that is to say the electronegativity decreases.
- the greater the radius of the nucleus, the weaker the charge of these nuclear particles is to attract electrons, that is to say the electronegativity increases.
- the more protons are in a nucleus, the stronger the charge to attract the electrons, that is to say the electronegativity decreases.
- the greater the radius of the nucleus, the weaker the charge of these nuclear particles is to attract electrons, that is to say the electronegativity increases.
Just realised my tired brain misread this.
Nuclear radius isn’t considered at A level. Yes, a larger nucleus generally has a greater pull on the outer electrons, but atomic radius (i.e the separation between the outer electrons and the nucleus) is the factor of interest. Larger atomic radii mean smaller electronegativities.
Original post by TypicalNerd
Just realised my tired brain misread this.
Nuclear radius isn’t considered at A level. Yes, a larger nucleus generally has a greater pull on the outer electrons, but atomic radius (i.e the separation between the outer electrons and the nucleus) is the factor of interest. Larger atomic radii mean smaller electronegativities.
Nuclear radius isn’t considered at A level. Yes, a larger nucleus generally has a greater pull on the outer electrons, but atomic radius (i.e the separation between the outer electrons and the nucleus) is the factor of interest. Larger atomic radii mean smaller electronegativities.
To be fair, I was taught in German education system, not in the British. I don't know the lessons for A levels in chemistry.
Isn't it what I said? if larger radii means smaller electronegativities, they are decreasing, aren't they?
Original post by Kallisto
To be fair, I was taught in German education system, not in the British. I don't know the lessons for A levels in chemistry.
Isn't it what I said? if larger radii means smaller electronegativities, they are decreasing, aren't they?
Isn't it what I said? if larger radii means smaller electronegativities, they are decreasing, aren't they?
You said “the greater the radius of the nucleus, the weakerthe charge of these nuclear particles is to attractelectrons, that is to say the electronegativityincreases.”, which isn’t quite the same thing.
The radius of the nucleus ≠ the atomic radius.
The atomic radius does the opposite to the nuclear radius - a larger atomic radius results in a decrease in electronegativity and this becomes especially important when one atom is much larger than another.
Original post by TypicalNerd
You said “the greater the radius of the nucleus, the weakerthe charge of these nuclear particles is to attractelectrons, that is to say the electronegativityincreases.”, which isn’t quite the same thing.
The radius of the nucleus ≠ the atomic radius.
The atomic radius does the opposite to the nuclear radius - a larger atomic radius results in a decrease in electronegativity and this becomes especially important when one atom is much larger than another.
The radius of the nucleus ≠ the atomic radius.
The atomic radius does the opposite to the nuclear radius - a larger atomic radius results in a decrease in electronegativity and this becomes especially important when one atom is much larger than another.
You are so right. So glad that you read my explanation again in order to edit the mistake by myself. If I can, I would give you a green thumb.
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