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    The electron arrangement of Manganese is [Ar] 3d54s2. I understand that when Manganese is ionised to form Mn+, the first electron is removed from the 4s orbital because, despite the fact that it's at a lower energy than the 3d orbital, the 3d54s1 arrangement is more stable because of electron repulsion. However, when another electron is removed to form Mn2+, according to my textbook, it is also removed from the 4s shell to form [Ar] 3d5.

    I do not understand why this is the most stable electron configuration. Vanadium has the exact same number of electrons as Mn2+ yet its electron configuration is [Ar] 3d34s2. Vanadium has this electron configuration so it must be the most stable configuration, so why does Mn2+ have a completely different arrangement despite having the same number of electrons? I know that Mn2+ has a stronger nuclear charge than Vanadium but I don't see why that would somehow make it easier to remove electrons from a lower energy level. Why are electrons being removed from the 4s shell by 'default' given that it has a lower energy level than the 3d shell? Any help would be greatly appreciated, I'm very confused!
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    (Original post by Chlorophile)
    The electron arrangement of Manganese is [Ar] 3d54s2. I understand that when Manganese is ionised to form Mn+, the first electron is removed from the 4s orbital because, despite the fact that it's at a lower energy than the 3d orbital, the 3d54s1 arrangement is more stable because of electron repulsion. However, when another electron is removed to form Mn2+, according to my textbook, it is also removed from the 4s shell to form [Ar] 3d5.

    I do not understand why this is the most stable electron configuration. Vanadium has the exact same number of electrons as Mn2+ yet its electron configuration is [Ar] 3d34s2. Vanadium has this electron configuration so it must be the most stable configuration, so why does Mn2+ have a completely different arrangement despite having the same number of electrons? I know that Mn2+ has a stronger nuclear charge than Vanadium but I don't see why that would somehow make it easier to remove electrons from a lower energy level. Why are electrons being removed from the 4s shell by 'default' given that it has a lower energy level than the 3d shell? Any help would be greatly appreciated, I'm very confused!
    I am not sure how to answer your question, but I can help in one point.

    You know electrons fill the orbitals having lower energy levels first.

    However, the 4s subshell after filling will have a slightly higher energy than the 3d subshell and therefore it loses the electron first.
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    (Original post by Daniel Atieh)
    I am not sure how to answer your question, but I can help in one point.

    You know electrons fill the orbitals having lower energy levels first.

    However, the 4s subshell after filling will have a slightly higher energy than the 3d subshell and therefore it loses the electron first.
    Thanks, and I understand that because of the electron repulsion, but after the first electron has been removed wouldn't the energy be lower than the 3d shell so shouldn't the second electron be removed from there? And regardless, why doesn't the electron arrangement change after it's removed to go to a more stable arrangement?
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    (Original post by Chlorophile)
    Thanks, and I understand that because of the electron repulsion, but after the first electron has been removed wouldn't the energy be lower than the 3d shell so shouldn't the second electron be removed from there? And regardless, why doesn't the electron arrangement change after it's removed to go to a more stable arrangement?
    Now I think it's the other way round. Perhaps when the 3d subshell is filled, the 4s energy becomes slightly higher. So even when the 4s loses electrons, it still have a higher energy as long the 3d is filled.

    You're welcome.
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    (Original post by Daniel Atieh)
    Now I think it's the other way round. Perhaps when the 3d subshell is filled, the 4s energy becomes slightly higher. So even when the 4s loses electrons, it still have a higher energy as long the 3d is filled.

    You're welcome.
    But why would the 4s energy increase when the 3d subshell is filled?
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    (Original post by Chlorophile)
    But why would the 4s energy increase when the 3d subshell is filled?
    Probably because of the slight shielding from the nucleus provided by the 3d when it is filled. I am not sure at all though.
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    (Original post by Daniel Atieh)
    Probably because of the slight shielding from the nucleus provided by the 3d when it is filled. I am not sure at all though.
    That sort of makes sense and could explain why the electron is removed from the 4s subshell but it still doesn't explain why the electron arrangement doesn't then flip and convert to Vanadium's arrangement?
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    The answer is really fiddly.

    The energy of a 4s orbital in atomic Vanadium is higher than a 3d. The 4s orbital is only actually lower in energy for potassium and magnesium (see this), after that while their energies remain really similar (which is important), the 3d is lower in energy.

    This holds true on ionisation - the highest energy, 4s electrons are removed first, therefore Mn becomes [Ar]3d5. 4s electrons sit predominantly very far away form the nucleus, with the 3d electrons in between them. The 3d screens the 4s, making 4s higher in energy.

    So then I guess the question is why Vanadium keeps its electrons in the 4s orbital. It's because adding them to the 3d shell increases electron electron repulsions enough to make [Ar]3d5 higher in energy than [Ar]3d54s2. There's been a reduction in nuclear charge of 2, and it turns out that's enough to tip the balance, and electron electron repulsions win out - the increased nuclear charge that the 3d sees isn't enough to make 3d5 in vanadium lower in energy, but it is for Mn2+
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    (Original post by KombatWombat)
    The answer is really fiddly.

    The energy of a 4s orbital in atomic Vanadium is higher than a 3d. The 4s orbital is only actually lower in energy for potassium and magnesium (see this), after that while their energies remain really similar (which is important), the 3d is lower in energy.

    This holds true on ionisation - the highest energy, 4s electrons are removed first, therefore Mn becomes [Ar]3d5. 4s electrons sit predominantly very far away form the nucleus, with the 3d electrons in between them. The 3d screens the 4s, making 4s higher in energy.

    So then I guess the question is why Vanadium keeps its electrons in the 4s orbital. It's because adding them to the 3d shell increases electron electron repulsions enough to make [Ar]3d5 higher in energy than [Ar]3d54s2. There's been a reduction in nuclear charge of 2, and it turns out that's enough to tip the balance, and electron electron repulsions win out - the increased nuclear charge that the 3d sees isn't enough to make 3d5 in vanadium lower in energy, but it is for Mn2+
    Thanks a lot for your very thorough answer! One more question though. You wrote that "adding them to the 3d shell increases electron electron repulsions". Why is that, since they're being added to different orbitals within the 3d subshell and I thought repulsion only happened when electrons share the same orbital? There would definitely be repulsion in the 4s orbital when you have two electrons in it, so why is this more stable than the lack of repulsion within 3d? Isn't the lack of repulsion in the 3d shell the whole reason why the structure of Chromium is [Ar] 3d54s1 rather than [Ar] 3d44s2?

    Also, why do all the school textbooks say that the energy level of 3d is higher than 4s when according to that graph that's incorrect for practically all elements?

    Edit: Don't worry, I've found out the answer. Thanks again
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    (Original post by KombatWombat)
    The answer is really fiddly.

    The energy of a 4s orbital in atomic Vanadium is higher than a 3d. The 4s orbital is only actually lower in energy for potassium and magnesium (see this), after that while their energies remain really similar (which is important), the 3d is lower in energy.

    This holds true on ionisation - the highest energy, 4s electrons are removed first, therefore Mn becomes [Ar]3d5. 4s electrons sit predominantly very far away form the nucleus, with the 3d electrons in between them. The 3d screens the 4s, making 4s higher in energy.

    So then I guess the question is why Vanadium keeps its electrons in the 4s orbital. It's because adding them to the 3d shell increases electron electron repulsions enough to make [Ar]3d5 higher in energy than [Ar]3d54s2. There's been a reduction in nuclear charge of 2, and it turns out that's enough to tip the balance, and electron electron repulsions win out - the increased nuclear charge that the 3d sees isn't enough to make 3d5 in vanadium lower in energy, but it is for Mn2+
    Thanks a bunch for the quality answer! I was waiting for your response.
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    (Original post by Chlorophile)
    That sort of makes sense and could explain why the electron is removed from the 4s subshell but it still doesn't explain why the electron arrangement doesn't then flip and convert to Vanadium's arrangement?
    I hope everything is clear to you now. All the best!
 
 
 
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