yazyj
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Could you explain this part to me? If not during tuition coz I’m not really clear about how a more effective shield is formed and what exactly penultimate 3d subshell means  
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BedokFunlandJC
CS Toh already explained it succinctly, and that's exactly what you'll have to write if Cambridge asks about it in Paper 2 or 3.

Remember to ask during tuition, I'll explain this to you statement by statement then.
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yazyj
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And also for elements Fe and below, how do we calculate max OS? For elements like Mn we just add up the number of electrons in 3d and 4s.
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BedokFunlandJC
Simply be familiar with the known OSes for the 3d metals as shown in the table, very easy to do so once you understand the underlying concept :

(And don't forget that the 3d metals Sc and Zn, are not transition metals).

From Sc to Mn, maximum OS increases from +3 to +7, due to increasing number of 3d electrons available to be lost (in both ionic and covalent compounds), but decreases from +7 to +2 from Mn to Zn, due to the increasing dominance of the opposing factor of increasing nuclear charge (going from left to right of the period), outweighing the previous factor of availability of 3d electrons.

With an additional proton in Fe (compared to Mn) resulting in stronger electrostatic attraction between the valence electrons and the positively charged nucleus, it no longer becomes thermodynamically feasible for Fe to lose 7 electrons (in both ionic and covalent compounds), hence the maximum OS decreases to +6, and so on for next few elements Co, Ni, etc.
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yazyj
Simply be familiar with the known OSes for the 3d metals as shown in the table, very easy to do so once you understand the underlying concept :

(And don't forget that the 3d metals Sc and Zn, are not transition metals).

From Sc to Mn, maximum OS increases from +3 to +7, due to increasing number of 3d electrons available to be lost (in both ionic and covalent compounds), but decreases from +7 to +2 from Mn to Zn, due to the increasing dominance of the opposing factor of increasing nuclear charge (going from left to right of the period), outweighing the previous factor of availability of 3d electrons.

With an additional proton in Fe (compared to Mn) resulting in stronger electrostatic attraction between the valence electrons and the positively charged nucleus, it no longer becomes thermodynamically feasible for Fe to lose 7 electrons (in both ionic and covalent compounds), hence the maximum OS decreases to +6, and so on for next few elements Co, Ni, etc.



You mentioned “outweighing the previous factor of availability of 3d electrons” is availability of 4s electrons considered too? 
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BedokFunlandJC
yazyj wrote:



You mentioned “outweighing the previous factor of availability of 3d electrons” is availability of 4s electrons considered too? 


Yes, but since all 3d metals have the same number of 4s electrons to be lost, that factor is a constant not a variable like the 3d electrons (which differs for each 3d element), hence irrelevant.
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yazyj
Where is the factor availability of 3d electrons used in TM? What is it used to describe? 
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BedokFunlandJC
yazyj wrote:
Where is the factor availability of 3d electrons used in TM? What is it used to describe? 


In the context we were just discussing, it's used to describe the trend of why going from Sc to Mn, there are an increasing number of possible oxidation states. Simply because going from Sc to Mn, there are an increasing number of 3d electrons available to be lost.
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