AQA electronic configuration and ionisation energy
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| Explain the general trends for the first ionisation energy that can be seen across the period | show 🗑
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| show | 1s2 2s2 2p6 3s2 3p6; Did you read the question correctly? It says ION!!!
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| show | FILL as you read the Periodic Table 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 but WRITE 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 (ie showing all the shell 3 orbitals first)
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| show | FILL as you read the Periodic Table 1s2 2s2 2p6 3s2 3p6 4s2 3d6 but WRITE 1s2 2s2 2p6 3s2 3p6 3d6 4s2 (ie showing all the shell 3 orbitals first); then you lose the “last” e- so 1s2 2s2 2p6 3s2 3p6 3d6 (ie you lose the 4s!)
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| What are isoelectronic ions? | show 🗑
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| show | ionisation energy increases across a period
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| How many electrons are there in an orbital? | show 🗑
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| show | Exam technique: the "SECOND" tells us that the product will have a +TWO charge; Na1+(g) -> Na2+(g) + e-; did you remember state symbols?
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| What are the trends of first ionisation energy down a group? | show 🗑
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| show | Start by writing electron config, then apply SCARED principle:Mg=[Ne]3s2 and Al=[Ne]3s2 3p1: new sub shell; outer electrons further away;less attracted;less energy needed to remove e-
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| Explain why the first ionisation energy is lower for Sulfur atoms than Phosphorus atoms | show 🗑
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| CHALLENGE: Explain HOW you can work out the group number from the successive ionisation energy graph | show 🗑
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| show | Write the equations: 1st=Mg-->Mg+ + e-; 2nd=Mg+ --> Mg+2 + e-;in 1st IE the Mg atom has as many e- than p+; in2nd IE Mg ion is already positive/there are more p+ than e- so the outer electrons are more attracted for the ion
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| show | Small increase from Na to Mg (increased nuclear charge); drop (new subshell); increase from Al to P (increased nuclear charge); drop from P to S (paired electrons); increase from S to Ar
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| show | Start by writing electron config, then apply RaDiSh ChAtE principle:Li=1s2 2s1 and Na=1s2 2s2 2p6 3s1; both have 1 e- in outer shell;Na has one more shell=increased shielding; outer electrons further away;less attracted;less energy needed to remove e-
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| show | small increase from 1 to 2 and 2 to 3 as in each case, less e- than p+ so e- more attracted; BIG jump from 3 to 4 as the fourth e- to be removed is in the shell 2 (the inner shell) so far more attracted
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| Why is chromium’s electron configuration different to what is expected? | show 🗑
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| show | Copper should have the configuration (Ar) 3d9 4s2, but one of the 4s electrons is donated to the 3d sub shell for Copper to become (Ar) 3d10 4s1. This is because it is more stable to have a completely filled sub shell, or a half filled subshell.
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| show | The 4s sub shell has a higher energy level than the 3d sub shell, hence it is easier to remove. To remember this, write the sub shells in ascending order e.g. Calcium: 1s2 2s2 2p6 3s2 3p6 4s2, not 1s2 2s2 2p6 3s2 4s2 3p6
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| show | Nuclear charge, distance from the nucleus (nuclear attraction), shielding, whether the orbital contains a pair of electrons leading to electron repulsion or a singularly occupied orbital.
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| CHALLENGE: Which is highest of 1st IE of Na or 2ndIE of Mg? | show 🗑
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| show | Magnesium: In the successive IE graph, an element STARTS the big jump on the group number. So Mg will have a big increase in IE between 2 and 3rd IE: the 3IE will be very high. Aluminium will not have had the big jump yet...
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| What is the electronic configuration of Copper +2 ions | show 🗑
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| show | always start with the electron configuration of the ATOM: FILL then WRITE [Ar] 3d6 4s2; then REMOVE the 4s first: [Ar] 3d6
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