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 | As you go across the periods, ionisation energies increase because , whilst shielding remains constant, the nuclear charge increases, atomic radii decrease, electrons are more attracted to nucleus so energy need to remove electrons increases
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| Write the electronic configuration of magnesium ion | 1s2 2s2 2p6 3s2 3p6; Did you read the question correctly? It says ION!!!
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| Write the electronic configuration of Bromine atom | 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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| Write the electronic configuration of Fe+2 ion | 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? | Ions with the same electronic configuration for example K+ or Cl-: they have the electronic configuration as Argon 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
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| What is the general trend in ionisation energy across a period? | ionisation energy increases across a period
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| How many electrons are there in an orbital? | 2
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| What is the equation for the second ionisation of sodium? | 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? | First IE decreases down a group as the number of shells increases, so shielding increases and the electrons are less strongly attracted to the atom, so less energy needed to remove the outer electron.
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| Explain why the first ionisation energy is lower for Aluminium atoms than Magnesium atoms | 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 | Start by writing electron config, then apply SCARED principle:P=[Ne]3s2 3p3 and S=[Ne]3s2 3p4: electrons paired up inS so will repel more; bigger effect than increase in nuclear charge; outer e less attracted;less energy needed to remove e-
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| CHALLENGE: Explain HOW you can work out the group number from the successive ionisation energy graph | There will inevitably be a BIG jump when an electron is removed from the “inner” shell; the number of the x-axis at the START of the jump is the group number
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| Explain why 2nd ionisation energy is always bigger than the 1st IE | 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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| describe and explain the shape of the the 1st IE graph for Period3 | 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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| Explain why ionisation energy is lower for Sodium atoms than Lithium atoms | 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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| CHALLENGE: describe and explain the shape of the successive IE graph for Aluminium | 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? | Chromium should have the configuration (Ar) 3d4 4s2, but one of the 4s electrons is donated to the 3d sub shell for Chomium to become (Ar) 3d5 4s1. This is because it is more stable so the ‘4s’ subshell is half filled, and the ‘3d’ is completely filled.
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| Why is copper’s electron configuration different to what is expected? | 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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| When a transition metal forms an ion, why are the 4s sub shell electrons lost before the 3d sub shell? | 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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| Name the four main factors which affect ionisation energy | 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? | No brainer! Magnesium: 1st IE of Mg is bigger than 1st IE of Na because there are more protons in Mg (same shielding, same distance, same shell etc...). And the 2nd IE of any element is always higher than the 1st one.
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| CHALLENGE: Which is highest of 3rd IE of Al or 3rd IE of Mg? | 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 | always start with the electron configuration of the ATOM: FILL then WRITE [Ar] 3d9 4s2; (copper is an exceptiononly as an ATOM); then REMOVE the 4s: [Ar] 3d9
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| What is the electronic configuration of Iron+2 ions | 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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