A level chemistry questions

to address my confusion on transition metals, I do not understand why the electron config of tellurium would be [Ke] 5s2 4d20 5p4. I would have left the 4d10 out but that was wrong

hi
a) Which one of the first, second or third ionisations of thallium produces an ion with
the electron configuration [Xe] 5d106s1? ANSWER) second
b) Q6.Which change requires the largest amount of energy?
A He+
(g) He2+(g) + e–
B Li(g) Li+
(g) + e–
C Mg+
(g) Mg2+(g) + e–
D N(g) N+
(g) + e–
ANSWER = option A
also this : question 7 on this link. I'm just getting very confused with ionisation energies of transition metals, and electron configurations too sometimes when dealing with fifth orbitals and above. https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FTopic-Qs%2FAQA%2FPhysical-I%2F1.1-Atomic-Structure%2FSet-F%2FElectron%20Configuration%201%20QP.pdf

First question:
Thallium is in group 3, so it will be [something] ns2 np1 (in this case, the [something] is [Xe] 5d10 - if you look at my above post and apply similar reasoning, you’ll see why).
As such, when electrons are removed, the highest energy electrons are usually lost first (the exceptions being when you have d and f electrons) and so it must have lost one p and one s electron (in that order) to have attained a configuration of the form [something] 6s1. This means a total of two electrons are lost and so it must have been the second ionisation energy.
Second question:
The first thing you should do is think in terms of which orbitals are involved.
He+ and Li^+ both lose an electron from the innermost shell (1s orbital)
Mg+ loses an electron from the 3s subshell
N+ loses an electron from the 2p subshell
So it must be either A or B, as these involve the removal of an electron from the shell nearest the nucleus.
A would be more endothermic as He^+ only has one electron. Li^+ has two paired electrons and these will repel against each other, which makes the removal of one of them comparatively easier than the removal of the single electron in He^+.

It should be [Kr] 5s2 4d10 5p4. The 4d20 is a typo, as you cannot put more than 10 electrons in a d-subshell (remember it’s 5 orbitals and one orbital can take a maximum of 2 electrons).
It is on period 5, so the noble gas it is assigned will be krypton as that is the noble gas on the period above it.
It must have a complete 5s subshell as it is in the p-block and as it is 4 squares along the p-block, it must have 4 electrons in the 5p subshell.
d-subshells are awkward. Essentially the principal quantum number assigned to a given d-subshell is the corresponding period minus 1 and as Te is on the fifth period, the d-subshell of interest has a principal quantum number of 5 - 1 = 4. It must also be full as Te is past the section of the d-block on period 5, so it will have 4d10 as part of its configuration.

First question:
Thallium is in group 3, so it will be [something] ns2 np1 (in this case, the [something] is [Xe] 5d10 - if you look at my above post and apply similar reasoning, you’ll see why).
As such, when electrons are removed, the highest energy electrons are usually lost first (the exceptions being when you have d and f electrons) and so it must have lost one p and one s electron (in that order) to have attained a configuration of the form [something] 6s1. This means a total of two electrons are lost and so it must have been the second ionisation energy.
Second question:
The first thing you should do is think in terms of which orbitals are involved.
He+ and Li^+ both lose an electron from the innermost shell (1s orbital)
Mg+ loses an electron from the 3s subshell
N+ loses an electron from the 2p subshell
So it must be either A or B, as these involve the removal of an electron from the shell nearest the nucleus.
A would be more endothermic as He^+ only has one electron. Li^+ has two paired electrons and these will repel against each other, which makes the removal of one of them comparatively easier than the removal of the single electron in He^+.

Nice explanation! Just going to point out that I believe it means to say Li -> Li+ + e- in the question, so even more simply a case of comparison of attraction between ion and electron vs atom and electron.

hiya, sorry about that but the typo was actually mine! I didn't understand why a 4d10 subshell would be present at all. does this mean wherever there is a p orbital (past period 3) there must be a full d orbital of the number below?