Group 13 elements (like B or Al) have the largest jump in ionisation energies (IEs) between the removal of the third and fourth electrons, because the first three electrons are removed from a p sub-level and s sub-level that are all within the same energy level (The electron configuration of Boron for example is 1s² 2s² 2p¹). The fourth electron, however, is removed from a new energy level which is lower in energy, closer to the nucleus with stronger nuclear attraction and less shielding making the electrons more difficult to remove.

Element X has gradually increasing 1st to 3rd IEs and then a dramatic jump (much greater in comparison to the differences between the first three IEs) in IE to the 4th, so element X is most likely to be in group 13.

X is the correct answer.

The graph showing the trend in first ionisation energies (1st IE) across the periodic table needs to be learned (at least up to calcium, atomic number 20) such that you are able to sketch the graph. The graph is shown below:

There are only two elements in the first period, H and He.

The best way to 'anchor' the position of a graph showing a section of the trend in ionisation energies is to look for the highest points which are always in group 18 (Noble gases) and the lowest points which are always group 1 (Alkali metals). Also, notice the similar pattern across the eight elements in period 2 and period 3. There are regular dips in 1st IE in the pattern 2, 3, 3 across both periods, before a big drop to the next period. The pattern is explained in the video: First ionisation energies

The 2, 3, 3 pattern is therefore seen across period 2 in the correct answer:

The electron configurations for the four possible elements are; aluminium 1s²2s²2p⁶3s²3p¹ ; silicon 1s² 2s² 2p⁶ 3s² 3p² ; phosphorus 1s² 2s² 2p⁶ 3s² 3p³ and argon 1s²2s²2p⁶3s²3p⁶.

The graph has the largest jump in ionisation energy (IE) between the 5th and 6th electron and a relatively smaller jump between the 3rd and 4th electron. The large jump in IE (5th to 6th) will correspond to a new energy level, and the smaller jump to a new sub-level.

Phosphorus, P, is the only element in the list with a change in energy level after the removal of five electrons (3p³ and 3s² removed) and it also has a change in sub-level (3p to 3s) after three electrons have been removed.

P is therefore the correct answer.

The electron configurations for the four possible elements are; oxygen 1s² 2s² 2p⁴ ; silicon 1s² 2s² 2p⁶ 3s² 3p² ; carbon 1s² 2s² 2p² and sulfur 1s²2s²2p⁶3s²3p⁴.

Element X has a similar difference of approx. 2000 units between each ionisation energy (IE), with no larger jumps in IE. For oxygen there will be no larger jumps in ionisation energy, since the first four electrons are all removed from the 2p sub-level, so element X is oxygen. (Carbon would show a slightly larger jump is IE between the second and third IE; 2p to 2s.)

Element Z has the largest difference of approx. 1700 units between the second and third IEs. For silicon the largest jump will be between the second and third IEs (3p to 3s), so element Z is silicon. (Sulfur would have a similar difference between all of the first four IEs as all electrons are removed from 3p.)

The correct answer is therefore X is oxygen; Z is silicon.

A hydrogen atom has only one electron. In the ground state, the electron is in the 1s sub-level (in the n=1 energy level). The first ionisation energy involves removing an electron from every atom in one mole of gaseous atoms, this can be represented as moving the ground state electron to the n=∞ energy level.

Therefore an electron transition from n=1 to n=∞ could represent the first ionisation energy of hydrogen.

A is therefore the correct answer.

This graph of ionisation energy does not begin with hydrogen. The best way to 'anchor' the position of a graph showing a section of the trend in ionisation energies is to look for the highest points which are always in group 18 (Noble gases) and the lowest points which are always group 1 (Alkali metals). Look also for the 2, 3, 3 pattern in elements that is seen across period 2 and period 3. See the graph below:

The graph showing the trend in first ionisation energies (1st IE) across the periodic table needs to be learned (at least up to calcium, atomic number 20) such that you are able to sketch the graph. The pattern is explained in the video: First ionisation energies

In this question element X is at a high point on the graph that corresponds to group 18. Element Z is in group 13.

The correct answer is therefore Element X is in group 18.

The graph has the largest jump in ionisation energy (IE) between the 4th and 5th electrons and a relatively smaller jump between the 2nd and 3rd electrons.

Large jumps in successive IEs will always correspond to new (lower) energy levels, and the smaller jumps will often correspond to a new (lower) sub-levels.

The 5th electron removed is in a lower energy level is therefore the correct answer.

The graph showing the trend in first ionisation energies (1st IE) across the periodic table needs to be learned (at least up to calcium, atomic number 20) such that you are able to sketch the graph. The pattern is explained in the video: First ionisation energies

The large decreases across the graph are due to an electron being removed from a new energy level of higher energy e.g. element 2 is Helium (1s²) to element 3, which is Lithium (1s² 2s¹).

The smaller decreases between elements 4 to 5 and 12 to 13 are due to an electron being removed from a new sub-level of higher energy e.g. element 4 is Berylium (1s²2s²) to element 5, which is Boron (1s² 2s² 2p¹).

The decreases in IE between elements 7 to 8 (N to O), and 15 to 16 (P to S) are due to the electron being removed from a p orbital in which the electrons have been required to pair-up, creating additional electron-electron repulsion (relative to the previous three elements in which the electron removed occupies the p orbital singularly).

The correct answer is therefore Element 16 has greater electron-electron repulsion than element 15 due to electron pairing.

The element X forms an ionic compound with oxygen, so must be a metal (ionic compounds are always metal and non-metal) and has formula X₂O, so X forms a 1+ ion. X is in period 2, so it must be Lithium.

Lithium has electron configuration 1s² 2s¹, so there will be a large relative increase in ionisation energy (IE) between the 1st and 2nd IEs, as the second electron is removed from a new energy level.

The only series that shows a large relative increase in IE between 1st and 2nd IE is 520, 7298, 11815. Therefore, this is the correct answer.

Large decreases in first ionisation energy (like 1521 to 419) across the periodic table always indicate a new period. The graph showing the trend in first ionisation energies (1st IE) across the periodic table needs to be learned (at least up to calcium, atomic number 20) such that you are able to sketch the graph. The pattern is explained in the video: First ionisation energies

In this question the 1st IEs 1251 and 1521 suggest the last two elements in a period, as there is a big drop to 419 suggesting a new period.

The correct answer is therefore The last two elements of one period and the first three elements of the next period.

Orbitals are plots of electron position (90% probability) around the nucleus. They are three dimensional (not planar). An s-orbital is a sphere; a p-orbital is a three dimensional figure-of-eight.

So the answer here that best decribes as s-orbital is .

The red figure-of-eight image represents a p-orbital.

For HL students only: The elongated blue image represents a sigma molecular orbital (two orbitals overlapping end on) and the elongated red image represents a pi molecular orbital (two p-orbitals overlapping side on).

The sub-levels of an atom are filled in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s. Remember that 's' sub-levels hold a maximum of 2 electrons, 'p' hold 6, and 'd' hold 10.

A copper atom has 29 electrons in total (atomic number is 29), but the configuration is not what would be expected as the electrons in the 4s and 3d energy levels are 4s¹ 3d¹⁰ instead of filling the sub-levels in the conventional manner (4s² 3d⁹). Chromium and copper have unusual configurations and need to be learned.

Remember that [Ar] is shorthand and indicates the electron configuration of argon: 1s²2s²2p⁶3s²3p⁶.

The correct answer is therefore 1s² 2s²2p⁶3s²3p⁶4s¹3d¹⁰

The upper number is the mass number, which is equal to protons plus neutrons. The lower number is the atomic number which is the number of protons. The number of electrons is equal to the number of protons in a neutral atom, but this is 2+ positive ion, so it has two fewer electrons than protons.

So the species has 20 protons, (44−20) 24 neutrons and 18 electrons (two fewer than protons).

The line absorption spectrum of hydrogen is composed of discrete lines that represent particular frequencies. Lines are produced by electron transitions from lower to higher energy levels (for absorption; from higher to lower for emission).

The lines converge at higher frequencies/energies which corresponds to lower wavelengths.

Any lines in the absorption spectrum are not seen as coloured lines, but as black lines (an absence of colour) as electromagnetic radiation is absorbed (not emitted).

Thus the answer is 2 only.

The upper number on the left is the mass number, which is equal to protons plus neutrons. The atomic number for zinc is 30 and for copper is 29; this is the number of protons, shown on the periodic table.

The number of electrons is equal to the number of protons in a neutral atom, but these species are positive ions so will have more protons than electrons.

⁶⁵Zn²⁺ therefore has 30 protons, 65−30=35 neutrons, and 30−2=28 electrons (as it is a 2+ ion).

⁶⁵Cu⁺ therefore has 29 protons, 65−29=36 neutrons, and 29−1=28 electrons (as it is a 1+ ion).

The correct answer is therefore Both have the same number of electrons.

The upper (superscript) number on the left is the mass number, which is equal to protons plus neutrons. The atomic number, which is the number of protons is shown on the periodic table. The number of electrons is equal to the number of protons in a neutral atom but negative ions will have correspondingly more electrons and positive ions correspondingly fewer.

⁸Li⁺ has 8−3=5 neutrons and 3−1=2 electrons

⁸Li has 8−3=5 neutrons and 3 electrons

³¹P³⁻ has 31−15=16 neutrons and 15+3=18 electrons

³¹P has 31−15=16 neutrons and 15 electrons

¹⁵N³⁻ has 15−7=8 neutrons and 7+3=10 electrons

¹⁵N has 15−7=8 neutrons and 7 electrons

The correct answer is therefore ⁸Li⁺, ³¹P, ¹⁵N as this is the only set in which all species contain more neutrons than electrons.

Longer wavelength corresponds to lower frequency/lower energy. So the longest wavelength will correspond to the electron transition of lowest energy.

The question also asks about emission - energy given out - so the transition must be from a higher to lower energy level.

The energy gaps between the levels get smaller at higher energy.

Therefore the smallest energy transition that is also an emission is n=6 to n=5, which is therefore the correct answer.

The electron configuration reflects an element's position on the periodic table.

The easiest way to solve this is to count the total number of electrons (23); as this is an atom, the number of electrons will be equal to the number of protons, so the element is vanadium.

Vanadium is in Group 5, Period 4.

The upper number is the mass number, which is equal to protons plus neutrons (19 sub-atomic particles in the nucleus). The lower number is the atomic number which is the number of protons. The number of electrons is equal to the number of protons in a neutral atom, but this is a 1− negative ion, so it has one more electron than protons.

So this fluoride ion has 9 protons, (19−9=)10 neutrons and 10 electrons (1s² 2s² 2p⁶).

The question states that there is only one stable isotope of fluorine so all stable fluoride ions will have 10 neutrons.

The correct answer is therefore The ion has the same electron configuration as an atom of argon as this is incorrect: Argon has an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶.

The upper number on the left is the mass number, which is equal to protons plus neutrons. The atomic number for krypton is 36, which is the number of protons; shown on the periodic table.

⁸²Kr therefore has 36 protons and (82−36=)46 neutrons.

The correct answer is therefore 36 protons, 46 neutrons.

The sub-levels of an atom are filled in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s.

Remember that 's' sub-levels hold a maximum of 2 electrons, 'p' hold 6, and 'd' hold 10, and that [Ar] is shorthand and indicates the electron configuration of argon: 1s²2s²2p⁶3s²3p⁶.

A vanadium atom has 23 electrons in total (atomic number is 23) so the condensed atomic electron configuration is [Ar] 4s² 3d³.

The V³⁺ ion has a 3+ charge so has lost three electrons. These electrons are lost from the 4s orbital before the 3d. Remember that '4s are gained first and lost first'! So two electrons are lost from 4s and one electron is lost from 3d.

The correct answer is therefore [Ar] 4s⁰ 3d².

Each and every orbital can hold two electrons. Any s sub-level contains one orbital and can therefore hold two electrons. Any p sub-level contains three orbitals and can therefore hold six electrons. Any d sub-level contains five orbitals and can therefore hold ten electrons. Any f sub-level contains seven orbitals and can therefore hold fourteen electrons.

The n=3 energy level consists of the 3s, 3p, and 3d sub-levels.

Thus the answer is 18 (2+6+10).