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Group 17 and group 18

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Total cards: 16•Content version: v4•Public deck•Updated: 8/1/2026

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#1
Front (Question)

What is the general electronic structure and bonding for group 17

Back (Answer)

• Electronic structure: s² p⁵ • The element’s behaviour is dominated by high electronegativity F>Cl>Br>I • Diatomic (which can give rise to high volatility for the halogens)

#2
Front (Question)

Which is the only halogen which can form chains

Back (Answer)

Iodine But not through covalent interactions Polyiodides from [I₃]⁻ up to [I₂₉]³⁻

#3
Front (Question)

How does F-F react compared to S-F?

Back (Answer)

• F-F is very reactive (direct combination with most elements) • F-F reactions are very exothermic (violent) • This is due to the low F-F bond energy as it is weak due to high electrostatic repulsion • High S-F bond energy due to large difference in electronegativity, hence electrostatic interactions

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#4
Front (Question)

True or false Only ionic compounds can be formed from fluorides

Back (Answer)

False Both ionic and covalent compounds are formed Will depend on the difference in electronegativity

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#5
Front (Question)

How easily will halogens react with hydrogen

Back (Answer)

Going down the group, more energy is required for halogens to react with hydrogen

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#6
Front (Question)

What happens to reactivity and acid strength of hydorgen halides going down group 17 Why?

Back (Answer)

• Reactivity decreases going down the group • Acidity increases going down the group • Both related to the strength of th H-X bond

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#7
Front (Question)

A wide range of oxidation states available for heavier halides (i.e. not F) Why?

Back (Answer)

• Due to how electronegative it is, as the valance shell is very close to the nucleus • We cannot form any high/positive oxidation states

#8
Front (Question)

Explain the trend of the following group 18 ionisation energies

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Back (Answer)

• He-Ar are particularly stable and don’t have any d-orbitals • As we go down to larger valence shell the electrons are further away from the nucleus and more easily ionised • As well as d-orbitals from Kr onwards

#9
Front (Question)

Xe has the most extensive chemistry Oxidation states ranging from +2 to +8 What does this depend on?

Back (Answer)

Must have very electronegative substituents (e.g. F, O or Cl) present to stabilise the compounds of the group 18 elements

#10
Front (Question)

The graph below shows the formation of different Xenon fluorides What are some key features of these Xenon fluorides?

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Back (Answer)

• All colourless, volatile solids • Very powerful oxidising agents and fluorinating agents • Fluorinating ability XeF₂ < XeF₄ < XeF₆

#11
Front (Question)

Why is XeF₆ not your typical octahedron?

Back (Answer)

• 14e-, gaseous structure is a monocapped octahedron i.e. with a lone pair projecting from one face • Lone pair distorts the octahedron by pushing back three of the F⁻ ligands • The structure is fluxional and lone pair shifts from face to face - average of all eight structures

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#12
Front (Question)

XeF₂ has 10 electrons Hence goes against the ocetet rule Use the Molecular orbital diagram below to rationalise where electrons go within this compound

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Back (Answer)

• Three-centre MO’s derived from a p-orbital on each atom • Combination of 3 AO’s gives 3 MO’s: 1x bonding, 1x non-bonding and 1x antibonding orbtial • 2e- from Xe, 2e- from F⁻ gives 4e⁻ to populate • (note: the 2 electrons in this bonding MO are spreadout over the entire molecule (not located entirely in one Xe-F bond) so the two bond are weak)

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#13
Front (Question)

Phosphorus will react with oxygen to form a phosphorus oxide complex e.g. P₄ + 5O₂ → P₄O₁₀ Does S react the same way as O?

Back (Answer)

• Yes and no • P₄ + ₙS₈ → P₄S₁₀ (isostructual with the oxygen analogue) when n is in excess • But can form P₄ + ₙS₈ → P₄S₃, when n is deficient • This compound does not have a stable oxygen equivalent

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#14
Front (Question)

• Carbon can have triplet ground states in carbenes (single unpaired electron) which enables the formation of alkenes - shown below • Silyenes, germylenes, stanylenes and plumbylenes have singlet ground states, how do they form double bonds

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Back (Answer)

• As shown there can’t have an interaction between the two filled orbitals due to repulsion • Instead two fragment tilt at 45° so there is two acid-base interactions • (the larger the element the bigger the tilt)

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#15
Front (Question)

Germylenes show that electropositive substituents (e.g. Si) can favour great double bond character seen by flatter conformation to carbon germylene equivalents Why?

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Back (Answer)

• Due to Si and Ge having a large difference in electronegativity (hence ΔΧ) • Si will release electron density towards germanium • This will affect bond polarity and relative energies of singlet and triplet states AND HOMO and LUMO

#16
Front (Question)

The following compound is Borazine Which is a boron-nitrogen analogue of benzene (isoelectronic and isostructural) What are some differences however been Borazine and Benzene?

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Back (Answer)

• Nitrogen is a lot more electronegative (3.04) and smaller (0.70Å) than boron (2.04, 0.88Å) • So, more localised charges than benzene (though both formally obey Huckel’s rule) in structures that are in resonance with the neutral representation

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