Sulfur hexafluoride is octahedral in shape, with bond angles of 90° and 180°. The name 'octahedral' comes from the number of faces of the shape (rather than the number of vertices/corners). As you may be able to see from the diagram below, the shape is two square-based pyramids base-to-base on top of each other making an octahedron.

Xenon tetrafluoride is square planar in shape, with bond angles of 90° and 180°.

The shape of the electron domains is octahedral, but with two vertices missing (the lone pairs) the bonds form the shape of a square. Imagine the octahedron shown below without the top and bottom vertices.

Sulfur tetrafluoride is seesaw (sawhorse) in shape, with bond angles of 90°, 120° and 180°.

The shape of the electron domains is trigonal bi-pyramidal, but with one equatorial (rather than axial) vertex missing (the lone pair) the bonds form the shape of a seesaw. Imagine the shape shown below with one equatorial vertex missing.

Iodine trifluoride is T–shaped, with bond angles of 90° and 180°.

The shape of the electron domains is trigonal bi-pyramidal, but with two equatorial (rather than axial) vertices missing (the lone pairs) the bonds form the shape of a 'T'. Imagine the shape shown below with two equatorial vertices missing.

Every bond in the diagram is one pair of electrons. Every sigma bond and every pi bond is one pair of electrons, so there must be a total of 5 bonds.

A single bond is just a sigma bond.

A double bond is one sigma bond and one pi bond.

A triple bond is one sigma and two pi bonds.

There are therefore 3 sigma bonds and 2 pi bonds present in ethyne.

Formal charge = normal valence electrons – (non-bonding electrons + (½ × bonding electrons))

This can be confusing (and appear in different formats): formal charge works out the number of electrons an atom 'has' compared to how many it 'should have'.

The normal valence electrons is equal to the number of electrons in an atom's valence shell when in a 'normal' (ground or isolated) state. E.g oxygen in group VI (16) has 6 electrons in its valence shell.

(non-bonding electrons + (½ × bonding electrons)) are the electrons that the atom actually has in its current state.

The electrons an atom 'has' (all of the non-bonding electrons and half of the bonding electrons are 'owned' by that atom) are being taken away from the electrons it 'should have'. Atoms that have a formal charge of zero are in a 'normal' (electronically stable) state those that have a +ve formal charge are electron deficient; –ve, electron rich.

The oxygen atoms have FC of (left-to-right first resonance structure below):

FC = 6 – (4 + (½ × 4)) = 6 – (6) = 0

FC = 6 – (2 + (½ × 6)) = 6 – (5) = +1

FC = 6 – (6 + (½ × 2)) = 6 – (7) = –1

(or vica versa) respectively.

The atoms of Lewis diagram A have FC of:

Sulfur FC = 6 – (2 + (½ × 8)) = 6 – (6) = 0

Oxygen (both the same) FC = 6 – (4 + (½ × 4)) = 6 – (6) = 0

The atoms of Lewis diagram B have FC of:

Sulfur FC = 6 – (2 + (½ × 6)) = 6 – (5) = +1

Oxygen (double bonded) FC = 6 – (4 + (½ × 4)) = 6 – (6) = 0

Oxygen (single bonded) FC = 6 – (6 + (½ × 2)) = 6 – (7) = –1

The first Lewis diagram, A, (in which the sulfur has expanded its octet) is therefore the most stable since all atoms in the structure have FC of zero.

'FC 0 (sulfur A); FC +1 (sulfur B); most stable diagram is A' is the correct answer.

Remember:

Formal charge = normal valence electrons – (non-bonding electrons + (½ × bonding electrons))

This can be confusing (and appear in different formats): formal charge works out the number of electrons an atom 'has' compared to how many it 'should have'.

The normal valence electrons is equal to the number of electrons in an atom's valence shell when in a 'normal' (ground or isolated) state. E.g sulfur in group VI (16) has 6 electrons in its valence shell.

(non-bonding electrons + (½ × bonding electrons)) are the electrons that the atom actually has in its current state.

The electrons an atom 'has' (all of the non-bonding electrons and half of the bonding electrons are 'owned' by that atom) are being taken away from the electrons it 'should have'. Atoms that have a formal charge of zero are in a 'normal' (electronically stable) state those that have a +ve formal charge are electron deficient; –ve, electron rich.

Ultra-violet light is needed to dissociate both oxygen and ozone. The wavelength of light needed to dissociate oxygen is lower (the light is therefore higher in energy) than ozone. This is because the double bond in oxygen is stronger than the '1½' bonds in ozone (normally shown with the resonance structures as alternating double/single bonds).

The chlorine radicals and nitrogen monoxide are both acting as catalysts in these reactions. Thus, despite the mechanisms both showing a 1:2 molar ratio overall of catalyst to ozone molecules, both catalysts are re-produced at the end of the reaction, and are not 'used up', so they can go on decomposing many thousands of ozone molecules before they come to the end of their lifetimes.

E = hν (energy = Planck constant × frequency) is given in the data book.

Also in the data book is \(c = νλ\) (speed of light = frequency × wavelength)

Rearrangement of \(c = νλ\) will give \(ν = {c \over λ}\) (frequency = speed of light / wavelength).

So using E = hν and \(ν = {c \over λ}\) by substituting frequency from the second expression into the first expression gives \(E = {hc \over λ}\).

Energy is given for one mole of electrons in the question, so 498 kJ mol^–1 must be divided by Avogadro's number (L) to give the value for one bond.

498/6.02×10²³=8.27...×10⁻²² kJ which is 8.27...×10⁻¹⁹ J (must be in Joules as Planck's constant is in Joule seconds, Js)

Thus, putting in all the known values, \(E = {hc \over λ}\) becomes:

\({8.2724252×10^{-19}} = {(6.63×10^{-34})(3.00×10^{8}) \over λ}\)

Rearranging:

\({λ} = {(6.63×10^{-34})(3.00×10^{8}) \over 8.2724252×10^{-19}}\)

Wavelength is therefore: 2.40×10⁻⁷ m (240nm)