Determine which of the frieze groups is the symmetry group of each of the following patterns. a.

Determine which of the frieze groups is the symmetry group...

Key Concepts

Frieze Groups

Frieze groups are the seven distinct types of symmetry groups that describe the possible patterns in one-dimensional, infinitely repeating designs. They encapsulate all combinations of translations, rotations, reflections, and glide reflections that are possible along a strip, serving as an important classification tool in both mathematics and art for analyzing patterned repetitions.

Translation Symmetry

Translation symmetry is a fundamental characteristic of frieze groups, where a pattern is invariant under shifts by a fixed distance along the direction of the strip. This basic repeating step forms the backbone of all frieze patterns, as every frieze group necessarily contains translation symmetry by definition.

Reflection Symmetry

Reflection symmetry in the context of frieze groups involves flipping a pattern over a line (which can be vertical or horizontal) such that the pattern remains unchanged. Depending on the placement of the reflection axes (e.g., along the center of motifs or at the boundaries between them), different types of reflection symmetry can be identified as part of the overall classification.

Rotation Symmetry

Rotation symmetry in frieze groups typically refers to half-turn (180°) rotational symmetry. This means that the pattern is invariant when rotated by 180° about a point. Such symmetry is less common than translation symmetry in frieze designs but plays a key role in distinguishing among the various frieze groups.

Glide Reflection Symmetry

Glide reflection symmetry combines a translation with a reflection. In frieze groups, this kind of symmetry means that a pattern remains invariant after it is translated along the strip by a fixed distance and then reflected across a line. Glide reflections are essential for understanding the full range of symmetries present in one-dimensional repeating patterns.

Transcript

00:01 In this question we have been given three different molecules so let us first of all write these molecules where we have ch3 oh then we have ch3 c double bone o and then we have ch3 and in the third molecule we have c h 3 c triple bond n so here in this question the oxygen atom contains two lone pairs in both a and b while the nitrogen atom in part c contains one loan pair of electron and we need to identify the orbitals where these lone pairs of electrons are situated or they exist so first of all we can say that this particular this particular oxygen atom is not forming any double bond right that is why it is going to be sp3 hybridized and since we have sp3 hybrid orbitals we can say that four of these hybrid orbitals would be there four of these hybrid orbitals would be there because it is sp3 hybridization however in this case the oxygen is sp2 hybridized and that is how we can say that there would be a total of three hybrid orbitals we can say that there would be three hybrid orbitals because from the principle of hybridization we can say that the atomic orbitals of almost same energy and similar size mix up together and give the hybrid orbitals in equal number and from sp2 we can deduce that there should be 1 s orbital and 2 p orbitals and these 1 s and 2 p orbitals would mix up together and then they would be giving three hybrid orbitals three hybrid orbitals because initially there were three atomic orbitals involved one s and two p and that is why you can say that these three hybrid orbitals would be called as sp two hybrid orbitals and similarly in the third case we have this nitrogen which forms two pi bonds which basically means that it forms a triple bond and that is why it would be s p hybridized which means that there would be two such hybrid orbitals there would be two such hybrid orbitals now in the case of this first molecule we can say that out of these four hybrid orbitals two would be used in forming two single bonds or we can say two sigma bonds one with this carbon and one with this hydrogen and then we can say that the remaining two hybrid orbitals would contain these lone pair of electrons so we can say that in the first molecule the lone pair of electrons would be there in the sp3 hybrid orbitals similarly in case of the second molecule we can say that there are three hybrid orbitals out of which out of which one is one is accommodating this sigma bond with carbon and now the remaining two sp2 hybridized orbitals the remaining two sp2 orbitals would contain the lone pair of electrons similarly in the third molecule we can just conclude that since two of the hybrid orbitals are there on nitrogen and one of them would be used in the formation of single bond or sigma bond with this carbon atom.

02:56 So we can say that the other sp hybrid orbitals, the other sp hybrid orbital would contain the one lone pair of electrons...

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