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College Algebra and Trigonometry1st EditionDonna Gerken, Julie Miller 9,697 solutions Algebra and Trigonometry9th EditionMichael Sullivan 10,535 solutions Trigonometry8th EditionCharles P. McKeague, Mark D. Turner 3,737 solutions Algebra and Trigonometry5th EditionRobert F. Blitzer 10,851 solutions When we travel, we often take a lot more stuff than we need. Trying to fit it all into a suitcase can be a real challenge. We may have to repack or just squeeze it all in. Atoms often have to rearrange where the electrons are in
order to create a more stable structure. The molecular geometries of molecules change when the central atom has one or more lone pairs of electrons. The total number of electron pairs, both bonding pairs and lone pairs, leads to what is called the
electron domain geometry. When one or more of the bonding pairs of electrons is replaced with a lone pair, the molecular geometry (actual shape) of the molecule is altered. In keeping with the A and B symbols established in the previous section, we will use E to represent a lone pair on the central atom (A). A subscript will be used when there is more than one lone pair. Lone pairs on the surrounding atoms (B) do not affect the geometry. The ammonia molecule contains three single bonds and one lone pair on the central nitrogen atom (see figure below). The domain geometry for a molecule with four electron pairs is tetrahedral, as was seen with \(\ce{CH_4}\). In the ammonia molecule, one of the electron pairs is a lone pair rather than a bonding pair. The molecular geometry of \(\ce{NH_3}\) is called trigonal pyramidal (see figure below). Recall that the bond angle in the tetrahedral \(\ce{CH_4}\) molecule is \(109.5^\text{o}\). Again, the replacement of one of the bonded electron pairs with a lone pair compresses the angle slightly. The \(\ce{H-N-H}\) angle is approximately \(107^\text{o}\). AB\(_4\)E: Sulfur Tetrafluoride, \(\ce{SF_4}\)The Lewis structure for \(\ce{SF_4}\) contains four single bonds and a lone pair on the sulfur atom (see figure below). Figure \(\PageIndex{6}\): Lone pair electrons in \(\ce{SF_4}\). (Credit: Joy Sheng; Source: CK-12 Foundation; License: CC BY-NC 3.0(opens in new window))The sulfur atom has five electron groups around it, which corresponds to the trigonal bipyramidal domain geometry, as in \(\ce{PCl_5}\) (see figure below). Recall that the trigonal bipyramidal geometry has three equatorial atoms and two axial atoms attached to the central atom. Because of the greater repulsion of a lone pair, it is one of the equatorial atoms that are replaced by a lone pair. The geometry of the molecule is called a distorted tetrahedron, or seesaw. Figure \(\PageIndex{7}\): Ball and stick model for \(\ce{SF_4}\). (Credit: Ben Mills (Wikimedia: Benjah-bmm27); Source: http://commons.wikimedia.org/wiki/File:Sulfur-tetrafluoride-3D-balls.png(opens in new window); License: Public Domain) Table \(\PageIndex{1}\): Geometries in Which the Central Atom Has One or More Lone Pairs
Summary
Review
How many bond and lone pairs are in tetrahedral?The tetrahedral geometry exists when there are 4 bonds and 0 lone pairs. This is one of the most important and common geometries, as many molecules will adopt this. For example, CH4 C H 4 adopts a tetrahedral geometry (left). The bond angle for tetrahedral molecules is approximately 109.5∘ .
Do tetrahedral shapes have lone pairs?For example; four electron pairs are distributed in a tetrahedral shape. If these are all bond pairs the molecular geometry is tetrahedral (e.g. CH4). If there is one lone pair of electrons and three bond pairs the resulting molecular geometry is trigonal pyramidal (e.g. NH3).
How many atoms have a tetrahedral shape?In a tetrahedral molecule, there is one central atom bonded to four surrounding atoms with no lone electron pairs. The bonds form angles of 109.5°.
How do you find lone pairs on the central atom?Find the number of lone pairs on the central atom by subtracting the number of valence electrons on bonded atoms (Step 2) from the total number of valence electrons (Step 1). Divide the number of VEs not in bonds (from Step 3) by 2 to find the number of LPs.
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