Beryllium bond angle

The structure of dimethylberyllium is similar to that of trimethylaluminum except for the fact that the beryllium compound forms chains, whereas the aluminum compound forms dimers. Dimethylberyllium has the structure shown in Figure 12.3. The bridges involve an orbital on the methyl groups overlapping an orbital (probably best regarded as sp3) on the beryllium atoms to give two-electron three-center bonds. Note, however, that the bond angle Be-C-Be is unusually small. Because beryllium is a Lewis acid, the polymeric [Be(CH3)2] is separated when a Lewis base is added and adducts form. For example, with phosphine the reaction is... 

Beryllium is in Group 2 and so has two outer electrons. The two Cl atoms contribute one electron each. This gives four electrons in two electron pairs. As beryllium chloride has two Be-Cl bonds, the two electron pairs are two bonding pairs there are two bonds around the central beryllium atom. Thus beryllium chloride will be a linear molecule, Cl-Be-Cl, with bond angles equal to 180°. 

According to this simple picture, beryllium hydride should have two different types of H-Be bonds —one as in 1 and the other as in 2. This is intuitively unreasonable for such a simple compound. Furthermore, the H-Be-H bond angle is unspecified by this picture because the 2s Be orbital is spherically symmetrical and could form bonds equally well in any direction. 

Draw the Lewis structure for beryllium hydride, BeH2. Draw the orbitals that overlap in the bonding of BeH2, and label the hybridization of each orbital. Predict the H—Be — H bond angle. 

This is the maximum possible separation for two electron pairs. Once we have determined the optimum arrangement of the electron pairs around the central atom, we can specify the molecular structure of BeCl2—that is, the positions of the atoms. Since each electron pair on beryllium is shared with a chlorine atom, the molecule has a linear structure with a bond angle of 180 degrees ... 

Any atom surrounded by only two groups is linear and has a bond angle of 180°. Two examples illustrating this geometry are BeH2 (beryllium hydride) and HC=CH (acetylene). We consider each carbon atom in acetylene separately. Because each C is surrounded by two atoms and no lone pairs, each H-C-C bond angle in acetylene is 180°, and therefore all four atoms are linear. 

Using this sp-hybridized beryllium, let us construct beryllium chloride. An ritremely important concept emerges here bond angle. For maximum overlap - rween the sp orbitals of beryllium and the p orbitals of the chlorines, the two. "iorine nuclei must lie along the axes of the sp orbitals that is, they must be ocated on exactly opposite sides of the beryllium atom (Fig. 1.6). The angle chlorine bonds must therefore be 180°. 

Because the major lobes of the hybrid orbitals are much more directed than the 2s orbital, a much stronger bond can be formed, and the extra energy released is more than sufficient to compensate for the energy required to promote an electron from the 2s to the 2p atomic orbital of beryllium. From the orientation of the two hybrid orbitals, the molecule is expected to be linear with a H-Be-H bond angle of 180°. 

The first such molecule is made from a beryllium atom (Be) and as many hydrogen atoms as needed. Since Be is a double-connector atom and each H is a single connector, then as shown in Scheme 7.1, the atoms click and generate the familiar molecule BeH2- What shape will BeH2 prefer In order to predict the shape, we simply count the number of pairs around the atoms. Since Be is surrounded by two pairs, the maximum distance between the pairs will be obtained when the molecule adopts a hnear shape with a bond angle of 180° as shown in Scheme 7.1a. 

The central beryllium atom of BeH2 has only two electron pairs around it both electron pairs are bonding pairs. These two pairs are maximally separated when they are on opposite sides of the central atom, as shown in the following structures. This arrangement of the electron pairs accounts for the linear geometry of the BeH2 molecule and its bond angle of 180°. 

PRACTICE PROBLEM 1.26 What do the bond angles of beryllium hydride surest about the hybridization state of... 

The structure of beryllium chloride is linear with a bond angle 180°. The central atom beryllium undergoes sp hybridization and determines the shape of the molecule formed. 

Consider the gaseous beryllium chloride molecule, BeCl2(g) The Lewis structure of the molecule shows there are only two electron pairs (two electron domains) in the valence shell of the beryllium atom (Figure 4-44). These two pairs of electrons try to separate as far as possible from each other so as to minimize electron repulsion. Thus, the beryllium chloride molecule adopts a linear shape with a bond angle of 180°, because the electron pairs are ferthest apart when they are on opposite sides of the beryllium atom. 

Each sp orbital of beryllium overlaps the s orbital of a hydrogen. To minimize electron repulsion, the two sp orbitals point in opposite directions, resulting in a bond angle of 180°. 

Write an equation involving the reaction of an aqueous beryllium ion with a strong base to form [Be(OH)4]. Draw structures that represent the Lewis, VSEPR, and VBT representations of the tetrahydroxoberyllate ion. Estimate the bond angles in the anion. 

A further point of interest is that in both the dimeric and trimeric species shown, the beryllium atom still has a vacant orbital available which may be used in adduct formation without disruption of the electron-deficient bond. This type of behavior leads to the formation of dimers with four-coordinate beryllium atoms, e.g., structure XX (86). This structure has been determined in the solid state and shows that the phenylethynyl-bridging group is tipped to the side, but to a much smaller extent than observed in the aluminum derivative (112). One cannot be certain whether the distortion in this case is associated with a it - metal interaction or is simply a result of steric crowding, crystal packing, or the formation of the coordination complexes. Certainly some differences must have occurred since both the Be—Be distance and Be—C—Be angle are substantially increased in this compound relative to those observed in the polymer chain. 

Comparison of the elements of the third and fifth groups, both having three unpaired electrons leads to an explanation of their stereochemistry. Boron has one s and two p electrons and thus forms three planar sp hybrid bonds, whereas nitrogen possesses three p electrons which form bonds at right angles to each other. The comparison is similar between elements of the sixth group and beryllium, in the former case the unpaired electrons are p electrons thus forming two bonds at 90°, whilst beryllium with one s and one p electron forms two linear sp hybrid bonds. 

Similar behaviour is found with other atoms. Boron in the trivalent state has one s and two p electrons and hybridization leads to the formation of three equivalent hybrid sp2 orbitals lying in the same plane and with a valency angle of 120°. Experimental data6 for B(CH3)3 are in agreement with this prediction. Beryllium and mercury in the excited state necessary for bond formation, have one s and one p electron which form two hybrid sp bonds at an angle of 180° to each other. 

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