Nitrogen forming four tetrahedral bonds

The optical activity of substituted ammonium ions (Nahcd) and of more complex ions and molecules (see Chapter 2) provided the earliest proofs of the essentially 

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The stereochemistry of boron is simple in many of its compounds with the halogens (but note BgClg), oxygen, nitrogen, and phosphorus, three coplanar or four tetrahedral bonds being formed. The more complex stereochemistry of the element in electron-deficient systems (elementary B, some borides, and the boranes) is dealt with separately. The planar arrangement of three bonds from a B atom has been demonstrated in many molecules BX3 and BR3 (Table 24.1), in cyclic molecules of the types noted in the previous section, in many oxy-ions (see later section), and in crystals such as the graphite-like form of BN (p. 847) and AlBj (p. 842). 

The element before carbon in Period 2, boron, has one electron less than carbon, and forms many covalent compounds of type BX3 where X is a monovalent atom or group. In these, the boron uses three sp hybrid orbitals to form three trigonal planar bonds, like carbon in ethene, but the unhybridised 2p orbital is vacant, i.e. it contains no electrons. In the nitrogen atom (one more electron than carbon) one orbital must contain two electrons—the lone pair hence sp hybridisation will give four tetrahedral orbitals, one containing this lone pair. Oxygen similarly hybridised will have two orbitals occupied by lone pairs, and fluorine, three. Hence the hydrides of the elements from carbon to fluorine have the structures... 

The Tetrahedral Carbon Atom.—We have thus derived the result that an atom in which only s and p eigenfunctions contribute to bond formation and in which the quantization in polar coordinates is broken can form one, two, three, or four equivalent bonds, which are directed toward the corners of a regular tetrahedron (Fig. 4). This calculation provides the quantum mechanical justification of the chemist s tetrahedral carbon atom, present in diamond and all aliphatic carbon compounds, and for the tetrahedral quadrivalent nitrogen atom, the tetrahedral phosphorus atom, as in phosphonium compounds, the tetrahedral boron atom in B2H6 (involving single-electron bonds), and many other such atoms. 

The four sp3 orbitals for these three atoms (i.e. N, O and Cl) form a tetrahedral arrangement having one or more of the hybridised orbitals occupied by a lone pair of electrons. For an isolated atom, nitrogen forms a pyramidal shape where the bond angles are slightly less than 109.5 (c. 107 )fig.(a). This compression of the bond angles is because of the orbital containing the lone... 

The experimentally measured H-N-H bond angle in methylamine is 107.1° and the C-N-H bond angle is 110.3°, both of which are close to the 109.5° tetrahedral angle found in methane. We therefore assume that nitrogen hybridizes to form four sp orbitals, just as carbon does. One of the four orbitals... 

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