Hybrid orbitals nitrogen

Unlike carbon, which has four valence electrons to be distributed among the four equal hybrid orbitals, nitrogen has five electrons to fill those orbitals. That means one of the orbitals has a complete pair of electrons. The extra repulsion that results from that complete pair of electrons pushes the other bonds slightly away. If there were no such extra repulsion, the angle between Fi H< e3.32 the hybrid orbitals would be 109.5 degrees (refer back to the discussion of water molecules in this chapter). 

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... 

Describe the bonding in ammonia assuming sp hybridization of nitrogen In what kind of orbital is the unshared pair What orbital overlaps are involved in the N—H bonds , ... 

An orbital hybridization description of bonding m methylamme is shown m Figure 22 2 Nitrogen and carbon are both sp hybridized and are joined by a ct bond The unshared electron pair on nitrogen occupies an sp hybridized orbital This lone parr IS involved m reactions m which amines act as bases or nucleophiles The graphic that opened this chapter is an electrostatic potential map that clearly shows the concentration of electron density at nitrogen m methylamme... 

The sp hybrid state of nitrogen is just like that of carbon except nitrogen has one more electron Each N—H bond in NH3 involves overlap of an sp hybrid orbital of N with a li orbital of hydrogen The unshared pair of NH3 occupies an sp orbital... 

The valence-bond concept of orbital hybridization described in the previous four sections is not limited to carbon compounds. Covalent bonds formed by-other elements can also be described using hybrid orbitals. Look, for instance, at the nitrogen atom in methylamine, CH3NH2, an organic derivative of ammonia (NH3) and the substance responsible for the odor of rotting fish. 

Phosphorus and sulfur are the third-row analogs of nitrogen and oxygen, and the bonding in both can be described using hybrid orbitals. Because of their positions in the third row, however, both phosphorus and sulfur can expand their outer-shell octets and form more than the typical number of covalent bonds. Phosphorus, for instance, often forms five covalent bonds, and sulfur occasionally forms four. 

Strategy The first step is to draw Lewis structures. To find the hybridization of nitrogen, place in hybrid orbitals—... 

Borazine, B3N3Hft, a compound that has been called inorganic benzene because of its similar hexagonal structure (but with alternating B and N atoms in place of C atoms), is the basis of a large class of boron—nitrogen compounds. Write its Lewis structure and predict the composition of the hybrid orbitals used by each B and N atom. 

In triaziridine 23, the o-relaxation is appreciable due to the lone pair effect (Sect. 2.1.4) as is the case with triphosphirane 14 which has almost the same SE as tetra-phosphetane 15. Triaziridine 23 has lower SE (31.8 kcal moF ) than tetraazetidine 24 (SE = 33.4 kcal moF ) (Scheme 12) [12]. Lone pairs on nitrogen atoms have high s-character. The hybrid orbitals for the N-N bonds have lower s-character (sp ) in 23 than that (sp ) in triazane 25 [12]. The geminal a-a interaction is less antibonding in 23 (IBP, = -0.004) than in 25 (IBP. = -0.014). The geminal... 

Answer First we need to ask how many atoms are connected to this nitrogen atom. There are three hydrogen atoms. Next we need to ask how many lone pairs the nitrogen atom has. It has 1 lone pair. Now, we take the sum. 3 + 1 = 4. If we need to have fonr hybridized orbitals, then the hybridization state must be sp. ... 

In general, lone pairs occupy hybridized orbitals. For example, consider the lone pair on the nitrogen atom in the following compound ... 

This nitrogen atom has three bonds and one lone pair, so it is sp hybridized, just as we would expect. The lone pair occupies an sp hybridized orbital, and the nitrogen atom has trigonal pyramidal geometry, just as we saw in the previous section. But now consider the nitrogen atom in the following compound ... 

The lone pair on this nitrogen atom does NOT occupy a hybridized orbital. Why not Because this lone pair is participating in resonance ... 

Ozone, which has 18 valehce electrohs, exemplifies bent molecules. Another example is the hitrite ahioh, the subject of Extra Practice Exercise. The bohdihg of NO2 can be represented using s p hybrid orbitals for the inner nitrogen atom and one set of delocalized n orbitals. 

Formally, the lone pairs on molecular nitrogen, hydrogen cyanide, and carbon monoxide are sp hybrid orbitals, whereas NLMO hybridizations calculated even lower p contributions. Hence, these lone pairs have low directionality, the electron density remains close to the coordinating atom and interaction between the lone pair and the Be2+ is comparatively weak. The Be-L bonds are easily disrupted and ligand exchange consequently can proceed with a low activation barrier. A high degree of p character, on the other hand, means that the lone pair is directed toward beryllium, with electron density close to the metal center, and thus well suited for coordination. 

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