Orbital Overlap; Hybrid Orbitals (Sections)(a) Which

Chapter 5, Problem 58E

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QUESTION:

(a) Which geometry and central atom hybridization would you expect in the series \(\rm{BH}_{4}^{-}\), \(\rm{CH}_4\), \(\rm{NH}_{4}^{+}\)?

(b)What would you expect for the magnitude and direction of the bond dipoles in this series?

(c) Write the formulas for the analogous species of the elements of period 3; would you expect them to have the same hybridization at the central atom?

Questions & Answers

QUESTION:

(a) Which geometry and central atom hybridization would you expect in the series \(\rm{BH}_{4}^{-}\), \(\rm{CH}_4\), \(\rm{NH}_{4}^{+}\)?

(b)What would you expect for the magnitude and direction of the bond dipoles in this series?

(c) Write the formulas for the analogous species of the elements of period 3; would you expect them to have the same hybridization at the central atom?

ANSWER:

Step 1 of 3

(a)

The most common compound of Boron with hydrogen is borane \(\left(\mathrm{BH}_{3}\right)\). This compound has three electrons in its outermost shell, which is \(\mathrm{sp}^{2}\) hybridised. But the compound, \(\mathrm{BH}_{4}^{-}\) has four electrons and so boron combines with four hydrogen atoms to form \(\mathrm{BH}_{4}^{-}\). Hence the central atom, boron has \(\mathrm{sp}^{3}\) hybridization and so it has a tetrahedral structure. 

\(\mathrm{CH}_{4}\) has four electrons and carbon combines with four hydrogen atoms to form \(\mathrm{CH}_{4}\). Hence the central atom, carbon has \(\mathrm{sp}^{3}\) hybridization and so it has a tetrahedral structure.

The most common compound of nitrogen with hydrogen is ammonia \(\left(\mathrm{NH}_{3}\right)\). The compound, \(\mathrm{NH}_{4}^{+}\)  has three electrons and so nitrogen combines with four hydrogen atoms to form \(\mathrm{NH}_{4}^{+}\). Hence the central atom, nitrogen has \(\mathrm{sp}^{3}\) hybridization and so it has a tetrahedral structure. 

\(\mathrm{BH}_{4}^{-}\), \(\mathrm{CH}_{4}\) and \(\mathrm{NH}_{4}^{+}\) are isoelectronic \(\mathrm{AX}_{4}\) systems. All the central atom will have \(\mathrm{sp}^{3}\) hybridization and will have a regular tetrahedral structure with H-A-H bond angle \(109.5^{\circ}\).

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