Hybridization of PH3


What is Hybridization?

Orbital Hybridization is one of the most important topics in modern Physical Chemistry. Orbital Hybridization generally referred to as Hybridization in chemistry, is a concept that narrates the combining of mixing of atomic orbitals to form new orbitals. These new orbitals are different in shape and energies when compared to the orbitals that are combined to form these orbitals. The new orbitals are thus called hybrid orbitals. The hybrid orbitals are suitable for the pairing of electrons in valence bond theory to form chemical bonds. 

Cause for Hybridization

Hybridization is a phenomenon that occurs when an atom makes a bond with the other atom with the help of the electrons that are from both ‘s’ and ‘p’ orbitals. This kind of chemical bonding creates an imbalance in the energy levels of the two electrons. To stabilize this variation in energy levels of the electrons from two different orbitals, the orbitals that hold the electrons involved in bond formation combine to form a hybrid orbital. 

Structure of PH3 

The common name of PH3 is phosphine. Phosphine does not have any characteristic colour. However, it is an inflammable and toxic gas. It is identified as a pnictogen hydride. Though phosphine in its purest form does not have any characteristic odour, the technical grade samples of phosphine stink with an unpleasant odour of garlic or rotten fish. It is found that the presence of substituted phosphine and diphosphine induces this unpleasant odour. The molecular formula of Phosphine is PH3. This indicates that the structure of phosphine should include one phosphorus and three hydrogen atoms bound together. The bond angle in PH3 is 930 C. The geometry of its structure describes phosphine as a trigonal pyramidal molecule. Since phosphine is gaseous at room temperature, it's boiling and melting points are comparatively low. The melting point of phosphine is -132.80 C and its boiling point is -87.70 C. Phosphine has a molar mass of 33.99758 g/mol and is highly soluble in water. 

Hybridization in Phosphine

It is quite surprising to describe hybridization in Phosphine. This is because it is a known fact that Phosphine has a well defined orbital structure and electron distribution. In other words, we can simply say that the process of hybridization is not valid in the case of a phosphine molecule. The subsequent sections of this page will give a brief overview of the absence of hybridization in Phosphine molecules. 

The detailed analysis of the structure and formation of the phosphine molecule gives an understanding that the electrons in pure ‘p’ orbital will take part in the formation of chemical bonds. This acts as a resistance for the orbitals to get hybridized. The lone pair of electrons is mainly in the ‘s’ orbital and ‘s’ orbital is the lone pair orbital. Phosphorus will thus have three bond pairs and one lone pair of electrons. A detailed explanation of the absence of PH3 hybridization is given by Drago’s rule. PH3 is regarded as a Drago molecule. For the pure ‘p’ orbitals that hold the electrons involved in bond formation, the bond angle is nearly 900. The Lewis dot structure of Phosphine enables us to understand that the Phosphine is trigonal and pyramidal. 

Drago’s Rule and Hybridization of Phosphine

Drago’s rule states that there is no need for considering the hybridization of an element in the following cases:

Case 1: At least one lone pair of electrons is present on the central atom of the molecule.

Case 2: Any of the elements from groups 13, 14, 15, 16 or from the 3rd to 7th period forms the central atom.

Case 3: The central atom has an electronegativity less than or equal to 2.5.

Case 4: Sigma bonds are absent and 4 lone pairs are there.

Let us consider the phosphine molecule. In this molecule, the central element is Phosphorus. Phosphorus is an element that belongs to the 15th group and the third period of the modern periodic table. The electronegativity of phosphorus is 2.19. Also, the phosphine molecule has one lone pair of electrons on the phosphorus atom. Considering all these facts and figures, the hybridization is absent in the phosphine molecule according to Drago’s rule. However, atomic orbitals in phosphine overlap on one another to form chemical bonds.

Let’s have further insight on hybridization in Phosphine. It can be calculated that, in the P - H bonds, only 6% of the s - character will be recorded. Considering that there are three P - H bonds in the phosphine molecule, the s - character taking all the three P - H bonds together is 6 x 3 = 18 %. With this calculation, we can infer that the lone pair of electrons is not in this orbital and it is present in the orbital which has 100 - 18 = 82% ‘s’ character. However, it is proven that none of the hybridized orbitals will have such a higher percentage of s- character. This indicates that the lone pair of electrons in the PH3 molecule is not in any of the hybridized orbitals. It is present in the pure s - orbital. 

Fun Facts

  • Orbital hybridization does not happen in the case of phosphine molecules. 

  • The bond angle PH3 is approximately equal to 900.

  • Pure ‘p’ orbital electrons are involved in P - H bond formation of phosphine molecules.

FAQs on Hybridization of PH3

1. What are the different types of Hybridization?

Orbital hybridization is the process of mixing or combining unstable atomic orbitals to form hybrid orbitals to balance the energy level of electrons in different orbitals. Intermixing or hybridisation of pure atomic orbitals is analysed before the bond formation to confer maximum stability of molecules. The different types of hybridization are:

  • sp hybridization- Beryllium Chloride (BeCl2), Acetylene (C2H2)

  • sp2 hybridization- Boron Trichloride (BCl3), Ethylene (C2H4)

  • sp3 hybridization- Methane (CH4), Ethane (C2H6)

  • sp3d hybridization- Phosphorus Pentachloride (PCl5)

  • sp3d2 hybridization- Sulphur Hexachloride (SF6)

  • sp3d3 hybridization- Iodine Heptafluoride (IF7)

2. How do you find whether the hybridization is sp2 or sp3 in hydrocarbons?

Mastering the techniques of determining hybridisation goes a long way not just in this chapter but the whole subject in general. All the alkanes have central carbon atoms that are sp3 hybridized with tetrahedral geometry. However, in the case of alkenes and alkynes, the carbon atoms with double and triple bonds respectively are sp2 and sp hybridized respectively. The alkenes have trigonal planar geometry. We can also use a method in which we calculate the atoms plus the lone pairs for any bond type and not the multiple bonds, ie., determining the number of groups that are in each atom. This is called the Steric Number. For example, if the steric number is 4 then it is sp3; if the steric number is 2 then it is sp. Therefore C1-SN= 3 (three atoms) so it is sp2.

3. What are the rules for the process of Hybridisation?

There are certain rules which are counted for hybridisation are as follows:

  • The number of atomic orbitals mixed always equals the number of hybrid orbitals

  • Mixing of the number of orbitals, in the hybridisation process, is as per requirement.

  • Hybrid bonds are stronger than non-hybrid bonds.

  • Only the orbitals of the central atom can undergo hybridisation.

  • The hybrid bonds tend to be farthest apart and are distributed in space.

  • The orbitals of similar energy can be mixed to form hybrid orbitals.

4. Is the Hybridisation of PH3 an important topic in IIT-JEE?

The topic of hybridization explained in the chapter ‘Chemical Bonding’  is very important for the IIT JEE syllabus. Nearly 5 questions are asked every year from this chapter. Hence students can score full marks if they pay careful attention to the subsequent topics. Because all the questions are in multiple-choice format (MCQs), the questions are unpredictable. Plus students might wish to skip past the chapter due to lack of time or interest. For the students to understand the pattern and the type of questions related to the chapter, they can follow Previous Years’ questions and analyze their preparation accordingly through Vedantu.

5. How do students study the Hybridisation of PH3 through NCERT?

Before following up with the recommended books, students must clear all the basics of difficult concepts with the help of the NCERT. NCERT provides a concise and clear explanation of chapters through well-drawn diagrams, elaborate examples and question banks of varying difficulty levels. Vedantu has prepared study material related to NCERTs which students can easily access through the official website and download the files for free in PDF format. These notes will help in the revision as well as analysis of the preparation for the exam.