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Dragos Rule

Last updated date: 21st Apr 2024
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Introduction to Dragos Rule

Have you ever wondered why NH3, PH3, AsH3, SbH3, and other molecules in the same group have different bond angles? The bond angle of ammonia is 107 °, but for PH3, there is a difference of 13°, i.e.94°. Similarly, for AsH3 and SbH3, the corresponding bond angles are 93° and 92°, respectively.

Now let's discuss why NH3 and PH3 have a difference in bond angle of 13°. To explain this anomalous property, the Dragos rule can be used. In Dragos molecules, hybridisation does not occur for compounds of elements of the third period or higher, bonded to a less electronegative element like hydrogen.

Dragos Rule

If the central atom of a molecule belongs to the 15th or 16th group and is in the 3rd period or below, and is bonded to a substituent having electronegativity less than or equal to 2.5, then the lone pair present in the s-orbital doesn't participate in hybridisation and bonds are formed by pure p-orbitals. As pure p-orbitals are at 90 ͦ to each other, the angle between bond pairs is close to 90 ͦ. A lone pair of s-orbitals is found in a stereo chemically inactive s-orbital.

Dragos Molecules

  • Dragos molecules have no hybridisation.

  • Hence, i = ∞.

  • The bond angle is nearly equal to 90 ͦ. It ranges between 90-92 ͦ.

  • As for molecules formed with elements of the 3rd period or above, they usually form weak bonds because of poor overlapping of atomic orbitals.

Conditions for Dragos Rule

  1. The central atom should be large. That is, the third period or higher.

  2. The side atom or the surrounding atom should be small and less electronegative, like hydrogen (EN≤2.5).

  3. There should be a lone pair present. Lone pairs are usually present in s-orbitals and do not participate in hybridisation. Hence, these pairs of electrons are stereo-chemically inert. Or the s-orbital is said to be stereo-chemically inactive.

Molecules satisfying these conditions fall under the class of Dragos molecules.

Elements forming Dragos Molecule

Elements forming Dragos Molecule

Phosphorus (P), arsenic (As), antimony (Sb), sulphur (S), selenium (Se), and tellurium (Te) of the 15th and 16th groups, respectively effectively obey Dragos' rule. Hence, being bismuth and polonium radioactive, they show deviation from obeying Dragos' rule. Examples of Dragos molecules include: PH3, AsH3, SbH3, H2S, H2Se, H2Te etc.

Applications of Dragos Rule

1. Basicity

The Dragos Rule can be used to predict the basicity of molecules. While considering NH3 and PH3, NH3 is more basic than PH3. It is because the lone pair of nitrogen atoms are present in their hybrid orbitals, whereas those of phosphorus are in unhybridised orbitals. Nitrogen is sp3 hybridised. Hence, these electrons are involved in hybridisation. In PH3, the lone pair is stereo-chemically inactive and not involved in hybridisation.

Thus, NH3 > PH3

hybridization of PH3

Hybridization of PH3

2. Reducing Character

Dragos molecules act as reducing agents. The size of the atom increases as it moves down a group. Hence, the bond strength will decrease because of the poor overlapping of atomic orbitals.

∴ SbH3 > AsH3 > PH3 > NH3

3. Percentage s-Character

The Dragos molecule has no hybridisation, hence the percentage of s-characters is zero. Also, it is found that these molecules have a very small bond angle. The bond angle has an inverse relationship with the percentage s-character. Some other relations of percentage s-character are derived as follows:

% s-character ∝ Bond angle

∴ % s-character ∝ Bond strength

Phosphines (PH3)

Phosphine is an example of a Dragos molecule. It has a chemical formula of PH3. In the phosphine molecule, phosphorus belongs to the third period and obeys all the conditions of the Dragos Rule. The phosphorus atom makes three sigma bonds with three hydrogen atoms. The s-orbital of phosphorus has a lone pair that is not involved in hybridisation.

Thus, the formation of PH3 is solely dependent on p-orbitals. Every hydrogen atom forms a 90-degree angle with a phosphorus atom. Thus, it takes on the shape of a trigonal pyramid.



Interesting Facts

  • The Dragos rule is limited to certain elements only.

  • This rule is set up such that only the third or higher period elements follow it. From this, it is clear that the Dragos rule is applicable only to p-block elements.

  • Not all elements of p-block obey this rule.

  • Dragos molecules have no hybridisation. It is because the lone pairs present in the s-orbital do not participate in hybridisation. Thus, only p-orbitals are involved in the formation of Dragos molecules.

  • The bond angle observed in Dragos molecules is considered relatively small, nearly equal to 90 degrees. Due to this, groups of 15 and 16 elements tend to form hydrides easily, as hydrogen is a less electronegative atom.

  • The solubility of Dragos molecules in water is slightly lower as compared to the compounds of the 2nd period.

Key Features

  • The bond angles of hydrides in groups 15 and 16 or elements in the third period or higher are explained by Drago's rule.

  • It only applies to PH3, AsH3, SbH3, H2S, H2Se, and other related compounds.

  • Drago's rule states that when the following criteria are met, there will be a significant energy difference between the involved atomic orbitals, preventing any orbital mixing or hybridisation.

  • Only p-orbitals are involved in the formation of Dragos compounds.

  • The bond angle of the Dragos molecule is nearly 90 ˚.

Competitive Exams after 12th Science

FAQs on Dragos Rule

1. Explain the hybridisation of hydrogen telluride.

Hydrogen telluride is a Dragos molecule. It has a chemical formula of H2Te. In the hydrogen telluride molecule, the tellurium belongs to the fifth period and obeys all the conditions of the Dragos rule. Like in water, the tellurium atom makes two sigma bonds with two hydrogen atoms. The s-orbital of tellurium has two lone pairs that are not involved in hybridisation.

Thus, the formation of H2Te only depends on p orbitals. Every hydrogen atom forms a 90-degree angle with a tellurium atom. Thus, hydrogen telluride acquires a bent shape or V-shape.

2. Predict the geometry of SF6 and SH6 molecules.

In the SF6 molecule, the sulphur atom has six electrons in its valence shell. These electrons form covalent bonds with the most electronegative atom, fluorine. SF6 forms six S-F bond pairs. There is no lone pair in this molecule. Therefore, the hybridisation of SF6 is sp3d2. Hence, SF6 acquires an octahedral geometry. In the SH6 molecule, there is no hybridisation.

Even if there are vacant d-orbitals, they do not get hybridised. Since these vacant d-orbitals do not participate in hybridisation due to the large s-d energy gap, a large amount of energy should be required for effective overlapping.

3. Predict the reducing power of H2S, H2Se, and H2Te molecules.

H2S, H2Se, and H2Te are examples of Dragos molecules formed by the 16th group. The reducing power of molecules depends on the size of the atoms from which they are formed. The size of an atom increases as we move down the group. The size is given in the order of Te > Se > S.  As the atom grows larger, its effective valence shells overlap more effectively, which reduces the atom's effective nuclear charge. As a result, the reducing character deteriorates, and the bond becomes weaker. It follows that H2Te >  H2Se > H2S is the order of reducing power.