DRAGO’S RULE:NO NEED OF HYBRIDIZATION:

Drago’s Rule: Sudden change in bond angle: 

(1) Keeping surrounding atom same if electronegativity of central atom increases bond angle increases.

(2) As par Drago’s rule If central atom is in third row of or below in periodic table then lone pair will occupy in stereo-chemically inactive S-orbital and bonding will be through p-orbital and near 90⁰ angles. If electronegativity of surrounding atom is less than or equal < 2.5.

(3) According to Drag’s rule six molecules (PH₃, AsH₃, SbH₃ H₂S, H₂Se and H₂Te) all belong to 14^th group and except NH3 as nitrogen has EN > 3 and does not follow Drago’s rule and rest all satisfy the condition and hence there are no Hybridisation having on their central atom.

ORBITAL ANALYSIS: 

Hence (theta) is bond angle between equivalent orbital % S and P-character is given by S and P

(1) Condition for Drago’s rule:

1- It is applicable if there is % S + % P = 100%

2- Number of sigma bond + lone pair = 4

3- At least one loan pair must be there on central atom

4- The Electronegativity of central atom is ≤2.5

5- It is applicable when

6- The central atom of a molecule belongs to the 13^th 14^th , 15^th  or 16^th group and is in the 3rd period

6- Drago’s satisfied, then the energy difference between participating atomic orbitals will be very high and thus no mixing of orbitals. Or we can say there is no need to consider hybridisation. All the orbital must

illustrative example: Phosphine (PH₃):  

About 600 k j mole^-1 more energy is required so hybridized the central phosphorous (P) atom. due to this energy factor more stable arrangement is, when pure orbital’s are involved in bonding and lone pair present into pure s-orbital.

Drago’s Rule explains:

1- The Sudden decreases in bond angle.

2- Basicity

3- The solubility of NH3 gas in water.

4- complex formation

illustrative example: Ammonia NH₃:

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 unhybridized 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

 

Related Questions:

Quest: (1) Explain why NH₃ is more soluble in water than PH₃?

Quest: (2) The formation of PH₄⁺ is very difficult but the formation of NH₄⁺ is easier than PH₄⁺?

Quest: (2) NH₃ is stronger Lewis base  in comparison PH₃?

Quest: (2) The complex formation Tendency of NH₃ is wisher than PH₃.

 

Paragraph for question nos. 5 & 6:

Drago suggested an empirical rule which is compatible with the energetic of hybridisation. It states that if the central atom is in the third row or below in the periodic table, the lone pair will occupy a stereo chemically inactive s-orbital, and the bonding will be through p-orbitals and bond angles with be nearly 90º if the electronegativity of the surrounding atom is  2.5.

Question (5): In which of the following molecule central atom has higher % s-character in its bond pair-

(A) AsH₃                    

(B) GeH₄                  

(C) P₄                 

(D) H₂Se

Question (6): Correct order of bond angle is-

(A) PH₄ ⁺ > OF₂ > SF₂> SbH₃ > H₂Te     

(B) OF₂ > SF₂ > PH₄ > SbH₃ > H₂Te

(C) PH₄ ⁺ > SF₂ > OF₂ > SbH₃ > H₂Te    

(D) SF₂ > OF₂ > PH₄ ⁺ > SbH₃ > H₂Te

 

Paragraph for question nos. 7 & 8:

The more ionic character (i.e. poorer covalency) in a bond (due to the more electronegative substituent to the central atom) leads to the utilization of hybrid orbitals containing more p-character (i.e. less s-character) of the central atom. In other words, the multiple bonds encourage the central atom to utilize its hybrid orbitals which contains more s-character.

The percentage of s- or p-character between two adjacent and equivalent hybrid orbitals can determine from the knowledge of the corresponding bond angle () as follows:

Question (7): If x₁, x₂ and x₃ are S – S bond lengths in S₂O₄^-2, S₂O₅^-2 and S₂O₆^-2 respectively, then correct order for S – S bond length is -

(A) x₃ > x₂ > x₁          

(B) x₁ > x₂ > x₃          

(C) x₃ > x₁ > x₂        

(D) x₁ > x₃ > x₂

Question (8): The percentage of p-character (approx) in hybrid orbital having the lone pair at central atom in SF₄ molecule will be:

 (Given F –S- F equatorial bond angle is 102º and cos102º = – 0.21)

(A) 65.5                    

(B) 34.4                     

(C) 50.5                       

(D) 82.6

  

Paragraph for question nos. 9 & 10:

In all expected compounds each case central atom only uses its s and p orbitals in hybridisation. The relation between bond angle theta and decimal fraction of s and p character present in the equivalent hybrid orbitals is given by:

Here theta is bond angle between equivalent hybrid orbital (S and P)

Question (9): The correct order of % P character in bond pairs of central atoms in the following compounds:

(A) P>T>S>Q>R      

(B) S>R>T>P>Q   

(C) P>Q>S>R>T   

(D) P>Q>S>T>R

Question (10): If Value of n is 2 for compound T, then number of lone pair present at central atom of compound T will be:

(A) 0                             

(B) 1                              

(C) 2                           

(D) 3

Question (11): The correct statement is :

(A) The ratio of % p character to % s character is less than four, for the bond pair of central atom of compound S  

 (B) Central atom uses three hybrid orbitals to form compound R

 (C) Central atom uses four hybrid orbitals to form compound S

 (D) There are three compounds present between point C to E. according to % s character  in bond pair of central atom.

 

Answers Key: