Chemical Bonding in Compounds
Let us understand the chemical bonding of compounds with an example. Take H3PO2 acid. The formula for the electron pair is
12 × (V + L + A – C)
where, V= valence electron
L = number of lone pair
In H3PO2 we know that the central atom is phosphorus. The central atom of a compound is decided by its electronegativity. Generally, the least electronegative atom other than hydrogen is chosen as the central atom. The valence electron in phosphorus is 5. So the value of V (valence electron) in the formula will be 5. The value of L(lone pair) will be 3. And according to the VSEPR theory we have to ignore the oxygen atom. So now putting the values of bond pair and lone pair in the formula and solving it we get 4 electron pairs.
12 × (5 + 3 + 0 – 0)
= 12 × 8
= 4 electron pair
We all know that for 4 electron pairs the molecular geometry is ‘tetrahedral ‘.
|Number of electron pairs in outer shell||Shape of molecule||Bond angles|
|5||Trigonal bipyramid||1200 and 900|
|7||Pentagonal bipyramid||720 and 900|
Now let us draw the tetrahedral geometry. So the central atom phosphorus is kept in the centre and the two oxygen atoms are kept at any two of the positions other than the center. After that on rest two positions, hydrogen atoms are placed. We will form the chemical bonds with phosphorus first and then we will see that which atom is left out . After this we observe that one hydrogen atom is remaining. We can substitute it with any of the oxygen atoms. The other oxygen will have a double bond with the phosphorus atom.
In this article we got to know about the chemical bonding in an acid with the example H3PO4. To know more about the structure of a chemical compound watch this video.