Prediction of Shape of Molecules by VSEPR Theory

Let us apply VSEPR theory to predict the shape of few molecules;

1.    Boron triflouride (BF₃)

    In BF₃ molecule, the Boron atom has atomic number 5 and has electronic configuration of 2, 3. Thus, Boron has 3 valence electrons. These electrons are shared mutually with three fluorine atoms to form three B–F bonds. So, in BF₃, Boron has no lone pair over it.

No. of bond pairs = 3

No. of lone pairs = 0

Total no. electron pairs = bp + lp

= 3 + 0 = 3

        So, according to VSEPR theory, BF₃ has triangular planar shape. This is because it has 3 electron pairs.

2.     Methane CH₄

In CH₄ molecule, the Carbon atom has atomic number 6 and has electronic configuration of 2, 4. Thus, Carbon has 4 valence electrons. These electrons are shared mutually with three hydrogen atoms to form four C–H bonds. So, in CH₄, Boron has no lone pair over it.

No. of bond pairs = 4

No. of lone pairs = 0

Total no. electron pairs = bp + lp

= 4 + 0 = 4

        So, according to VSEPR theory, CH₄ has Tetrahedral shape. This is because it has 4 electron pairs.

3.     Ammonia NH₃

        In NH₃ molecule, the Nitrogen atom has atomic number 7 and has electronic configuration of 2, 5. Thus, Nitrogen has 5 valence electrons. Out of these 3 electrons are shared mutually with three hydrogen atoms to form three N – H bonds and the remaining 2 electrons exist as a lone pair. So, in NH₃, Nitrogen has one lone pair over it.

No. of bond pairs = 3

No. of lone pairs = 1

Total no. electron pairs = bp + lp

= 3 + 1 = 4

            So, according to VSEPR theory, NH₃ should have tetrahedral shape due to presence of 4 electron pairs. The presence of lone pair causes greater repulsion to the bond pairs, as a result of which the three N–H bonds move slightly closer, thereby decreasing the normal tetrahedral angle of 109.5⁰ to 107⁰. So, Ammonia has irregular geometry.

    Since one of the tetrahedral position is occupied by lone pair, the shape of ammonia molecule is said to be Pyramidal.

4.     Water H₂O

    In H₂O molecule, the Oxygen atom has atomic number 8 and has electronic configuration of 2, 6. Thus, Oxygen has 6 valence electrons. Out of these 2 electrons are shared mutually with two hydrogen atoms to form two O–H bonds and the remaining 4 electrons exist as two lone pairs. So, in H₂O, Oxygen has two lone pairs over it.

    No. of bond pairs = 2

    No. of lone pairs = 2

    Total no. electron pairs = bp + lp

     = 2 + 2 = 4

        So, according to VSEPR theory, H₂O
should have tetrahedral shape due to presence of 4 electron pairs. The presence of two lone pairs causes greater repulsion to the bond pairs, as a result of which the two O–H bonds move closer, thereby decreasing the normal tetrahedral angle of 109.5⁰ to 104⁰. So, Water molecule has irregular geometry.

    Since two of the tetrahedral positions are occupied by lone pairs, the shape of water molecule is said to be Bent or V-shaped.