00:01
Dear students, we have to draw the lewis dot structure of the given compounds chcl3, sf6, ibr3 and xco2 and you have to determine their geometries using vse pair theory.
00:12
So, let us see the first one chcl3, chcl3.
00:24
So, here carbon we know is having 4 electrons, 4 valence electrons and it forms 4 bonds with 4 atoms, 1 with hydrogen and 3 with 3 chlorine atoms.
00:43
So, this is the lewis dot structure of chcl3.
00:48
Now, it is having 4 bond pairs of electrons, therefore the shape is tetrahedral.
00:59
Now, coming to the next one sf6, sf6, sulfur is having we know in its outer most cell 6 electrons, so it forms 6 bonds with 6 chlorine atoms, 1, 2, 3, 4, 5, 6 with 6 chlorine atoms.
01:22
Now, each chlorine atom is having 7 valence electrons, so here each chlorine atom forms 1 bond sulfur, therefore these are the electrons of sulfur, this is the lewis dot structure.
01:55
Here sf6 is having 6 bond pairs, therefore 6 bond pair means the shape is octahedral, the shape is octahedral.
02:10
Now, coming to ibr3, next compound is ibr3.
02:17
Now, ibr3 iodine is having 7 electrons with 3 bromine atoms, it shares 3 electrons, so these are the 3 electrons it shares and 4 remains unshared.
02:33
Now, here bromine atoms, 3 bromine atoms having sharing 1 electron each with iodine, so it is having 3 bond pairs and 2 lone pairs, so 5 pairs of electrons, 5 pairs of electrons means out of 5 pairs 3 are bond pairs and 2 are lone pairs, so it is t shape, it is t shapes because triagonal bipyramidal geometry it will have ibr, br, br because it has 2 lone pairs, so 2 lone pairs will be in the equatorial position and here 1 bromine and here 1 bromine, so this will be the structure which is t shape...
