TRICK for finding Hybridisation and Molecular Shapes - Chemical Bonding Video Lecture | Chemistry for GRE Paper II

hello and welcome to commentor in this
video we are going to learn a very
important trick which will help you in
deducing the
shapes of molecules in very short time
suppose you are given these species clo3
- you have been asked to find out hide
reduction of chlorine here and the shape
of this species in order to solve this
problem in very short time this approach
should be followed first you have to
identify the central atom in the given
species the central atom is chlorine
next we have to see the total number of
valence electrons present on the central
atom chlorine belongs to group 17 and it
does have 7 electrons in its valence
shell so those 7 electrons we are going
to write as 7x
then we have to form bonds Sigma or by
with the surrounding atoms the
surrounding atoms in the species are
three oxygen atoms so valency of often
atom is two so each oxygen atom will
form two bonds three bonds okay and here
it will form 1 pi 1 Sigma you know that
double bonded oxygen forms 1 pi n1
signals so similarly till form 1 pi here
then what Sigma N 1 PI 1 Sigma and if
you see carefully we have one negative
charge for this negative charge we are
going to add one more electron here and
these two making one lone pair and now
you calculate how many Sigma bonds we
have and how many lone pairs we have we
have 1 2 3 3 Sigma and we have one lone
pair so the summation gives us 4 and it
corresponds to sp3 hybridization meaning
if the submission of Sigma and lone
pairs comes up to be for it cause fours
2 SP 3 hybridisation for sp3
hybridization we have four possibilities
it can be either tetrahedral or
pyramidal or bank or linear depending
upon number of lone pairs present in the
species we have one lone pair 2 SP 3
with one lone pair will be pyramidal in
shape so the hybridization of CL in this
is sp3 and shape is pyramidal let's get
another species
CL f 4 plus touring is the central atom
it has seven electrons in the valence
shell and it has four fluorine atoms in
the surroundings for me four Sigma bonds
and if you observe carefully there's one
positive charge because of one positive
charge this electron will be removed now
we have one lone pair now let's
calculate how many Sigma bonds we have
how many lone lone pairs we have we have
four Sigma's and we have one lone pair
that is five and summation of Sigma and
lone pairs giving five corresponds to SP
3d hybridization sp3 hybridization has
four possibilities either it can be
trigonal by pyramidal or seesaw or
t-shaped or linear
since we have one lone pair SP kiti with
one low death will have seesaw shape
so this heavy - Liz SP 3d and the shape
of this molecule is P so so very easily
we could find this group now let's let's
take another example st o+ this
hydronium ion the central atom here is
oxygen oxygen has six valence electrons
so we are going to put 6x
one two three four five six okay and we
have three hydrogen atoms hydrogen
balance is y so single single bonds of
that is one positive charge on oxygen so
this electron is getting removed because
of positive charge and we have now three
sigma bonds one lone pair Sigma bonds
three lone pairs one so it is sort core
this is - this corresponds to SP 3
hybridisation SP 3 hybridization with
one lone pair is pyramidal so hydronium
ion will have pyramidal shape it is
similar to that of ammonia let's take
another specious
I have seven
what is the hydration of iodine and what
is the molecular shape of this special
sensory Thomas I didn't I didn't have
seven valence electrons and outer shells
we are going to put 7x and then there
are seven fluorine atoms each fluorine
atom has valency one so one bond
okay now let's count the number of Sigma
bonds and lone pair we have seven Sigma
bonds and we have zero lone pair so
summation gives up seven summation of
Sigma and lone pair giving seven means
it is SP 3 D 3 hybridization and SP 3 D
3 hybridization with zero lone pair
corresponds to pentagonal bi-pyramidal
so iodine has hybridization SP 3 D 3
because it is 7 summation and the
molecular shape is pentagonal
bi-pyramidal
now 60 species pcl5
phosphorus is the central atom here and
it has five valence electrons one two
three four five five gold atoms are
bonded each chlorine atom has valency
one so forming five Sigma bonds the
summation of Sigma and lone pairs five
plus zero five this corresponds to SP 3d
hybridization SP 3d hybridization with
zero lone pair with zero role pair have
trigonal bipyramidal shape so phosphorus
has fabulous ition SP 3d and the
molecular shape of PCL 5 is trigonal
bi-pyramidal let's draw the structure of
trigonal bi-pyramidal geometry
phosphorus atom at the center
this four atoms are in one plane one
phosphorus atom a big top of it one
little bit behind of that since in
two-dimensional board I cannot draw one
like this so while we're putting these
bodies like this one has swung over here
and there is other ball behind touching
like this CL here so this is trigonal
bipyramidal point bifida Vidal is like
and this is triangle one pyramid that
another pyramid so two pyramids of their
trigonal bi-pyramidal and one very
important thing has to be noted here is
that the PCL bonds bond lengths are
elongated then we see an axial bonds it
is because of repulsion see these three
Touring's are in one plane these
chlorines over here and it is suffering
a 90 degree pulse from three chlorines
so in order to reduce the repulsion the
bond length gets elongated hope this
video was helpful to you one more video
will be uploaded regarding hypertension
and finding molecular shapes in detail
in PPT thank you