Skip to main content

shape of compounds due hybridisation

shape of compounds due hybridisation
1.Shape of sncl2 (stannois chloride)
Ground state
Sn
(50)  5s^2 5px^1 5py^1 5pz^0
since sp^2 hybridisation takes place so sncl2 molecule should be triangular or triginal planar but actually sncl2 molecule is Bent because ane position of triangle is occupied by lone pair of electron.
2 shape of ClO4(perchlorate)
Ground state
Cl
(17)   3s^2 3px^2 3py^2 3pz^1 3d^0
excited state
          3s^2 3px^1 3py^1 3pz^1   3d^3
           { sp3 hybridisation }      {π bond}

orbital formed π bond don't take part in hybridisation .since sp3 hybridisation take place . so ClO4 is tetrahedral.

3.Shape of ClO3^-(chlorate ion)
Ground state
Cl
(17)  3s^2  3px^2 3py^2 3pz^1 3d^0
excited state
          3s^2  3px^1 3py^1 3pz^1 3d^2
          { sp3 hybridisation}    {2π bond}
since sp3 hybridisation takes place .so clo3^-should be tetrahedral. but actually clo3^-
is pyramidal .because one position of tetrahedral occupied by lone pair of electron.

4.Shape of So4^2-(sulphate ion)
ground state
S
(16)     3s^2  3px^2 3py^1 3pz^1 3d^0
excited state
             3s^1  3px^1 3py^1 3pz^1 3d^2
               (sp3 hybridisation)   (2πbond)
since sp3 hybridisation takes place. so   So4^2-
is tetrahedral.

5.Shape of No3^-(nitrate ion)
ground state
N
(7)    1s^2  2s^2 2px^1 2py^1    3pz^1
                  ( sp hybridisation)   ( form c-o-b)
since sp^2 hybridisation takes place NO3^- is triangular planar.  bond angle 120°


6.Shape of Icl2^-(Inter halogen)
ground state
I
(53)    5s^2  5px^2 5py^2 5pz^1    5d^0
            {sp3d hybridisation}  ( form c-o-b)
since sp3d hybridisation takes place so Icl2^- ion should be trigonal bypyramidal but  actually Icl2^- is linear. because three positions of triangle occupied by lone pair of electron.




Direct Determine Hybridisation

formula
1/2[V+X-C+A]
here
V- no. of electron in valence shell of                       central atom.
X-no.  of monovalent atom
C- charge on cation(+)
A- charge on anion(-)


Labels:

Comments

Post a Comment

Popular posts from this blog

Radial probability distribution curves

Radial probhjjhajajbability distribution curves:  The probability of finding the electron is given by the quantity  sie^2.By radial probability us probability of finding the electron within small Radial space around the nucleus. volume of spherical shell between radius r and r+dr =4πr^2 and probability of finding the electron will be 4πr^2dr sie^2. radial probability distribution curves are obtained by plotting radial probability at various distance from the nucleus . 1.Radial probability distribution curves for S orbital n=1, l= 0. distance from nucleus here A° is called angstrom. the probability of finding of electron in a shell is maximum at distance r=r° (r not)  which is 0.529 A° for H atom. diagram shows that probability plot for 2s has(two region of high probability)  or (two peaks)  separated by node. we conclude that  no.  of high probability region in S orbital = n no.  of node = ( n-1) 
Post a Comment
Read more

Radial probability distribution for p and d orbital, radial wave functions

Radial probability distribution for p and d orbital Radial and Angular wave function:  the radial part of wave function depends upon quantum no.  n and  l  and gives the distribution of electron w. r. t distance .it is governed mainly exponential term     e^-Zr/na°(a not)   here  e   based on natural log.  Z   Atomic number r     distance from nucleus  n    principal quantum no. or radial quantum           no.  a°    0.529A° for hydrogen  ( Bohr radii)  the exact mathematical expression for radial part of wave function for 1s or 2s and 2p orbitals. n= 1 ,l=0 s orbital    R(r) =2× (z/a°) ^3/2 ×(2-zr/a°) ×e^-zr/2a° n= 2 ,l=0 2s orbital    R(r) = (z/a°) ^3/2 ×(2-zr/a°) ×e^-zr/2a° n= 2 ,l=1 2p orbital    R(r) = 1√3×(z/2a°) ^3/2 ×(zr/a°) ×e^-zr/2a ° the rad...
Post a Comment
Read more

Electronegativity scales and disadvantage of scales and its nunericals

Electronegativity scales: 1.Pauling scale of electronegativity: in diatomic molecule (A-B) ,the bond formed between two atoms A and  B will be intermediate between  pure Covalent (A-B)  and pure ionic character,  the bond between A and B will be  strong than bond energy increased. if bond (A-B) has been purily covalent than bond energy can be calculated as average bond energy of bond (A-A)  and bond (B-B).     that means it is equal to bond energy of bond (A-B) E(A-B)  = 1/2[ E(A-A) +E(B-B) ] however experiemental value of bond (A-B)  more than this value  because of difference in electronegativity  of A and B the difference ∆ is given by simple expression ∆ = E(A-B) - 1/2 [E(A-A) +E(B-B) ] where E(A-B)  is experimental value of bond energy. if Ҳ(A)  and  Ҳ(B) are the electronegative of elements A and B than Ҳ(A) - Ҳ(B) = 0.18√∆ for numerical ∆ = E(A-B)  -[√(E(A-A) ×E(B-B) )] Disadvan...
Post a Comment
Read more