Advertisements
Advertisements
प्रश्न
Explain VSEPR theory. Applying this theory to predict the shapes of IF7 and SF6.
Advertisements
उत्तर
Lewis’s concept of the structure of molecules deals with the relative position of atoms in the molecules and sharing of electron pairs between them. However, we cannot predict the shape of the molecule using Lewis concept. Lewis theory in combination with VSEPR theory will be useful in predicting the shape of molecules.
IF7:
Iodine has 7 valence electrons in the valence shell. In an excited state it has 6 valency electrons by 1st and 2nd excited state & under sp3d2, hybridization and combines with 7 Fluorine to get pentagonal bipyramidal shape. There are no lone pairs.
I = ns2 np5 nd0


pentagonal bipyramidal
SF6:
Sulphar attain six valence electron in 1st and 2nd excited states undergoes sp3d2 hybridization and combines with six ‘F’ atoms, as there no lone pair electrons it geometry in octahedral.
S = 16 = ns2 np4 nd0


Octahedral
APPEARS IN
संबंधित प्रश्न
The H-N-H bond angle in NH3 molecule is ____________.
Identify the molecule with linear geometry?
Stable form of A may be represented by the formula:
Explain the non-linear shape of \[\ce{H2S}\] and non-planar shape of \[\ce{PCl3}\] using valence shell electron pair repulsion theory.
Elements \[\ce{X, Y}\] and \[\ce{Z}\] have 4, 5 and 7 valence electrons respectively. Write the molecular formula of the compounds formed by these elements individually with hydrogen.
Which of the possible molecule/species is having maximum values for dipole moment. (where "A" is the central atom)?
Number of lone pair(s) of electrons on central atom and the shape of BrF3 molecule respectively are ______.
The number of lone pairs of electrons on the central I atom in `"I"_3^-` is ______.
Which of the following molecules have regular geometry without lone pair of electrons in the valence shell of central atom?
In XeF2, XeF4, XeF6 the number of lone pairs on Xe are respectively ______.
