Question

26.On the basis of VSEPR theory, predict
the shapes of the following molecules
ions-
1) CIF7 2)BrF5. 3)IF7 4)H30+​

Answer 1

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The p-Block Elements

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Using VSEPR theory, predict the structures of SO3,2– IF7–, XeF2, ClO4–, ICl4– and IBr2–.

(i) Structure of SO3: In gaseous state SO3 has a plane triangular structure with O–S–O bond angle 120°. Solid SO3 exists in three types of structures. One form of structure is ice like and is a cyclic trimer (SO3)3 or S3O9. The other two forms have asbestos like structures. The chain polymer of (SO3)x is made of SO4 tetrahedron where each tetrahedron shares two O-atoms.

(ii) Structure of IF7: Total number of electrons in the valency shell of the central atom, i.e., I = 7.

No. of electrons provided by the seven F atoms = 7 x 1 = 7.

Total no. of electrons around the cental atoms i.e., = 7 + 7 = 14.

Therefore, total no. of electron pair around the central atom i.e., = 14/2 = 7.

But the total no. of bond pairs = 7.

(Because there are seven I—F bonds)

Therefore, total no. of lone pairs = 7 – 7 = 0 on the basis of VSEPR theory, a molecule with seven bond pairs and no of lone pair must have pentagonal bipyramidal geometry.

Fig. Shapes of IF7 molecule.

Fig. Structure of xenon difluoride.

Structure of XeF2: Xenon difluoride molecule possesses a trigonal bipyramidal structure. The xenon and fluorine atoms lie in a straight line (linear position) while the three lone pairs of xenon occupy the equatorial positions.

Structure of ClO–4: It has tetrahedral shape

Fig. Shapes of ClO4–.

Structure ICl4–

centre iodine contain 8 electron (7 from iodine and one form negtive charge) In which 4 bond pair and 2 lone pair total 6 pair of electron

thus strycture of ICl4– is square planer

centre iodine contain 8 electron (7 from iodine and one form negtive charge) In which 2 bond pair and 3 lone pair total 5 pair of electron

thus strycture of IBr2– is linear structure IBr2–.