On the basis bo VSEPR theory, prodict the shapes the shapes of the following :(i) CIF_(3) (ii) BrF_(5) (iii) NH_(2)^(-) (iv) H_(3) O^(+)

On the basis bo VSEPR theory, prodict the shapes the shapes of the fol

1.	On the basis bo VSEPR theory, prodict the shapes the shapes of the following :(i) CIF_(3) (ii) BrF_(5) (iii) NH_(2)^(-) (iv) H_(3) O^(+)
Answer» Solution :(i) Shape of `CIF_(3)`<BR>No, of valence electrons of the central CL atom = 7 No. of atoms linked to it by single bonds = 3`therefore`Total no. of electron pairsaround`Cl=(7+3)/(2) = 5 ` No. of bond pairs = No. of atoms linked to Cl = 3 ` therefore ` No. of lone pairs = 5 - 3 = 2 Thus, the molecule is of the type `AB_(3) L_(2)` Hence, it is T-shaped. (ii) Shape of ` BrF_(5)` No. of valence electrons of centrons of central Br atom = 7 No. of atoms linked to it by single bonds = 5 ` therefore ` Total no. of electrons pairs around ` Br = (7+5)/(2) = 6 ` No. of bond pairs = No. of atoms linked to Br = 5 `therefore`Noof lone pairs = 6 - 5 = 1 Thus, the molecule is of the type ` AB_(5) L`. Hence,it has square pyramidal shape. (iii)Shape of ` NH_(2)^(-)` No . of valence electrons of central N atom = 5 + 1 (due to one unit -ve charge) = 6 No. of bond pairs = 2 ` therefore `Total no. of electron pairs around ` N = (6 +2)/(2) = 4 ` No, of bond pairs = 2 ` therefore ` No, of lone pairs = 4 - 2 = 2 Thus, the ion is of the type ` AB_(2) L_(2)`. Hence , it has a bent shape (V-shape). (iv) Shape of ` H_(3) O^(+)` ion No. of valenceelectrons of contral 'O' atom = 6 - 1(Due to one unit + ve charge ) = 5 No. of atoms linked 'O' atom = 3