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Discuss the shpaes of the following molecules using the VSEPR mode : BeCl_(2), BCl_(3),SiCl_(4), AsF_(5), H_(2)S,PH_(3). |
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Answer» SOLUTION :(i)`BeCl_(2) = Cl :Be : C: `. The central atom has only two bond PAIRS and no lone pair,i.e., it is of the TYPE `AB_(2)` . Hence, shape is linear (ii) `BCl_(3) = Cl :overset(cdotcdot)B:Cl`. The central atom has only 3 bondpairs and no lone pair ,i.e., it of the type ` AB_(3)` .Hence, shape is TRIANGULAR planar (iii)` SiCl _(4)= Cl :overset(Cl)overset(cdotcdot )underset( Cl)underset(cdotcdot )Si:Cl` Bond pairs = 4 ,lone pairs = 0 ,i.e. it is of the type `AB_(4)`Shape = Tetrahedral (iv)  Bond pairs = 5 . lone pair= 0 , i.e., it is of the type `AB_(5)`. Shape = Trigonal bipyramidal(v) ` H_(2) S = H:underset (cdotcdot )overset(cdotcdot )S:H` . Bond pairs = 2, lone pairs = 2 , i.e., it is of the type ` AB_(2) L_(2)`. Shape = Bent/V-shaped (vi) `PH_(3) = H:underset (H )underset (cdotcdot )overset(cdotcdot )(P):H`Bond pairs = 3, lone pair = 1, i.e., it is of the type ` AB_(3) L`. Shape = Trigonal pyramidal |
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