In $I C l_{4}^{-}$ , the shape is square planar, the number of bond pair-lone pair repulsion at $90^{\circ}$ are:
$A .$ $6$
$B .$ $8$
$C .$ $12$
$D .$ $4$

Answer

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Hint:We can predict the number of bond pair-lone pair repulsions at

90^{\circ}

by drawing its structure. The shape of

I C l_{4}^{-}

is square planar because it has four bond pairs and lone pairs. The geometry is octahedral.

Complete step-by-step answer:Before solving the question we have to look at the structure to determine their number of bond pair-lone pairs.
Now, the structure of

I C l_{4}^{-}

is,

Iodine

(I)

has

7

electron it its valence shell in

I C l_{4}^{-}

eight electron participate in formation of compound
Shape of

I C l_{4}^{-}

is square planar. Since its geometry is octahedral, lone pairs reside above and below the square planar. In

I C l_{4}^{-}

each lone pair interacts with the

4

bond pair.
So, Number of bond pair-lone pair repulsion=Total number of lone pairs multiplied by number of bond pairs by which they interact. We have two lone pairs and four bond pairs
Number of bond pairs-lone pairs repulsion

= 2 \times 4

= 8

Thus, there is

8

bond pairs-lone pairs repulsion at

90^{\circ}

. So, the correct option is

B .

Note: Valence Shell Electron Pair Repulsion Theory

(V S E P R)

.This is a very useful theory to predict the geometry or shape of a number of polyatomic molecules or ions on a non-transition element. This theory says that shapes of a species depend on the number of and nature of electron pairs surrounding the central atom of a species.
Table summarizes the relationship between the number of electron pairs that are bond pairs and lone pairs and the shape or geometry.