Question

The shape of $I_3^ - $ is ----------------
A.tetrahedral
B.linear
C.T-shape
D.trigonal

Answer 1

Hint:We use VSEPR theory and Lewis theory to predict the shape and geometry of a compound. In this theory we use the number of bond and lone pairs of electrons to predict the geometry.

Complete step by step answer:
We know that I has the atomic number 53 and the electronic configuration of I is $\left[ {{\rm{Kr}}} \right]{\rm{4}}{{\rm{d}}^{{\rm{10}}}}{\rm{5}}{{\rm{s}}^{\rm{2}}}{\rm{5}}{{\rm{p}}^{\rm{5}}}$. The ${\rm{I}}_3^ - $molecule is formed by the bonding of the ${{\rm{I}}^ - }$with that of the ${{\rm{I}}_{\rm{2}}}$. The central
The atom gains one negative charge when the combination of the iodine atom takes place. In this compound, the ${{\rm{I}}^ - }$donates electron pairs and the ${{\rm{I}}_2}$molecule accepts the electrons. The electrons get accommodated inside the empty d-orbital. The ${\rm{I}}_3^ - $molecule is ${\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$ hybridized and it has a linear geometry. This compound has a bond angle of ${\rm{10}}{{\rm{8}}^{\rm{0}}}{\rm{C}}$.
We can draw the structure of ${\rm{I}}_3^ - $ in the following way.

Therefore, out of the given four options, B is the correct option.

Additional information:
VSEPR helps in predicting the geometry of a compound taking in account the arrangement of electron pairs. VSEPR theory states that the electrons present around repel each other and it tends to take up an arrangement that will have minimum repulsion. The number of the valence shell present in the central metal atom is determined by drawing a Lewis structure of that atom. The number of electrons in the valence shell determines how and how many numbers of other atoms can find it and construct a structure.


Note:
 VSEPR theory considers the repulsion by the lone pair to be much greater or effective than the repulsion by the bonding pair. Based on this theory, structures are classified as linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral etc.