Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# ${I_3}^ -$ and ${I_3}^ +$ have same:(A) Geometry(B) No. of lone pair(s)(C) Bond angle(D) None of these

Last updated date: 16th Jun 2024
Total views: 393.3k
Views today: 5.93k
Verified
393.3k+ views
Hint: ${I_3}^ +$ and ${I_3}^ -$ possess different numbers of total electrons on them. The difference in number of electrons is due to different charges on the two ions making them dissimilar.

Complete Step- by –step solution:
Let us calculate the number of electron pairs on each of the ions and find out the hybridization of the given species.
(1)For ${I_3}^ -$(triiodide ion) the number of electrons in the valence of the central atom of I (V ) = 7. As there is a negative charge on the given ion, the negative charge gets added to the central atom. One negative charge represents one electron hence we need to add 1electron on the central atom as a negative charge, anionic charge (A) =1. The central atom is again bonded to two other iodine which are monovalent. So the number of monovalent atoms (H) = 2. There is no anionic charge (-ve charge). Let us substitute these values in the hybridization formula$(V + H + C - A)\frac{1}{2}$. We get, $\begin{gathered} \frac{1}{2}\left( {7 + 2 + 1} \right) \\ \frac{1}{2}(10) \\ = 5 \\ \end{gathered}$
5 orbitals involved in hybridization are 1s, 3p, 1d thus we get $s{p^3}d$ hybridization and the five orbitals align themselves in trigonal bipyramidal geometry. In these five two are bond pairs and the other three are non-bonding lone pairs. Lone pair electrons align along the plane of the trigonal at equal angles of ${120^ \circ }$ and the bond pairs form bonds with two iodine atoms along axial positions at an angle of${180^ \circ }$. Hence triiodide ions are linear in shape.
(2) for  ion the number of valence electrons in the central atom (V)=7.The ion consists of one positive charge on it so the cationic charge(C) =1.There is no anionic charge on the molecule so 0. The number of monovalent atoms bonded to the central atom (H) = 2. Let us put these values in the hybridization formula. We get, $\begin{gathered} \left( {V + H + C - A} \right)\frac{1}{2} \\ (7 + 2 - 1)\frac{1}{2} \\ \frac{{\left( 8 \right)}}{2} \\ = 4 \\ \end{gathered}$
The number of electron pairs are four and the orbitals involved in hybridization are 1s,3p so the hybridization is $s{p^3}$. The geometry adapted for this hybridization is tetrahedral. Two are bond pair electrons which are involved in bonding and the other two are lone pair of electrons which are non-bonded electrons. Hence the shape of ${I^{3 + }}$ is bent shape or angular.Upon comparing both ${I^{3 - }}$ and ${I^{3 + }}$ ions they are dissimilar in geometry (trigonal bipyramidal, tetrahedral), dissimilar in number of lone pairs (3, 2), dissimilar in bond angle$\left( {{{180}^ \circ },{{104}^ \circ }} \right)$.

The solution for this question is (D) none of these.

Note: From the hybridization of the species one can get to know the geometry of the given atom thus knowing the shape and bond angle also. Iodine is a halogen belonging to the 17th group of the periodic table. Therefore the number of electrons in the valence shell of iodine is 7.