Answer
Verified
405.3k+ 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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE