
The hybridization states of central atoms of the ion \[I_3^ - ,ICl_4^ - \] and \[ICl_2^ - \]are respectively:
A) \[s{p^2},ds{p^2},s{p^3}\]
B) \[s{p^3}d,s{p^3}{d^2}\] and \[s{p^3}d\]
C) \[s{p^3}d,s{p^3}d,ds{p^2}\]
D) \[sp,sp,ds{p^2}\]
Answer
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Hint: As we know that the central atom which is iodine possess a total of seven electrons and an electronic configuration of $[Kr]4{d^{10}}5{s^2}5{p^5}$ and thus we can say that iodine possess two paired electrons and one unpaired electron in its valence shell.
Complete solution:
As we know that iodine contains two paired electrons and one unpaired electron in its outermost valence shell.
Now, we know that \[I_3^ - \] is a linear anion where iodine contains a total of seven electrons and two monovalent atoms and during the combining of iodine with other two iodine atoms the central atom acquires a negative charge of $1$. Thus the hybridization number becomes equivalent to $5$ and therefore, the hybridization becomes $s{p^3}d$.
Similarly, the \[ICl_4^ - \] being a neutral compound and according to VSEPR theory, \[ICl_4^ - \] is a \[A{B_4}{E_2}\] molecule where A is the central atom, B is the bond pair and V is the electron pair. It possesses a total of six electrons with four bonding pairs and two lone pairs of electrons. According to VSEPR theory, it will have \[s{p^3}{d^2}\] hybridization with octahedral geometry.
Lastly we have \[ICl_2^ - \] and according to VSEPR theory \[ICl_2^ - \] is a \[A{B_2}{E_3}\] molecule where A is the central atom, B is the bond pair and V is the electron pair. And we know that iodine possesses three lone pairs of electrons and two bond pairs. According to VSEPR theory it will have \[s{p^3}d\] hybridization with trigonal bipyramidal geometry.
Therefore,the correct option is (B).
Note: Always remember that the hybridization of any compound is given by mixing of its orbitals of two different atoms and iodine in this case possess a total of seven electrons which are combining with chloride in different manner to form different compounds possessing different hybridization states.
Complete solution:
As we know that iodine contains two paired electrons and one unpaired electron in its outermost valence shell.
Now, we know that \[I_3^ - \] is a linear anion where iodine contains a total of seven electrons and two monovalent atoms and during the combining of iodine with other two iodine atoms the central atom acquires a negative charge of $1$. Thus the hybridization number becomes equivalent to $5$ and therefore, the hybridization becomes $s{p^3}d$.
Similarly, the \[ICl_4^ - \] being a neutral compound and according to VSEPR theory, \[ICl_4^ - \] is a \[A{B_4}{E_2}\] molecule where A is the central atom, B is the bond pair and V is the electron pair. It possesses a total of six electrons with four bonding pairs and two lone pairs of electrons. According to VSEPR theory, it will have \[s{p^3}{d^2}\] hybridization with octahedral geometry.
Lastly we have \[ICl_2^ - \] and according to VSEPR theory \[ICl_2^ - \] is a \[A{B_2}{E_3}\] molecule where A is the central atom, B is the bond pair and V is the electron pair. And we know that iodine possesses three lone pairs of electrons and two bond pairs. According to VSEPR theory it will have \[s{p^3}d\] hybridization with trigonal bipyramidal geometry.
Therefore,the correct option is (B).
Note: Always remember that the hybridization of any compound is given by mixing of its orbitals of two different atoms and iodine in this case possess a total of seven electrons which are combining with chloride in different manner to form different compounds possessing different hybridization states.
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