
In which one of the following species the central atom has the type of hybridization which is not the same as that present in the other three:
(A) $PC{{l}_{5}}$
(B) $S{{F}_{4}}$
(C) ${{I}_{3}}^{-}$
(D) $SbC{{l}_{5}}^{2-}$
Answer
576.3k+ views
Hint: The concept of intermixing of different orbitals of atoms and then redistribution of energy to give equivalent energy of new orbitals, identical shape, and symmetrical orientation in space is hybridization. As a result of hybridization, new orbitals are formed known as hybrid orbitals, and bonds formed by hybrid orbitals are called hybrid bonds.
Complete step by step solution: To predict the hybridization of the molecule by using the below formula,
$H=\dfrac{1}{2}[V+M-C+A]$
Where H = number of orbitals in hybridization
V = valence electrons in central atom
C = charge on cation
A = charge on the anion
M = number of monovalent atom
(i) For $PC{{l}_{5}}$ ,
-V = 5 and M = 5
-Then number of orbitals in hybridization, $H=\dfrac{1}{2}[5+5]=5$
-H =5, then $PC{{l}_{5}}$ undergoes $s{{p}^{3}}d$ hybridization.
(ii) For $S{{F}_{4}}$ , V = 6 and M =4
-The number of orbitals in hybridization, $H=\dfrac{1}{2}[6+4]=5$
-H =5, then $S{{F}_{4}}$ undergoes $s{{p}^{3}}d$ hybridization.
(iii) For$I_{3}^{-}$, V = 7, A =1 and M=2
-The number of orbitals in hybridization, $H=\dfrac{1}{2}[7+2+1]=5$
-H =5, then \[I_{3}^{-}\]undergoes $s{{p}^{3}}d$ hybridization.
(iv) For$SbC{{l}_{5}}^{2-}$, V = 5, A = 2, and M =5
-The number of orbitals in hybridization, $H=\dfrac{1}{2}[5+5+2]=6$
-Hence, H =6, then $SbCl_{5}^{-2}$ undergoes $s{{p}^{3}}{{d}^{2}}$ hybridization.
-Among all given species, $SbC{{l}_{5}}^{-2}$ differs in hybridization.
The correct answer is option D.
Note: Hydrogen, alkali metals, and group 17 elements will be considered as monovalent atoms. The characteristics of hybridization are the number of hybrid orbitals formed is equal to the number of atomic orbitals which orbitals undergo hybridization and these hybrid orbitals are equivalent in energy with shape. The stability of hybrid orbitals more than atomic orbitals and these hybrid orbitals have different orientations in space.
Complete step by step solution: To predict the hybridization of the molecule by using the below formula,
$H=\dfrac{1}{2}[V+M-C+A]$
Where H = number of orbitals in hybridization
V = valence electrons in central atom
C = charge on cation
A = charge on the anion
M = number of monovalent atom
(i) For $PC{{l}_{5}}$ ,
-V = 5 and M = 5
-Then number of orbitals in hybridization, $H=\dfrac{1}{2}[5+5]=5$
-H =5, then $PC{{l}_{5}}$ undergoes $s{{p}^{3}}d$ hybridization.
(ii) For $S{{F}_{4}}$ , V = 6 and M =4
-The number of orbitals in hybridization, $H=\dfrac{1}{2}[6+4]=5$
-H =5, then $S{{F}_{4}}$ undergoes $s{{p}^{3}}d$ hybridization.
(iii) For$I_{3}^{-}$, V = 7, A =1 and M=2
-The number of orbitals in hybridization, $H=\dfrac{1}{2}[7+2+1]=5$
-H =5, then \[I_{3}^{-}\]undergoes $s{{p}^{3}}d$ hybridization.
(iv) For$SbC{{l}_{5}}^{2-}$, V = 5, A = 2, and M =5
-The number of orbitals in hybridization, $H=\dfrac{1}{2}[5+5+2]=6$
-Hence, H =6, then $SbCl_{5}^{-2}$ undergoes $s{{p}^{3}}{{d}^{2}}$ hybridization.
-Among all given species, $SbC{{l}_{5}}^{-2}$ differs in hybridization.
The correct answer is option D.
Note: Hydrogen, alkali metals, and group 17 elements will be considered as monovalent atoms. The characteristics of hybridization are the number of hybrid orbitals formed is equal to the number of atomic orbitals which orbitals undergo hybridization and these hybrid orbitals are equivalent in energy with shape. The stability of hybrid orbitals more than atomic orbitals and these hybrid orbitals have different orientations in space.
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