Question

Which has a trigonal planar shape?
(a)- $C{{H}_{3}}^{+}$
(b)- $Cl{{O}_{2}}^{-}$
(c)- ${{H}_{3}}{{O}^{+}}$
(d)- $Cl{{O}_{3}}^{-}$

Answer 1

The shape or structure of the compound can be predicted by calculating the hybridization of the compound. The hybridization can be calculated with the number of valence electrons of the central atom, the number of monovalent atoms/ groups surrounding the central atom, charge on the cation, and charge on the anion.

Complete step by step answer:
To predict the shape of the molecule by:
If the number of hybrid orbitals is equal to the number of surrounding groups, it has a regular geometry and the shape is the same as that predicted by hybridization. If the number of surrounding groups is less than the hybrid orbitals, the difference gives the number of lone pairs present. So, the molecule will have irregular geometry and the shape is predicted by leaving the hybrid orbitals containing the lone pairs.

For calculating the number of hybrid orbital or hybridization of the central atom we can use the formula:
$X=\dfrac{1}{2}\left[ \begin{aligned} & \\{\text{no}\text{. of valence electrons of central atom }\\!\\!\\}\\!\\!\text{ + }\\!\\!\\{\\!\\!\text{ no}\text{. of monovalent atoms }\\!\\!\\}\\!\\!\text{ } \\\ & \text{ - }\\!\\!\\{\\!\\!\text{ charge on cation }\\!\\!\\}\\!\\!\text{ + }\\!\\!\\{\\!\\!\text{ charge on the anion }\\!\\!\\}\\!\\!\text{ } \\\ \end{aligned} \right]$
$X=\dfrac{1}{2}\left[ VE+MA-c+a \right]$

(a)- $C{{H}_{3}}^{+}$
The central atom has 4 valence electrons. There are 3 monovalent atoms in $C{{H}_{3}}^{+}$and +1 cationic charge.
So, the hybridization will be,
$X=\dfrac{1}{2}\left[ 4+3-1+0 \right]=\dfrac{6}{2}=3$
The value of X is 3, therefore, the hybridization is $s{{p}^{2}}$
Since the number of surrounding atoms is equal to hybrid orbitals, therefore, there are no lone pairs in $C{{H}_{3}}^{+}$. So, $s{{p}^{2}}$hybridization has a trigonal planar shape.

(b)- $Cl{{O}_{2}}^{-}$
The central atom has 7 valence electrons. There are 0 monovalent atoms in$Cl{{O}_{2}}^{-}$ and -1 anionic charge.
So, the hybridization will be,
$X=\dfrac{1}{2}\left[ 7+0-0+1 \right]=\dfrac{8}{2}=4$
The value of X is 4, therefore, the hybridization is $s{{p}^{3}}$
Since the difference in surrounding atoms and hybrid orbitals is 2, therefore, there are 2 lone pairs of electrons. So, the $s{{p}^{3}}$with 2 lone pairs have a Bent (V-shape) structure.

(c)- ${{H}_{3}}{{O}^{+}}$
The central atom has 6 valence electrons. There are 3 monovalent atoms ${{H}_{3}}{{O}^{+}}$and a +1 cationic charge. So, the hybridization will be,
$X=\dfrac{1}{2}\left[ 6+3-1+0 \right]=\dfrac{8}{2}=4$
The value of X is 4, therefore, the hybridization is $s{{p}^{3}}$
Since the difference in surrounding atoms and hybrid orbitals is 1, therefore, there are 1 lone pair of electrons. So, the $s{{p}^{3}}$ with lone pairs has a Pyramidal structure.

(d)- $Cl{{O}_{3}}^{-}$
The central atom has 7 valence electrons. There are 0 monovalent atoms in $Cl{{O}_{2}}^{-}$ and -1 anionic charge. So, the hybridization will be,
$X=\dfrac{1}{2}\left[ 7+0-0+1 \right]=\dfrac{8}{2}=4$
The value of X is 4, therefore, the hybridization is $s{{p}^{3}}$

Since the difference in surrounding atoms and hybrid orbitals is 1, therefore, there are 1 lone pair of electrons. So, the $s{{p}^{3}}$ with 1 lone pair has a Pyramidal structure.
So, the correct answer is “Option A”.

Note: For predicting the shape of the molecule, the only number of atoms or groups is considered. Even if they are monovalent or divalent atoms or molecules.