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**Hint:**Hybridisation is a process in which two or more atomic orbitals combine mathematically of an atom to form a new set of orbitals. Hybridization number is used to predict the structure of molecules. Hybridization of a central atom can be determined by the formula .Formula is as shown below,

$\text{ H = GA +}\dfrac{\left( \text{VE}-\text{V}-\text{C} \right)}{2}\text{ }$

Where H is the hybridization number, GA is the number of groups attached to the central atom, VE is the valence electrons of the central atom, V denotes the valency of the central atom and C is a charge on the molecule.

**Complete answer:**

Hybridization is a process in which two or more atomic orbitals combine mathematically of an atom to form a new set of orbitals. This is known as the degenerated hybrid orbitals. Orbitals taking part in the hybridization process are similar in energy and on hybridization forms degenerate orbitals which have the same energy.

Hybridization number is used to predict the structure of molecules. Hybridization of a central atom can be determined by the formula .Formula is as shown below,

$\text{ H = GA +}\dfrac{\left( \text{VE}-\text{V}-\text{C} \right)}{2}\text{ }$

Where H is the hybridization number, GA is the number of groups attached to the central atom, VE is the valence electrons of the central atom, V denotes the valency of the central atom and C is a charge on the molecule.

In the $\text{ ClO}_{3}^{+}\text{ }$ chlorine atom is a central atom .electronic configuration of chlorine is as follows,

$\text{ Cl = 1}{{\text{s}}^{\text{2}}}\text{ 2}{{\text{s}}^{\text{2}}}\text{ 2}{{\text{p}}^{\text{6}}}\text{ 3}{{\text{s}}^{\text{2}}}\text{ 3}{{\text{p}}^{\text{5}}}\text{ }$

The valence shell (third shell) of the chlorine atom contains 7 valence electrons. Therefore valence electrons of central atom chlorine are 7.

Chlorine atom is attached to the three oxygen atoms. Therefore the number of groups attached to chlorine atoms is 3.Chlorine atoms are attached to the three oxygen atoms through three double covalent bonds. Thus central atom chlorine has a valency equal to 6. $\text{ ClO}_{3}^{+}\text{ }$ Molecule has a positive charge. It is a $\text{ +1 }$ positive charge. Let’s substitute all values in the equation stated above. We have,

$\begin{align}

& \text{ H = }\left( \text{3 +}\dfrac{\left( 7-6-1 \right)}{2} \right)\text{ } \\

& \Rightarrow \text{ H = 3 + 0 = 3 } \\

\end{align}$

Thus $\text{ ClO}_{3}^{+}\text{ }$ molecule has a hybridization number of 3. Thus in $\text{ ClO}_{3}^{+}\text{ }$ the molecule chlorine atom is $\text{ s}{{\text{p}}^{\text{2}}}\text{ }$ hybridized. $\text{ s}{{\text{p}}^{\text{2}}}\text{ }$ hybridized orbitals show trigonal planar geometry. This is placed at the three corners of the triangle and has a bond angle equal to $\text{ 12}{{\text{0}}^{\text{0}}}\text{ }$. Thus the given statement is correct.

**Hence, (A) is the correct option.**

**Note:**Note that, the hybridization of a molecule can also be found by looking through the formula.in simpler terms hybridization number is equal to the sum of sigma bonded atoms bonded to the central and lone pair of electrons on the atom (if available). Thus here the hybridization number $\text{ ClO}_{3}^{+}\text{ }$ would be the number of oxygen bonds i.e. 3.

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