Question

Among the molecules ${\mathrm{SO}}_2,{\mathrm{SF}}_4,{\mathrm{ClF}}_3,{\mathrm{BrF}}_5\mathrm{and}{\mathrm{XeF}}_4$. Which of the following shapes does not describe any of these molecules?

• A. V shape
• B. Linear
• C. square pyramidal
• D. Trigonal bipyramidal

Answer 1

The correct option is B

Linear

Explanation for correct option:

B. Linear

(i) ${\mathrm{SO}}_2$

Step 1: Hybridization:

• “Hybridization is defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.”

Step 2: Formula used for calculation:

• $\mathrm{Steric}\mathrm{number}=\frac{1}{2}\left(\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}\right)$
• Where V is the number of valence electrons, M is the monovalent electron, C is the positive charge and A represents the negative charge.

Step 3: Calculation for Hybridization and shape of ${\mathrm{SO}}_2$:

• The total number of valence electrons in ${\mathrm{SO}}_2$ is 6 and there is 0 monovalent atom.
• $\mathrm{Steric}\mathrm{number}=\frac{1}{2}\left(6\right)=3$
• The steric number is 3, thus the hybridization of ${\mathrm{SO}}_2$ is ${\mathrm{sp}}^2$.
• As per VSEPR, the shape of ${\mathrm{SO}}_2$is V-shaped or bent shaped.

(ii) ${\mathrm{SF}}_4$

Calculation for Hybridisation and shape of ${\mathrm{SF}}_4$:

• The total number of valence electrons in ${\mathrm{SF}}_4$ is 6 and there are 4 monovalent atoms.
• $\mathrm{Steric}\mathrm{number}=\frac{1}{2}\left(6+4\right)=\frac{10}{2}=5$
• The steric number is 5, thus the hybridization of ${\mathrm{SF}}_4$ is ${\mathrm{sp}}^3\mathrm{d}$.
• As per VSEPR, the shape of ${\mathrm{SF}}_4$is a Trigonal bipyramidal shape.

(iii) ${\mathrm{ClF}}_3$

Calculation for Hybridisation and shape of ${\mathrm{ClF}}_3$:

• The total number of valence electrons in ${\mathrm{ClF}}_3$ is 7 and there are 3 monovalent atoms.
• $\mathrm{Steric}\mathrm{number}=\frac{1}{2}\left(7+3\right)=\frac{10}{2}=5$
• The steric number is 5, thus the hybridization of ${\mathrm{ClF}}_3$ is ${\mathrm{sp}}^3\mathrm{d}$.
• As per VSEPR, the shape of ${\mathrm{ClF}}_3$is a Trigonal bipyramidal shape.

(iv) ${\mathrm{BrF}}_5$

Calculation for Hybridisation and shape of ${\mathrm{BrF}}_5$:

• The total number of valence electrons in ${\mathrm{BrF}}_5$ is 7 and there are 5 monovalent atoms.
• $\mathrm{Steric}\mathrm{number}=\frac{1}{2}\left(7+5\right)=\frac{12}{2}=6$
• The steric number is 6, thus the hybridization of ${\mathrm{BrF}}_5$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$.
• As per VSEPR, the shape of ${\mathrm{BrF}}_5$is a Square bipyramidal shape.

(v) ${\mathrm{XeF}}_4$

Calculation for Hybridisation and shape of ${\mathrm{XeF}}_4$:

• The total number of valence electrons in ${\mathrm{XeF}}_4$ is 8 and there are 4 monovalent atoms.
• $\mathrm{Steric}\mathrm{number}=\frac{1}{2}\left(8+4\right)=\frac{12}{2}=6$
• The steric number is 6, thus the hybridization of ${\mathrm{XeF}}_4$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$.
• As per VSEPR, the shape of ${\mathrm{XeF}}_4$is a Square planar shape.

Thus, out of the given molecules, there is no linear shape.

Explanation for incorrect options:

Since linear shape does not describe any molecules. Thus, options A, C, and D are incorrect.

Hence, the correct option is B i.e. linear shape.