Which of the following conversions involves change in both shape and hybridisation?

To identify the conversion that shows a change in both geometric shape and hybridisation, we examine every option one by one.

Option A : $$H_2O \;\rightarrow\; H_3O^+$$

For the water molecule $$H_2O$$ the central atom is oxygen. Oxygen has the ground-state configuration $$1s^2\,2s^2\,2p^4$$. In $$H_2O$$ it forms two σ-bonds with hydrogen and possesses two lone pairs. The steric number is therefore

$$\text{Steric number}= \text{number of σ-bonds}+ \text{lone pairs}=2+2=4.$$

For steric number $$4$$ we recall the rule:

$$\text{Steric number }4 \;\longrightarrow\; sp^3 \text{ hybridisation.}$$

Hence $$H_2O$$ is $$sp^3$$ hybridised. With two bonding and two lone pairs, the observed shape (after taking lone-pair repulsion into account) is bent.

In the ion $$H_3O^+$$ oxygen forms three σ-bonds with hydrogen and carries one lone pair. The steric number again is $$3+1=4$$, so the hybridisation remains $$sp^3$$. With one lone pair the shape is now trigonal pyramidal. Thus the shape changes (bent → pyramidal) but the hybridisation stays the same ($$sp^3$$). Therefore Option A does not satisfy the condition.

Option B : $$BF_3 \;\rightarrow\; BF_4^-$$

For $$BF_3$$ the central atom boron has three σ-bonds to fluorine and no lone pair. The steric number is $$3+0=3$$, and we use the rule:

$$\text{Steric number }3 \;\longrightarrow\; sp^2 \text{ hybridisation.}$$

The three $$sp^2$$ orbitals lie in one plane at $$120^\circ$$, so the molecular shape is trigonal planar.

In the tetrafluoroborate ion $$BF_4^-$$ boron forms four σ-bonds with fluorine and again has no lone pair. The steric number is $$4+0=4$$, leading to

$$\text{Steric number }4 \;\longrightarrow\; sp^3 \text{ hybridisation.}$$

The four $$sp^3$$ orbitals point toward the corners of a regular tetrahedron, giving a tetrahedral shape.

Here we observe:

Shape changes:  trigonal planar → tetrahedral.

Hybridisation changes:  $$sp^2 \;\rarr\; sp^3.$$

Both properties change, so Option B satisfies the requirement.

Option C : $$CH_4 \;\rightarrow\; C_2H_6$$

In methane $$CH_4$$ carbon makes four σ-bonds and has zero lone pairs, giving steric number $$4$$ and hybridisation $$sp^3$$ with a tetrahedral shape.

In ethane $$C_2H_6$$ each carbon still forms four σ-bonds (three to H and one to the other carbon) with no lone pairs, so each carbon remains $$sp^3$$ and the local shape around each carbon is still tetrahedral. Neither the hybridisation nor the shape changes, so Option C is eliminated.

Option D : $$NH_3 \;\rightarrow\; NH_4^+$$

For $$NH_3$$, nitrogen has three σ-bonds and one lone pair, giving steric number $$4$$ ⇒ $$sp^3$$ hybridisation. One lone pair leads to a trigonal pyramidal shape.

In $$NH_4^+$$ nitrogen forms four σ-bonds and has no lone pair, steric number $$4$$ ⇒ still $$sp^3$$ hybridised, with a tetrahedral shape. Thus the shape changes but the hybridisation remains $$sp^3$$, so Option D also fails the “both change” condition.

Among all four possibilities, only Option B ($$BF_3 \rightarrow BF_4^-$$) exhibits a simultaneous change in both hybridisation (from $$sp^2$$ to $$sp^3$$) and molecular shape (from trigonal planar to tetrahedral).

Hence, the correct answer is Option B.