1. We need the conversion in which both:
   • shape changes, and
   • hybridisation changes.

So we check each option one by one.

2. Option A: $NH_3 \to NH_4^+$

• In $NH_3$:

  • Central atom = N
  • Bond pairs = 3, lone pairs = 1
  • Hybridisation = $sp^3$
  • Electron pair geometry = tetrahedral
  • Molecular shape = trigonal pyramidal

• In $NH_4^+$:

  • Central atom = N
  • Bond pairs = 4, lone pairs = 0
  • Hybridisation = $sp^3$
  • Shape = tetrahedral

Result: Shape changes, but hybridisation remains $sp^3$.

So, A is not correct.

3. Option B: $CH_4 \to C_2H_6$

• In $CH_4$:

  • Carbon is $sp^3$
  • Shape around C = tetrahedral

• In $C_2H_6$:

  • Each carbon is also $sp^3$
  • Shape around each C = tetrahedral

Result: Neither shape nor hybridisation changes around carbon.

So, B is not correct.

4. Option C: $H_2O \to H_3O^+$

• In $H_2O$:

  • Central atom = O
  • Bond pairs = 2, lone pairs = 2
  • Hybridisation = $sp^3$
  • Shape = bent (V-shaped)

• In $H_3O^+$:

  • Central atom = O
  • Bond pairs = 3, lone pairs = 1
  • Hybridisation = $sp^3$
  • Shape = trigonal pyramidal

Result: Shape changes, but hybridisation remains $sp^3$.

So, C is not correct.

5. Option D: $BF_3 \to BF_4^-$

• In $BF_3$:

  • Central atom = B
  • Bond pairs = 3, lone pairs = 0
  • Hybridisation = $sp^2$
  • Shape = trigonal planar

• In $BF_4^-$:

  • Central atom = B
  • Bond pairs = 4, lone pairs = 0
  • Hybridisation = $sp^3$
  • Shape = tetrahedral

Result:

• Shape changes: trigonal planar $\to$ tetrahedral
• Hybridisation changes: $sp^2 \to sp^3$

So, D is correct.

6. Final Answer

The conversion involving change in both shape and hybridisation is:

$\boxed{D\; :\; BF_3 \to BF_4^-}$

7. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.