1. Find oxidation state of Cr in chromyl chloride

   Chromyl chloride is $\mathrm{CrO_2Cl_2}$.

   Let oxidation state of Cr be $x$.

   We know:

   • Oxygen $= -2$ each, so $2 \times (-2) = -4$
   • Chlorine $= -1$ each, so $2 \times (-1) = -2$

   Therefore, $x - 4 - 2 = 0$ $x = +6$

   So chromium is in +6 oxidation state in $\mathrm{CrO_2Cl_2}$.

2. Find number of d-electrons on Cr in $\mathrm{Cr^{6+}}$

   Atomic number of Cr $= 24$

   Ground-state electronic configuration of Cr: $\mathrm{Cr} : [Ar]3d^5 4s^1$

   For $\mathrm{Cr^{6+}}$, remove 6 electrons:

   • first 1 from $4s$
   • next 5 from $3d$

   Hence, $\mathrm{Cr^{6+}} : [Ar]3d^0$

   So number of $d$-electrons $= 0$.

3. Check each option

   A: $\mathrm{Mn(VII)}$

   Mn: $[Ar]3d^5 4s^2$

   $\mathrm{Mn^{7+}}$ means removal of 7 electrons: $\mathrm{Mn^{7+}} : [Ar]3d^0$ Number of $d$-electrons $= 0$

   This matches.

   B: $\mathrm{Ti(III)}$

   Ti: $[Ar]3d^2 4s^2$

   $\mathrm{Ti^{3+}}$: remove 2 from $4s$ and 1 from $3d$ $\mathrm{Ti^{3+}} : [Ar]3d^1$ Number of $d$-electrons $= 1$

   Does not match.

   C: $\mathrm{Fe(III)}$

   Fe: $[Ar]3d^6 4s^2$

   $\mathrm{Fe^{3+}}$: remove 2 from $4s$ and 1 from $3d$ $\mathrm{Fe^{3+}} : [Ar]3d^5$ Number of $d$-electrons $= 5$

   Does not match.

   D: $\mathrm{V(IV)}$

   V: $[Ar]3d^3 4s^2$

   $\mathrm{V^{4+}}$: remove 2 from $4s$ and 2 from $3d$ $\mathrm{V^{4+}} : [Ar]3d^1$ Number of $d$-electrons $= 1$

   Does not match.

4. Conclusion

   The number of $d$-electrons in chromium of chromyl chloride is same as in: $\boxed{\mathrm{Mn(VII)}}$

So the correct option is A.