1. Identify the complex and oxidation state

Given complex: $[Cr(NH_3)_6]Cl_3$

This can be written as: $[Cr(NH_3)_6]^{3+} + 3Cl^-$

Since $NH_3$ is a neutral ligand, the oxidation state of chromium is: $x + 6(0) = +3 \Rightarrow x = +3$

So the metal ion is $Cr^{3+}$.

2. Find the electronic configuration of $Cr^{3+}$

Atomic number of Cr = 24

Neutral Cr: $Cr = [Ar]3d^5 4s^1$

For $Cr^{3+}$, remove 3 electrons:

• first from $4s$: $1$ electron
• then from $3d$: $2$ electrons

So, $Cr^{3+} = [Ar]3d^3$

3. Magnetic nature

A $d^3$ configuration has 3 unpaired electrons in an octahedral field.

Thus the complex is paramagnetic.

So Option A is correct.

4. Hybridization and type of orbital complex

For an octahedral $d^3$ ion like $Cr^{3+}$, two vacant $3d$ orbitals are available for bonding. Hence the complex uses inner $(n-1)d$ orbitals, giving: $d^2sp^3$

This corresponds to an inner orbital complex, not an outer orbital complex.

So:

• Option D is correct: $d^2sp^3$ hybridization and octahedral shape.
• Option B is wrong: it is not an outer orbital complex.

5. Reaction with silver nitrate

In $[Cr(NH_3)_6]Cl_3$, the three chloride ions are outside the coordination sphere: $[Cr(NH_3)_6]^{3+} + 3Cl^-$

These free $Cl^-$ ions react with $AgNO_3$ to give white precipitate of $AgCl$: $Ag^+ + Cl^- \rightarrow AgCl \downarrow$

Therefore Option C is correct.

6. Conclusion

The wrong statement is: $\boxed{\text{B}}$

because $[Cr(NH_3)_6]Cl_3$ is an inner orbital octahedral complex, not an outer orbital complex.