Magnetic moment (spin-only) formula

The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}\ \text{BM}$ where $n$ is the number of unpaired electrons.

For $\mu = 1.73\ \text{BM}$, $\sqrt{n(n+2)} = 1.73 \approx \sqrt{3}$ which gives $n=1$ So we need the species that does not have exactly one unpaired electron.

Check option A: $\mathrm{O_2^+}$

Neutral $\mathrm{O_2}$ has 2 unpaired electrons in antibonding $\pi^*$ orbitals.

For $\mathrm{O_2^+}$, one electron is removed from a $\pi^*$ orbital, so it has 1 unpaired electron.

Hence, $\mu = 1.73\ \text{BM}$

Check option B: $\mathrm{CuI}$

In $\mathrm{CuI}$, $\mathrm{Cu} = +1$ so copper is $\mathrm{Cu^+}$.

Electronic configuration of Cu: $\mathrm{Cu}: [Ar]3d^{10}4s^1$ Therefore, $\mathrm{Cu^+}: [Ar]3d^{10}$

This has no unpaired electron.

So, $\mu = 0\ \text{BM}$ and not $1.73\ \text{BM}$.

Check option C: $[Cu(NH_3)_4]Cl_2$

The complex ion is $[Cu(NH_3)_4]^{2+}$ so copper is in $+2$ oxidation state: $\mathrm{Cu^{2+}}: 3d^9$

A $d^9$ configuration has 1 unpaired electron.

Hence, $\mu \approx 1.73\ \text{BM}$

Check option D: $\mathrm{O_2^-}$

Neutral $\mathrm{O_2}$ has 2 unpaired electrons. Adding one electron gives $\mathrm{O_2^-}$ (superoxide), so one $\pi^*$ orbital gets paired and 1 unpaired electron remains.

Hence, $\mu = 1.73\ \text{BM}$

Conclusion

The only species that does not have magnetic moment $1.73\ \text{BM}$ is $\boxed{\mathrm{CuI}}$

Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

They match.