A compound of vanadium chloride has spin only magnetic moment of 1.73 BM. Its formula is:

The spin only magnetic moment ($$\mu$$) is given by the formula $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons. We are given $$\mu = 1.73$$ BM. So, we set up the equation:

$$\sqrt{n(n+2)} = 1.73$$

Squaring both sides to solve for $$n$$:

$$n(n+2) = (1.73)^2$$

Calculating $$(1.73)^2$$:

$$1.73 \times 1.73 = 2.9929 \approx 3.00$$

So,

$$n(n+2) = 3$$

Rearranging into a quadratic equation:

$$n^2 + 2n - 3 = 0$$

Factoring the quadratic equation:

We need two numbers that multiply to $$-3$$ and add to $$2$$. These are $$3$$ and $$-1$$. So,

$$(n + 3)(n - 1) = 0$$

Thus, $$n = -3$$ or $$n = 1$$. Since the number of unpaired electrons cannot be negative, $$n = 1$$. Therefore, the vanadium ion in the compound has one unpaired electron.

Next, we determine the oxidation state of vanadium in each compound and find the number of unpaired electrons. Vanadium (atomic number 23) has the electron configuration $$[Ar] 4s^2 3d^3$$. When forming ions, electrons are removed first from the $$4s$$ orbital and then from the $$3d$$ orbital.

• For VCl$$_2$$: Oxidation state of vanadium is $$+2$$ (since $$x + 2(-1) = 0 \rightarrow x = +2$$). The ion is V$$^{2+}$$, with configuration $$[Ar] 3d^3$$. In a free ion, the three electrons in the $$d$$ orbitals occupy separate orbitals with parallel spins (Hund's rule), so there are three unpaired electrons ($$n = 3$$).
• For VCl$$_5$$: Oxidation state of vanadium is $$+5$$ (since $$x + 5(-1) = 0 \rightarrow x = +5$$). The ion is V$$^{5+}$$, with configuration $$[Ar] 3d^0$$. There are no unpaired electrons ($$n = 0$$).
• For VCl$$_4$$: Oxidation state of vanadium is $$+4$$ (since $$x + 4(-1) = 0 \rightarrow x = +4$$). The ion is V$$^{4+}$$, with configuration $$[Ar] 3d^1$$. There is one unpaired electron ($$n = 1$$).
• For VCl$$_3$$: Oxidation state of vanadium is $$+3$$ (since $$x + 3(-1) = 0 \rightarrow x = +3$$). The ion is V$$^{3+}$$, with configuration $$[Ar] 3d^2$$. In a free ion, the two electrons occupy separate orbitals with parallel spins, so there are two unpaired electrons ($$n = 2$$).

Only VCl$$_4$$ (vanadium in $$+4$$ oxidation state) has one unpaired electron, matching the given magnetic moment of 1.73 BM.

Verifying the magnetic moments:

• VCl$$_2$$: $$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$$ BM (not 1.73 BM).
• VCl$$_5$$: $$\mu = \sqrt{0(0+2)} = 0$$ BM (not 1.73 BM).
• VCl$$_4$$: $$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$$ BM (matches).
• VCl$$_3$$: $$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$$ BM (not 1.73 BM).

Hence, the correct answer is Option C.