The spin-only magnetic moment value of B$$_2^+$$ species is _________ $$\times 10^{-2}$$ BM (Nearest integer)
[Given: $$\sqrt{3}$$ = 1.73]

We begin with the electronic details of a single boron atom. A boron atom has atomic number 5, so its ground-state electronic configuration is $$1s^2\,2s^2\,2p^1$$.

In the diatomic ion B$$_2^{+}$$ we have two boron atoms, but one electron is missing because of the positive charge. Hence the total number of electrons present in the molecule is

$$2 \times 5 \;-\; 1 \;=\; 9.$$

For homonuclear diatomic molecules of the first-row elements up to nitrogen (Z ≤ 7), the well-established order of molecular orbitals is

$$\sigma(1s),\;\sigma^{\*}(1s),\;\sigma(2s),\;\sigma^{\*}(2s),\; \pi(2p_x)=\pi(2p_y),\;\sigma(2p_z).$$

Now we start filling these molecular orbitals with the nine electrons of B$$_2^{+}$$, always obeying the Pauli exclusion principle (maximum two electrons per orbital with opposite spins) and Hund’s rule (maximum multiplicity):

$$ \begin{aligned} \sigma(1s) &:& 2 \text{ electrons} \\ \sigma^{\*}(1s) &:& 2 \text{ electrons} \\ \sigma(2s) &:& 2 \text{ electrons} \\ \sigma^{\*}(2s) &:& 2 \text{ electrons} \\ \pi(2p_x) \text{ or } \pi(2p_y) &:& 1 \text{ electron (unpaired)} \end{aligned} $$

After placing eight electrons in the four lowest orbitals, one electron still remains. The degenerate $$\pi(2p_x)$$ and $$\pi(2p_y)$$ orbitals are next in energy, so the ninth electron enters either of these two orbitals singly. Consequently, there is exactly

$$n = 1$$

unpaired electron in B$$_2^{+}$$.

To find the spin-only magnetic moment, we explicitly state the formula first:

$$\mu_\text{spin only} = \sqrt{n(n+2)}\;\text{BM},$$

where $$n$$ is the number of unpaired electrons and “BM” stands for Bohr magneton.

Substituting $$n = 1$$ gives

$$ \mu = \sqrt{1(1+2)} \\ \mu = \sqrt{3}\;\text{BM}. $$

We are given $$\sqrt{3} = 1.73$$. Therefore,

$$\mu = 1.73\;\text{BM}.$$

The problem asks for the value in the form “ ______ $$\times 10^{-2}$$ BM” and instructs us to give the nearest integer. Writing $$1.73\;\text{BM}$$ in that format, we have

$$1.73\;\text{BM} = 173 \times 10^{-2}\;\text{BM}.$$

Because 173 is already an integer, no further rounding is required.

Hence, the correct answer is Option 173.