According to MOT, which statement is correct about the Boron molecule?
A. It is diamagnetic in nature
B. It is paramagnetic in nature having magnetic moment $2.8BM$ by using spin only formula
C. It is paramagnetic but having magnetic moment $1.7BM$
D. Its bond order is $2$.

Hint: The attractive type of a substance can be dictated by inspecting its electron setup: in the event that it shows unpaired electrons, at that point the substance is paramagnetic; if all electrons are combined, the substance is diamagnetic.

Complete step by step answer:
Paramagnetism alludes to the attractive condition of a particle with at least one unpaired electron. The unpaired electrons are pulled in by an attractive field because of the electrons' attractive dipole minutes. Hund's Rule expresses that electrons must involve each orbital independently before any orbital is doubly involved. This may leave the molecule with numerous unpaired electrons.
Diamagnetic substances are described by matched electrons—aside from in the already talked about instance of progress metals, there are no unpaired electrons. As per the Pauli Exclusion Principle which expresses that no two indistinguishable electrons may take up a similar quantum state simultaneously, the electron turns are situated in inverse ways. This makes the attractive fields of the electrons offset; consequently there is no net attractive second, and the particle can't be pulled in into an attractive field.
The Boron particle has $2{s^2}2{p^1}$ as the electron arrangement. Which means only the $p$ electron is unpaired. Since it has one unpaired electron, it is paramagnetic.
Hence, number of unpaired electrons in boron = $n = 1$
Now spin only magnetic moment of boron = $\mu = \sqrt {n(n + 2)} $
Spin only magnetic moment of boron = $\mu = \sqrt {1(1 + 2)} $
Spin only magnetic moment of boron = $\mu = \sqrt 3 $
Spin only magnetic moment of boron = $\mu = 1.7BM$

Therefore option (C) is the correct answer.

Note:
In the ground state, just one. Yet, for holding, every one of the three valence electrons are utilized, and the molecule procures a somewhat energized state.