Let's analyze each option for the compound NO[BF\u2084]:

Concept: Ionic compounds, molecular orbital theory, bond order, magnetic properties, and bond energy.

The compound NO[BF\u2084] is an ionic compound consisting of the nitrosonium ion (NO\u207a) and the tetrafluoroborate anion (BF\u2084\u207b).

Option Analysis:

A) Magnetic moment of the species is zero

The species are NO\u207a and BF\u2084\u207b.

• For NO\u207a: Nitrogen has 7 electrons, Oxygen has 8 electrons. So, NO has 15 electrons. NO\u207a has 15 - 1 = 14 electrons. According to molecular orbital theory, 14-electron species like N\u2082 and CO are diamagnetic (all electrons are paired). Thus, NO\u207a is diamagnetic.
• For BF\u2084\u207b: Boron has 3 valence electrons, each Fluorine has 7 valence electrons. Total valence electrons = 3 + 4(7) + 1 (for the negative charge) = 3 + 28 + 1 = 32 electrons. All electrons are paired in BF\u2084\u207b, making it diamagnetic.

Since both ions are diamagnetic, the compound NO[BF\u2084] is diamagnetic, and its magnetic moment is zero. Therefore, statement A is correct.

B) B \u2013\u2013 F bond energy is lower in this compound than in BF\u2083

• In BF\u2083, Boron is sp\u00b2 hybridized and has an incomplete octet (6 valence electrons). It acts as a Lewis acid.
• In BF\u2084\u207b, Boron is sp\u00b3 hybridized and has a complete octet (8 valence electrons). The B-F bonds in BF\u2084\u207b are single bonds.
• In BF\u2083, there is some degree of p\u03c0-p\u03c0 backbonding from Fluorine to Boron, giving the B-F bonds partial double bond character. This backbonding increases the bond strength and thus the bond energy in BF\u2083 compared to the pure single bonds in BF\u2084\u207b.

Therefore, the B-F bond energy is higher in BF\u2083 than in BF\u2084\u207b. Statement B is incorrect.

C) It has 5 \u03c3 and 2\u03c0 bond

• NO\u207a: The bond order is 3 (as discussed below), which means it has one \u03c3 bond and two \u03c0 bonds.
• BF\u2084\u207b: It is a tetrahedral molecule with four B-F single bonds. Each single bond is a \u03c3 bond. So, it has four \u03c3 bonds and zero \u03c0 bonds.

Total bonds in NO[BF\u2084] = (1 \u03c3 + 2 \u03c0) from NO\u207a + (4 \u03c3 + 0 \u03c0) from BF\u2084\u207b = 5 \u03c3 bonds and 2 \u03c0 bonds. Therefore, statement C is correct.

D) NO\u207a bond order is same as in N\u2082

• NO\u207a has 14 electrons (7 from N + 8 from O - 1 for charge).
• N\u2082 has 14 electrons (7 from N + 7 from N).

Both NO\u207a and N\u2082 are isoelectronic with 14 electrons. According to molecular orbital theory, 14-electron species have a bond order of 3. For N\u2082, the electronic configuration is ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^2$. Bond order = $\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-4)=3$. Similarly, for NO\u207a, the bond order is also 3.

Therefore, statement D is correct.

The question asks to select the correct statement(s). Options A, C, and D are correct. However, in typical MCQ format, if only one option is to be selected, there might be an issue with the question or options provided, or it might be a multiple-correct question. Assuming it's a single correct answer question and given the provided correct answer is A, let's re-evaluate if there's any subtlety. All three A, C, and D are factually correct statements based on chemical principles.

If this is a single-choice question and 'A' is the designated answer, it implies that the other correct statements (C and D) might be considered less 'correct' or there's a nuance missed. However, based on standard chemical understanding, A, C, and D are all correct. If forced to choose only one, and given the provided answer is A, we stick to the analysis that A is correct.

Final check:
A) NO\u207a (14e\u207b) is diamagnetic. BF\u2084\u207b (32 valence e\u207b) is diamagnetic. So, the compound is diamagnetic. Magnetic moment is zero. (Correct)
B) BF\u2083 has partial double