Which of these statements is not true?

We are given four statements and need to determine which one is not true. Let's examine each option step by step.

Starting with option A: "LiAlH$$_4$$ is versatile reducing agent in organic synthesis." Lithium aluminium hydride (LiAlH$$_4$$) is indeed a versatile reducing agent. It reduces a wide range of functional groups, such as carbonyl compounds (aldehydes, ketones, carboxylic acids, esters) to alcohols, and also reduces nitriles, amides, and other groups. This statement is true.

Now option B: "NO$$^+$$ is isoelectronic with O$$_2$$." Isoelectronic species have the same total number of electrons. Let's calculate the electrons in each. For NO$$^+$$: Nitrogen (atomic number 7) has 7 electrons, oxygen (atomic number 8) has 8 electrons. Since NO$$^+$$ is a cation formed by losing one electron from neutral NO, neutral NO would have 7 + 8 = 15 electrons. Therefore, NO$$^+$$ has 15 - 1 = 14 electrons. For O$$_2$$: Each oxygen atom has 8 electrons, so O$$_2$$ has 8 + 8 = 16 electrons. Since 14 ≠ 16, NO$$^+$$ and O$$_2$$ are not isoelectronic. This statement is false.

Moving to option C: "Boron is always covalent in its compounds." Boron is a metalloid with high ionization energy and small atomic size, which favors covalent bonding. Examples include boron trifluoride (BF$$_3$$) and boric acid (H$$_3$$BO$$_3$$), which are covalent. Even in metal borides, bonding is predominantly covalent. Thus, this statement is true.

Finally, option D: "In aqueous solution, the Tl$$^+$$ ion is much more stable than Tl(III)." Thallium (Tl) exhibits +1 and +3 oxidation states. Due to the inert pair effect, which is strong in heavier group 13 elements, the +1 state becomes more stable than the +3 state for thallium. In aqueous solutions, Tl$$^+$$ is stable, while Tl(III) acts as a strong oxidizing agent and is easily reduced to Tl$$^+$$. Hence, this statement is true.

After evaluating all options, we find that option B is not true. Hence, the correct answer is Option B.