What is the percent of Boron in Sr3(BO3)2

5.68%

Explanation:

Percent Composition of Boron = Atomic Mass of Boron / Molar mass of Sr3(BO3)2  * 100%

Atomic mass of B = 10.811 g/mol

Molar mass of Sr3(BO3)2 = 380.4784 g/mol

Percent composition = 2 (10.811) / 380.4784  * 100

Percent composition = 21.622 / 380.4784  * 100

Percent composition = 5.68%

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.