Question

electronegativity < 1.0 1.0 ≤ electronegativity < 2.0 2.0 ≤ electronegativity < 3.0 3.0 ≤ electronegativity < 4.0 1 h 2.20 3 li 0.98 4 be 1.57 11 na 0.93 12 mg 1.31 19 k 0.82 20 ca 1.00 21 sc 1.36 22 ti 1.54 23 v 1.63 24 cr 1.66 25 mn 1.55 26 fe 1.83 27 co 1.88 28 ni 1.91 40 select the correct answer. which element can bond with titanium (ti) to form a polar covalent bond? a. beryllium (be) b. fluorine (f) c. manganese (mn) d. nitrogen (n) e. oxygen (o)

Explanation:

Step1: Recall polar - covalent bond condition

A polar covalent bond forms when there is an electronegativity difference between the two atoms in the bond, typically in the range of 0.5 - 1.9.

Step2: Find electronegativity of Ti

From the given data, the electronegativity of titanium (Ti) is 1.54.

Step3: Check electronegativity of each option

• For beryllium (Be), electronegativity = 1.57. The difference $|1.57 - 1.54|= 0.03$.
• For fluorine (F), electronegativity is high (around 3.98). The difference $|3.98 - 1.54| = 2.44$.
• For manganese (Mn), electronegativity = 1.55. The difference $|1.55 - 1.54|=0.01$.
• For nitrogen (N), electronegativity is around 3.04. The difference $|3.04 - 1.54| = 1.5$.
• For oxygen (O), electronegativity is around 3.44. The difference $|3.44 - 1.54| = 1.9$.

Among these, fluorine (F) has an electronegativity difference with Ti that is within the range for a polar - covalent bond.

Answer:

B. fluorine (F)