Question

which of the following elements has the lowest electronegativity value?
a cl
b s
c p
d si
e i

Explanation:

To determine which element has the lowest electronegativity, we use the trend of electronegativity in the periodic table. Electronegativity generally increases from left to right across a period and decreases from top to bottom within a group.

Step 1: Identify the period and group of each element

• Cl (Chlorine): Period 3, Group 17 (Halogens)
• S (Sulfur): Period 3, Group 16 (Chalcogens)
• P (Phosphorus): Period 3, Group 15 (Pnictogens)
• Si (Silicon): Period 3, Group 14 (Carbon Group)
• I (Iodine): Period 5, Group 17 (Halogens)

Step 2: Analyze the trend across Period 3

In Period 3 (elements from left to right: Na, Mg, Al, Si, P, S, Cl), electronegativity increases as we move from left to right. So, the order of electronegativity for Period 3 elements (Si, P, S, Cl) is:
$\text{Si} < \text{P} < \text{S} < \text{Cl}$

Step 3: Compare Iodine (I) with Chlorine (Cl)

Iodine is in Group 17, same as Chlorine, but in Period 5 (below Period 3). Since electronegativity decreases down a group, $\text{I} < \text{Cl}$

Step 4: Compare Si with I

Now, we compare Si (Period 3, Group 14) and I (Period 5, Group 17). Let's recall the electronegativity values (approximate Pauling scale):

• Si: ~1.8
• I: ~2.5

So, $\text{Si} < \text{I}$

Combining all, the order from lowest to highest electronegativity among the options is:
$\text{Si} < \text{P} < \text{S} < \text{Cl}$ and $\text{Si} < \text{I}$

Thus, Si has the lowest electronegativity among the given elements.

Answer:

d. Si