Question

Electronegativity difference between a group 15 element and $\text{P}$ is less than electronegativity difference between another group 15 element and $\text{P}$. Those group 15 elements respectively are :

• A. $\text{Bi}, \text{N}$
• B. $\text{Sb}, \text{As}$
• C. $\text{Sb}, \text{Bi}$
• D. $\text{N}, \text{As}$

Hint:

Electronegativity decreases down a group. The element closest to $\text{P}$ in the group ($\text{As}$) will have the smallest $\text{EN}$ difference, while the element farthest ($\text{N}$) will have the largest $\text{EN}$ difference.

Answer 1

The Correct Option is A

The Group 15 elements are $\text{N}, \text{P}, \text{As}, \text{Sb}, \text{Bi}$. Electronegativity ($\text{EN}$) generally decreases down the group.
Approximate Pauling EN values: $\text{N} (3.0)$, $\text{P} (2.1)$, $\text{As} (2.0)$, $\text{Sb} (1.9)$, $\text{Bi} (1.9)$.
We calculate $\text{EN}$ difference with $\text{P} (2.1)$:
1. $|\text{EN}(\text{P}) - \text{EN}(\text{As})| = |2.1 - 2.0| = 0.1$.
2. $|\text{EN}(\text{P}) - \text{EN}(\text{Sb})| = |2.1 - 1.9| = 0.2$.
3. $|\text{EN}(\text{P}) - \text{EN}(\text{Bi})| = |2.1 - 1.9| = 0.2$.
4. $|\text{EN}(\text{P}) - \text{EN}(\text{N})| = |2.1 - 3.0| = 0.9$.
Let $E_1$ be the element with smaller EN difference, and $E_2$ be the element with larger EN difference.
Smallest difference with $\text{P}$ is $\text{As}$ (0.1). Second smallest is $\text{Sb}$ or $\text{Bi}$ (0.2).
Largest difference with $\text{P}$ is $\text{N}$ (0.9).
We need a pair (Element 1, Element 2) such that $\Delta \text{EN}(\text{P}, E_1)<\Delta \text{EN}(\text{P}, E_2)$.
Checking Option (1): $(\text{Bi}, \text{N})$. $\Delta \text{EN}(\text{P}, \text{Bi}) = 0.2$. $\Delta \text{EN}(\text{P}, \text{N}) = 0.9$.
$0.2<0.9$. The first element ($\text{Bi}$) has a smaller $\Delta \text{EN}$ than the second element ($\text{N}$). (1) is Correct.