Question

The common oxidation states of elements of 15th group elements are:

• A. −3, +3, +5
• B. −3, +2, +5
• C. −1, +3, +5
• D. −2, +3, +5

Hint:

Group 15 shows −3, +3, +5 due to 5 valence electrons and inert pair effect in heavier elements.

Answer 1

The Correct Option is A

Step 1: Concept: Understanding the Concept.
Group 15 elements (N, P, As, Sb, Bi) all have the outer configuration $ns^2np^3$, meaning 5 electrons in the valence shell. Their oxidation states come from how many of these electrons they share or lose.

Step 2: Key Formula or Approach.
A half filled $np^3$ configuration is extra stable, so these elements can gain 3 electrons to complete the octet, giving the $-3$ state, or they can share electrons in bonding to show positive states, most commonly $+3$ (using only the $p$ electrons) or $+5$ (using both $s$ and $p$ electrons).

Step 3: Detailed Explanation.
Nitrogen and phosphorus readily show all three: $-3$, $+3$, and $+5$. Going down the group, the inert pair effect grows stronger, meaning the pair of $s$ electrons becomes less willing to take part in bonding. Because of this, heavier members like antimony and bismuth increasingly favor the lower $+3$ state over $+5$, but $+3$, $+5$, and $-3$ remain the three states recognized as common across the group.

Step 4: Rule out the other options.
A $-2$ state would mean gaining only 2 electrons, which does not complete the stable octet for these atoms, and a $+2$ state does not match the $s^2p^3$ electron count either, so neither is a typical oxidation state here.

Step 5: Final Answer.
The common oxidation states of group 15 elements are $-3, +3, +5$. \[ \boxed{-3,\ +3,\ +5} \]