Question

which of the following radii comparisons is correct?
answer:
a $n^{3 - } < n$
b $ti^{4 + } > ti^{3 + }$
c $in^{3 + } < in^{ + }$
d $c^{4 - } > c^{4 + }$
e $as^{3 - } < as^{3 + }$

Explanation:

• For \(N^{3 +}\) and \(N\): Cations have smaller radii than their parent atoms. When \(N\) loses 3 electrons to form \(N^{3+}\), the effective nuclear charge per electron increases (since there are fewer electrons being attracted by the same number of protons in the nucleus). So, \(N^{3+}
• For \(Ti^{4 +}\) and \(Ti^{2+}\): As the positive charge on the ion increases (more electrons are lost), the ionic radius decreases. \(Ti^{4+}\) has lost more electrons than \(Ti^{2+}\), so \(Ti^{4+}Ti^{2+}\).
• For \(In^{3+}\) and \(In^{+}\): \(In^{3+}\) has lost more electrons than \(In^{+}\). As the positive charge increases (more electrons removed), the ionic radius decreases. So \(In^{3+}
• For \(C^{4+}\) and \(C^{4 -}\): Anions (\(C^{4-}\)) have larger radii than cations (\(C^{4+}\)) of the same element. When electrons are added (to form an anion), electron - electron repulsions increase and the electron cloud expands. When electrons are removed (to form a cation), the electron cloud contracts. So \(C^{4+}C^{4}\).
• For \(As^{3-}\) and \(As^{5+}\): Anions (\(As^{3-}\)) have larger radii than cations (\(As^{5+}\)). \(As^{3-}\) has more electrons (due to the negative charge), and electron - electron repulsions cause the electron cloud to expand. \(As^{5+}\) has lost more electrons (more positive charge), so its electron cloud is more contracted. So \(As^{3-}>As^{5+}\), not \(As^{3 -}

Answer:

A. \(N^{3+}