Question

b_18 ionic radii
which would have the
largest radius? smallest rad
largest smallest
a. s²- cl-
b. ca²+ s²-
c. k+ ca²+
d. s²- ca²+
uk
kentucky

Explanation:

Step1: Consider the isoelectronic series concept

Isoelectronic species have the same number of electrons. \(S^{2 -}\), \(Cl^{-}\), \(K^{+}\), \(Ca^{2+}\) are isoelectronic (all have 18 electrons).

Step2: Apply the rule for ionic radius in isoelectronic species

For isoelectronic species, the greater the nuclear charge (\(Z\), number of protons in the nucleus), the smaller the ionic radius. The nuclear charges are:

• For \(S^{2-}\): \(Z = 16\)
• For \(Cl^{-}\): \(Z=17\)
• For \(K^{+}\): \(Z = 19\)
• For \(Ca^{2+}\): \(Z=20\)

As \(Z\) increases, the electrons are more strongly attracted to the nucleus, reducing the ionic radius. So, \(S^{2-}\) has the largest radius (smallest \(Z\) among them) and \(Ca^{2+}\) has the smallest radius (largest \(Z\) among them)

Answer:

D. \(S^{2-}\) (Largest), \(Ca^{2+}\) (Smallest)