Question

The correct order of the decreasing ionic radii
among the following isoelectronic species are :
Ca2+ >K+ >52->c1
O
ci > 52- > Ca2+ >K+
5
52->C1 >K+ > Ca2+
O
K+ > Ca2+ > Cr > 52-​

Answer 1

Answer:

The correct order of the decreasing ionic radii among the following isoelectronic species are:

Option 1) Ca2+ > K+ > S2- > Cl- Option 2) Cl- > S2- > Ca2+ > K+ Option 3) S2- > Cl- > K+ > Ca2+ Option 4) K+ > Ca2+ > Cl- > S2-

Answers (1)

A Amit Kumar

Answered 2 years ago

First, we note that this is an isoelectronic series of ions, with all ions having 18 electrons. In such a series, size decreases as the nuclear charge (atomic number) of the ion increases. The atomic numbers of the ions are S (16), Cl (17), K (19), and Ca (20). Thus, the ions decrease in size in the order S2–> Cl–> K+> Ca2+.

Answer 2

Answer:

Explanation:

For isoelectronic species, anions will be larger than cations.

Anions are formed when an electron is gained. As a result there is greater electron-electron repulsion and the effective nuclear charge experienced by an electron decreases. The size becomes bigger to account for these changes. Hence an electron will be larger than a neutral atom.

Similarly, cations are formed when electrons are lost. Thus there is a greater nuclear charge experienced by each electron. The electrons are pulled closer towards the nucleus. Hence the cations are smaller than the neutral atom.

For isoelectronic species,

Anion > Neutral atom > Cation.

In this case, S

2−

will be the largest and Ca

2+

will be the smallest.

The order is S

2−

>Cl

−

>K

+

>Ca

2+

.

Hence, option A is the right answer.