The correct order of the decreasing ionic radii among the following isoelectronic species is :
Answers
C.
S2- > Cl- > K+ > Ca2+
Ionic space radii space proportional to space charge space on space anion space proportional to space fraction numerator 1 over denominator charge space on space cation end fraction
During the formation of a cation, the electrons are lost from the outer shell and the remaining electrons experience a great force of attraction by the nucleus, ie, attracted more towards the nucleus. In other words, nucleus holds the remaining electrons more tightly and result in decreased radii.
However, in the case of anion formation, the addition of an electron (s) takes place in the same outer shell, thus the hold of the nucleus on the electrons of outer shell decreases and this result in increased ionic radii.
Thus, the correct order of ionic radii is
S2- > Cl- > K+ > Ca2+
Answer:
S
−2
>Cl
−
>K
+
>Ca
+2
Explanation:
In iso-electronic species, if a positive charge is more then they will have a very less ionic radius. Because here protons are more than electrons and protons attract electrons to come close to the nucleus. But in anionic iso-electronic species, electrons are more than protons. Protons which are in less number can't attract all the electrons to come closer. So, ionic radius increases in the case of anionic species, if negative charge increases. Therefore the correct order is
S
−2
>Cl
−
>K
+
>Ca
+2