In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions)
1 answer:
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Answer : The correct option is, (D)
Explanation :
Effective nuclear charge : It is defined as the attraction of the protons present in the nucleus of an atom to the outermost electrons.
For ions, the effective nuclear charge changes than the neutral atom.
There are two types of ions:
Cations: They are formed when an atom looses its valence electrons. They are positive ions.
Anions: They are formed when an atom gain electrons in its outermost shell. They are negative ions.
For positive ions, the removal of electron increases the nuclear charge for an outermost electron because the outermost electrons are more strongly attracted by the nucleus. Thus, the effective nuclear charge increases for cations.
From this we conclude that, the size of the cation is smaller than their neutral atom because it has less number of electrons while its nuclear charge remains the same. So, the nucleus attracts the electron more towards itself and leads to the decrease in size.
For negative ions, the addition of electron decreases the nuclear charge for an outermost electron because the outermost electrons are less strongly attracted by the nucleus. Thus, the effective nuclear charge decreases for anions.
From this we conclude that, the size of the anion is greater than their neutral atom because it has more number of electrons while its nuclear charge remains the same. So, the nucleus attracts the electron less towards itself and leads to the increase in size.
Thus, the increasing order of radius of ions will be:
Hence, the smallest radius of ion is,