Amongst the metal Be, Mg, Ca, and Sr of group 2 of the periodic table, the least ionic chloride would be formed by
A. Mg
B. Be
C. Ca
D. Sr
Last updated date: 16th Mar 2023
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Answer
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Hint: Be, Sr, Ca, and Mg are alkaline earth metals and are from group-2 of the periodic table. These metals in reaction with chlorine form metal chlorides that are ionic. The ionic character of the chlorides can be explained based on Fajan's rule.
Complete Step by Step Solution:
The chemical bonds in chemistry are halved into two types, i.e., ionic and covalent bonds.
An ionic bond or electrovalent bond is formed by the complete gain or loss of electrons.
A covalent bond is formed by the sharing of electrons.
Fajan’s rule states that the magnitude of polarisation of an anion relies on the polarising power of the cation and the polarizability of the anion.
The polarising power of a cation is the capacity of the cation to polarise an anion.
For an ionic bond, the polarising power of cation rises with a rise in charge density, charge on cation, and size of cation and lowers with the growing size of the anion.
For the given question, the anion is the same i.e., chloride ion. As Be, Sr, Ca, and Mg are from the second group their cations have a +2 charge. We have to compare the size of the cations i.e., \[{\rm{B}}{{\rm{e}}^{{\rm{2 + }}}}{\rm{,M}}{{\rm{g}}^{{\rm{2 + }}}}{\rm{,C}}{{\rm{a}}^{{\rm{2 + }}}}{\rm{,S}}{{\rm{r}}^{{\rm{2 + }}}}\].
On moving down a group, atomic radii increase. The valence electrons are stored in higher energy levels. The distance between the nucleus and the outermost valence electron increases. \[{\rm{B}}{{\rm{e}}^{{\rm{2 + }}}}\] has the smallest size. So, polarising power is also insufficient to polarise the chloride anion effectively.
So, beryllium chloride is the least ionic.
So, option B is correct.
Note: The polarizability of an anion is the ability of an anion to go through polarisation. If an anion is distorted easily by the action of a cation, the polarizability of the anion is the highest. The polarizability of an anion rises with an increase in size and negative charge of the anion.
Complete Step by Step Solution:
The chemical bonds in chemistry are halved into two types, i.e., ionic and covalent bonds.
An ionic bond or electrovalent bond is formed by the complete gain or loss of electrons.
A covalent bond is formed by the sharing of electrons.
Fajan’s rule states that the magnitude of polarisation of an anion relies on the polarising power of the cation and the polarizability of the anion.
The polarising power of a cation is the capacity of the cation to polarise an anion.
For an ionic bond, the polarising power of cation rises with a rise in charge density, charge on cation, and size of cation and lowers with the growing size of the anion.
For the given question, the anion is the same i.e., chloride ion. As Be, Sr, Ca, and Mg are from the second group their cations have a +2 charge. We have to compare the size of the cations i.e., \[{\rm{B}}{{\rm{e}}^{{\rm{2 + }}}}{\rm{,M}}{{\rm{g}}^{{\rm{2 + }}}}{\rm{,C}}{{\rm{a}}^{{\rm{2 + }}}}{\rm{,S}}{{\rm{r}}^{{\rm{2 + }}}}\].
On moving down a group, atomic radii increase. The valence electrons are stored in higher energy levels. The distance between the nucleus and the outermost valence electron increases. \[{\rm{B}}{{\rm{e}}^{{\rm{2 + }}}}\] has the smallest size. So, polarising power is also insufficient to polarise the chloride anion effectively.
So, beryllium chloride is the least ionic.
So, option B is correct.
Note: The polarizability of an anion is the ability of an anion to go through polarisation. If an anion is distorted easily by the action of a cation, the polarizability of the anion is the highest. The polarizability of an anion rises with an increase in size and negative charge of the anion.
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