Boron does not form \[{B^{3 + }}\] ions whereas Al forms \[A{l^{3 + }}\] ions. This because:
A.The size of B atom is smaller than that of $Al$
B.The sum of \[I{E_1} + I{E_2} + I{E_3}\] of boron is equal to that of $Al$
C.The sum of \[I{E_1} + I{E_2} + I{E_3}\] of $Al$is much higher than that of B
D.Both A and B
Answer
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Hint: We need to remember that the boron is a chemical element having the symbol B with atomic number five and the aluminium is the chemical element having the symbol Al with atomic number \[13\]. And the ionization energy is the energy required to remove the electron from the isolated gaseous atom. It will depend on the size of the atom. Because, the ionization decreases with increasing size.
Complete answer:
We need to know that the boron does not form \[{B^{3 + }}\]ions, but it is not only due to the size of the boron atom. Hence, option (A) is incorrect.
Boron does not form \[{B^{3 + }}\] ions whereas Al forms \[A{l^{3 + }}\] ions and it not only depends on the ionization energy. Hence, the option (B) is incorrect.
The sum \[I{E_1} + I{E_2} + I{E_3}\] of aluminium is less than boron. of Hence, option (C) is incorrect.
Boron does not form \[{B^{3 + }}\] ions whereas Al forms \[A{l^{3 + }}\] ions. Because, the size of boron is less than aluminium, and therefore the sum of \[I{E_1} + I{E_2} + I{E_3}\] of boron is higher than that of $Al$ .
The electronic configuration of B \[ = 1{s^2}1{s^2}2{p^1}\]
And after the removal of one electron, the electronic configuration of B will become \[1{s^2}1{s^2}\]. And it is completely filled s – orbital which is highly stable. And it is not easy to remove the electron from the isolated gaseous atom. Hence, the second and third ionization energy of boron become high. Therefore, the sum of \[I{E_1} + I{E_2} + I{E_3}\];;is very high. Thus, the formation of \[{B^{3 + }}\] is not possible.
Hence, the option (D) is correct.
Note:
We have to remember that the oxidation state of an atom is equal to the number of losses of electrons by the atom or the number of gains of electrons by the atom. And it mainly depends on the size of an atom and the ionization energy of the atom. Because, when the size is increasing, the removal of electrons becomes difficult and accordingly the ionization energy become very high.
Complete answer:
We need to know that the boron does not form \[{B^{3 + }}\]ions, but it is not only due to the size of the boron atom. Hence, option (A) is incorrect.
Boron does not form \[{B^{3 + }}\] ions whereas Al forms \[A{l^{3 + }}\] ions and it not only depends on the ionization energy. Hence, the option (B) is incorrect.
The sum \[I{E_1} + I{E_2} + I{E_3}\] of aluminium is less than boron. of Hence, option (C) is incorrect.
Boron does not form \[{B^{3 + }}\] ions whereas Al forms \[A{l^{3 + }}\] ions. Because, the size of boron is less than aluminium, and therefore the sum of \[I{E_1} + I{E_2} + I{E_3}\] of boron is higher than that of $Al$ .
The electronic configuration of B \[ = 1{s^2}1{s^2}2{p^1}\]
And after the removal of one electron, the electronic configuration of B will become \[1{s^2}1{s^2}\]. And it is completely filled s – orbital which is highly stable. And it is not easy to remove the electron from the isolated gaseous atom. Hence, the second and third ionization energy of boron become high. Therefore, the sum of \[I{E_1} + I{E_2} + I{E_3}\];;is very high. Thus, the formation of \[{B^{3 + }}\] is not possible.
Hence, the option (D) is correct.
Note:
We have to remember that the oxidation state of an atom is equal to the number of losses of electrons by the atom or the number of gains of electrons by the atom. And it mainly depends on the size of an atom and the ionization energy of the atom. Because, when the size is increasing, the removal of electrons becomes difficult and accordingly the ionization energy become very high.
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