Boron does not form B^{3+} cation easily. It | vedclass

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(A) The atomic number of Boron is $5$ and its electronic configuration is $1s^2 2s^2 2p^1$. Due to the small size of the Boron atom,the ionization ent

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energy required to form $B^{3+}$ ion is far more than that which would be compensated by lattice energies or hydration energies of such ion

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(A) The atomic number of Boron is $5$ and its electronic configuration is $1s^2 2s^2 2p^1$. Due to the small size of the Boron atom,the ionization enthalpies required to remove three electrons are extremely high. The sum of the first,second,and third ionization enthalpies is so large that it cannot be compensated by the lattice energy (in case of ionic solids) or hydration energy (in case of aqueous solutions). Therefore,Boron does not form $B^{3+}$ cations easily.

Boron does not form $B^{3+}$ cation easily. It is due to

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