Question

explain breifly why the 1st ionization energy of "B" is less than of "Be" less than of the atomic number of "B" is greater than that of "Be"​

Answer 1

Explanation:

B → Boron → Atomic number 5

Be → beryllium → Atomic number 4

B - Hybridization → 1S² 2S² 2P¹

$\sf Boron - \boxed{↿↾} \: \boxed{↿↾} \: \boxed{ \small↾ }\huge\boxed{ }\huge\boxed{ }$

Be - Hybridization → 1S² 2S²

$\sf Beryllium →\boxed{↿↾} \: \boxed{↿↾} \: \huge\boxed{ }\huge\boxed{ }\huge\boxed{ }$

from above data we can see that the outermost orbital of Beryllium i.e. 2S is fully filled & that of boron have 1 electron in 2P orbital. It will require more energy to remove one electron from beryllium Because it will disturb the stability, on other hand the boron will lose one electron easily to attain full filled outermost electron hence it will require less energy for 1st ionization.

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