Question

Second I.E. amongest $Li,Be,B$ and $C$ is lowest for:

• A. $Li$
• B. $Be$
• C. $B$
• D. $C$

Answer 1

Answer: (B)

$Be$

$Li\to 1s^{2}2s^1$

$Be\to 1s^{2}2p^2$

$B\to 1s^{2}2s^{2}2p^1$

$C\to 1s^{2}2s^{2}2p^2$

The ionization energy increases when we move from left to right along a period.

So I.E of $Li$ will be least, but since in $Li$ the second electron will have to be removed from the $1^{st}$ shell which is very closely bound to the nucleus, hence large amount of energy will be required for the same.

But in case of $Be$, the second I.E will be minimum, as removal of electrons will give a stable fully filled quantum shell.

$Be$ has the lowest $2^{nd}$ I.E.

Electronegativity is a chemical property that says how well an atom can attract electrons towards itself. The electronegativity of an atom is influenced by the atom's atomic number and the distance between the atom's valence electron.

The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.