Question

Among the elements $A,B,C$ and $D$ having atomic numbers $7,8,9$ and $12$ respectively, the element with the smallest size and highest $I.E$ is:

• A. $A$
• B. $B$
• C. $C$
• D. $D$

Answer 1

Answer: (C)

$C$

Here atomic number of $A,B,C,$ and $D$ are $7,8,9,$ and $12$ respectively.

Thus element $A,B,C,$ and $D$ are $N,O,F,$ and $Mg$.

Now the atomic size decreases as we move left to right in any period.

Also, the ionization enthalpy also increases as we move from left to right in any period.

Here element $C$ i.e, $F$ is present in the extreme right of the periodic table.

Thus the element with the smallest size and highest ionization enthalpy is $C$.