Question

The ${IP}_1,{IP}_2,{IP}_3,{IP}_4$ and ${IP}_5$ of an element are $7.1,14.3,34.5,46.8,162.2eV$ respectively.

The element is likely to be:

• A. $Na$
• B. $Si$
• C. $F$
• D. $Ca$

Answer 1

Answer: (B)

$Si$

${IP}_1,{IP}_2,{IP}_3,{IP}_4$ and ${IP}_5$ of an element are $7.1,14.3,34.5,\mathrm{46.8.162.2}eV$ respectively. The element is likely to be $Si$. The jump in $IP$ values exist in ${IP}_5$ and thus, removal of fifth electron occurs from inner shell. Thus, the element contains four electrons in its valence shell.

The electronic configuration of silicon is {$Ne$}$3s^2,3p^2$. These 4 electrons need lesser $IP$ than the fifth electron which has to be released from $2p^6$, requires very high $IP$.

Hence, option B is correct.