Question

An element $X$ (atomic number $17$) reacts with an element $Y$ (atomic number $20$) to form a divalent halide. Where in the periodic table is element, $X$ placed?

• A. Group $17$ and $3^{rd}$ period
• B. Group $2$ and $4^{th}$ period
• C. Group $17$ and $4^{rd}$ period
• D. Group $2$ and $5^{th}$ period

Answer 1

The correct option is A Group $17$ and $3^{rd}$ period

As atomic number of $X$ is $17$ it will have $17$ electrons if we write there electronic configuration that is $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$.

From this we could understand that it will be in the period $3$ because of outer most shell is $3$ and as it has only one unpaired electron and has to accept only one electron to get octet configuration it will in the group of halogens that is $17$th group.

In case of $Y$ even it has $20$ electrons that is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^2$. It must be in $4$th period and as it has to donate $2$ electrons to get configuration it must be in $2$nd group.

Hence,

$X$ belongs to Group $17$ and $3^{rd}$ period
$Y$ belongs to Group $2$ and $4^{th}$ period.

Hence the correct option is A.