Question

Four elements $P,Q,R,S$ have atomic numbers $Z-1,Z,Z+1$ and $Z+2$ respectively. If $Z$ is 17, then bond between which pair of elements will be most ionic?

• A. $S$ and $Q$
• B. $P$ and $R$
• C. $S$ and $R$
• D. $S$ and $P$

Answer 1

The correct option is A $S$ and $Q$

Solution:- (A) $S$ and $Q$

Given:-

$Z=17$

Therefore,

Atomic no. of $P=Z-1=16$

Electronic configuration $=2,8,6$

Atomic no. of $Q=Z=17$

Electronic configuration $=2,8,7$

Atomic no. of $R=Z+1=18$

Electronic configuration $=2,8,8$

Atomic no. of $S=Z+1=19$

Electronic configuration $=2,8,1$

The element $P$ belongs to oxygen family and requires $2$ electrons to complete it's octet.

The element $Q$ belongs to halogen family and requires only $1$ electron to complete its octet.

The element $R$ belongs to the inert gas family.

The element $S$ belongs to alkali metal family requires to lose $1$ electron to get its complete octet.

Hence the bond between $S$ and $Q$ will be most ionic.