Question

Column-I: Property; Column-II: Variation of property; Column-III: Magnitude
$\begin{array}{cccccc}& \text{Column-I}& & \text{Column-II}& & \text{Column-III}\\ \text{(I)}& \text{Electron a affinity}\left(E{A}_1\right)& \left(i\right)& \text{Decreases along the period}& \left(P\right)& \text{Highest in halogen in their respective periods}\\ \left(II\right)& \text{Ionization energy}\left(I{E}_1\right)& \left(ii\right)& \text{Directly proportional to}{Z}_{eff}& \left(Q\right)& \text{Highest in noble gases in their respective periods}\\ \left(III\right)& \text{Electronegativity}& \left(iii\right)& \text{Decreases down the group}& \left(R\right)& \text{Highest in alkali metals in their respective periods}\\ \left(IV\right)& \text{Electropositive character}& \left(iv\right)& \text{Inversly proportional to size}& \left(S\right)& \text{Moderate in noble gases in their respective periods}\end{array}$
The only correct combination for the energy required to knock out most loosely bounded electron from isolated gaseous atom is:

• A. (I) (ii) (P)
• B. (I) (i) (R)
• C. (II) (iii) (Q)
• D. (II) (iv) (R)

Answer 1

The correct option is C

(II) (iii) (Q)

We need to patch the parameter with the trend and an observation of magnitude. This is a slightly tricky one than what we are used to. And since the question indirectly points only to ionization energy, that is the parameter we pick from Column I.

From the trends, it is clear that $I{E}_1$ decreases as we go down a group. Also, the maximum $I{E}_1$ for a period peaks at the respective noble gas! Hence the answer.