(solved)Question 3.5 of NCERT Class XI Chemistry Chapter 3

Write the electronic configuration first

Remember the noble gases:

He(2), Ne(10), Ar(18),
Kr(36),Xe(54), Rn(86),
Og(118) oganesson

$\displaystyle Rn(86) \lt Z(114) \lt Og(118)$

but Rn is in P6

$\displaystyle \therefore $ Z = 114 lies in P7

by aufbau $\displaystyle 7s \lt 5f \lt 6d \lt 7p$

$\displaystyle \therefore $ configuration is:

$\displaystyle \text{[Rn]}7s^25f^{14}6d^{10}7p^2$

[Please watch the video below for good explanation]

$\displaystyle 7p^2 \implies $ group is 14

Hence Z = 114 is Group 14, Period 7 $\displaystyle \underline{Ans}$