Question

If the difference in atomic sizw of Na-Li=x ,Rb-K=y ; Fr-cs=z then correct order will be:-
1) x=y=z
2)x>y>z
3)x And how please answer

Answer 1

x > y > z
Li = 1s² 2s¹
Na = 1s² 2s² 2p⁶ 3s¹
K = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
Rb = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹
Cs = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 6s¹
Fr = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 5f¹⁴ 6s² 6p⁶ 7s¹

Now going down the group, electrons began to fill in d and f orbitals.
Due to poor shielding by d and f orbitals , the outer s orbitals are dragged in by nucleus a bit . As shielding due to f orbital is lesser than shielding due to d orbital , there is much contraction in size of Fr than contraction in size of Cs.
Hence the difference between the atomic size of Fr - Cs isn't much.

Similarly in case of Rb , presence of d orbital contracts its size . But not in potassium .Thus the difference in atomic size of Rb - K is greater than Fr - Cs.

While talking about Li - Na , there is increase in size as usual. There's no contraction due to absence of d-f orbitals. So difference in atomic size is greatest.

Hope you'll get my solution.