Supposing that the pauli exclusion principal is not correct orbit can accammodate three electrons when are the respective atomic number of the second number of alkali metal and the first number of balogen family ?

Options

A. `16,14`
B. `11,9`
C. `16,9`
D. `34,17`

Explanation

Sodium is the second number of alkali metal family `Ne^(11) = 1s^(2) 2s^(2)2p^(6)3s^(1)`
We know that the orbit of sodium are falls filled and the outermost orbit orbit has one electron .If ioner orbitals can accommodate three electrons each the configuration will be as follows
Therefore three be `16 ` electron in all .Hence the atomic number will be `16`
ii. The first number of halogen family is flonrine `F^(9)` whose configuration is `1s^(2)2s^(2)2p^(5)`
Halogen has one electron less than the rest of inert elctron each ,the configuration will be as follows
`1s^(3) 2s^(3)2p^(8)`
Therefore , total number of electrons will be i.e. and thus the atomic number will be `14`