Q7 / 110 · VITEEE 2024 · Chemistry

Arrange in decreasing order, the energy of $2 s$ orbital in the following atoms $\mathrm{H}, \mathrm{Li}, \mathrm{Na}$ and K .

  • Correct answer
    A
    $E_{2 s}(\mathrm{H}) \gt E_{2 s}(\mathrm{Li}) \gt E_{2s}(\mathrm{Na}) \gt E_{2 s}(\mathrm{~K})$
  • B
    $E_{2 s}(\mathrm{H}) \gt E_{2 s}(\mathrm{Na}) \gt E_{2 s}(\mathrm{Li}) \gt E_{2 s}(\mathrm{~K})$
  • C
    $E_{2 s}(\mathrm{H}) \gt E_{2 s}(\mathrm{Na})=E_{2 s}(\mathrm{~K}) \gt E_{2 s}(\mathrm{Li})$
  • D
    $E_{2 s}(\mathrm{~K}) \gt E_{2 s}(\mathrm{Na}) \gt E_{2 s}(\mathrm{Li}) \gt E_{2 s}(\mathrm{H})$

Solution

The correct order (decreasing) of energy of $2 s$ orbital for given atom is $E_{2 s}(\mathrm{H}) \gt E_{2 s}(\mathrm{Li}) \gt E_{2 s}(\mathrm{Na}) \gt E_{2 s}(\mathrm{~K})$ The reason is that as atomic number (nuclear charge) increases, all the orbitals are pulled closer to the nucleus. Closes in an orbital to the nucleus, less is its energy.

Hence, the correct answer is option: 'A'.

Relevant Questions