For a one–electron system, the energy of the stationary state is given by:
$$ E_n = -2.18 \times 10^{-18} \frac{Z^2}{n^2}\ \text{J} $$
where Z is the atomic number and n is the principal quantum number.
For Li2+ ion, Z = 3 and for the third orbit, n = 3.
$$ E_3 = -2.18 \times 10^{-18} \times \frac{3^2}{3^2} $$
$$ E_3 = -2.18 \times 10^{-18}\ \text{J} $$
The negative sign indicates that the electron is in a bound state.
Hence, the correct statement is −2.18 × 10−18 J for third orbit of Li2+ ion.
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