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ChemMedMain 2026 • 2nd Apr - Shift 1

Calculate Energy Required for Ionization of Li2 Plus Ion

What is the energy (in J atom−1\text{J atom}^{-1}) required for the following process ? Li2+(g)→Li3+(g)+e−\text{Li}^{2+}(g) \rightarrow \text{Li}^{3+}(g) + e^{-} (Take the ionization energy for the H\text{H} atom in the ground state as 2.18×10−18 J atom−12.18 \times 10^{-18}\text{ J atom}^{-1})

Options

A

8.72×10−188.72 \times 10^{-18}

B

1.962×10−181.962 \times 10^{-18}

C

1.962×10−171.962 \times 10^{-17}

Correct
D

6.54×10−176.54 \times 10^{-17}

Topics & Concepts

Structure of AtomBohr ModelIonization Energy

Step-by-Step Solution

To calculate the energy required for the process: Li2+(g)→Li3+(g)+e−\text{Li}^{2+}(g) \rightarrow \text{Li}^{3+}(g) + e^{-}

we use Bohr's model for hydrogen-like single-electron species.

The ionization energy (IEIE) of a hydrogen-like species is given by the formula: IE=IEH×Z2n2IE = IE_{\text{H}} \times \frac{Z^2}{n^2}

where:

  • IEHIE_{\text{H}} is the ionization energy of the hydrogen atom in its ground state (2.18×10−18 J atom−12.18 \times 10^{-18} \text{ J atom}^{-1}).
  • ZZ is the atomic number of the species.
  • nn is the principal quantum number of the ground state orbit (n=1n = 1).

For lithium (Li\text{Li}), the atomic number is Z=3Z = 3.

Substituting the given values into the formula: IE(Li2+)=2.18×10−18 J atom−1×3212IE(\text{Li}^{2+}) = 2.18 \times 10^{-18} \text{ J atom}^{-1} \times \frac{3^2}{1^2}

IE(Li2+)=2.18×10−18×9 J atom−1IE(\text{Li}^{2+}) = 2.18 \times 10^{-18} \times 9 \text{ J atom}^{-1}

IE(Li2+)=19.62×10−18 J atom−1IE(\text{Li}^{2+}) = 19.62 \times 10^{-18} \text{ J atom}^{-1}

Expressing this in scientific notation: IE(Li2+)=1.962×10−17 J atom−1IE(\text{Li}^{2+}) = 1.962 \times 10^{-17} \text{ J atom}^{-1}

Therefore, the correct option is C.

Calculate Energy Required for Ionization of Li2 Plus Ion | Chemistry PYQ Solution - JEE Challenger