Question

Arrange the following species in the correct order of their stability:
$C_2$, $Li_2$, $O_2^+$, $He_2^+$

• A. Li\textsubscript{2} $<$ He\textsubscript{2}\textsuperscript{+} $<$ O\textsubscript{2}\textsuperscript{+} $<$ C\textsubscript{2}
• B. C\textsubscript{2} $<$ O\textsubscript{2}\textsuperscript{+} $<$ Li\textsubscript{2} $<$ He\textsubscript{2}\textsuperscript{+}
• C. He\textsubscript{2}\textsuperscript{+} $<$ Li\textsubscript{2} $<$ C\textsubscript{2} $<$ O\textsubscript{2}\textsuperscript{+}
• D. O\textsubscript{2}\textsuperscript{+} $<$ C\textsubscript{2} $<$ Li\textsubscript{2} $<$ He\textsubscript{2}\textsuperscript{+}
• E. C\textsubscript{2} $<$ Li\textsubscript{2} $<$ He\textsubscript{2}\textsuperscript{+} $<$ O\textsubscript{2}\textsuperscript{+}

Hint:

For species with atomic number $Z \leq 7$, the $\pi 2p$ orbitals are lower in energy than the $\sigma 2p_z$ orbital. For $Z > 7$, the $\sigma 2p_z$ orbital is lower. Correct MO ordering is crucial for accurate bond order calculations!

Answer 1

The Correct Option is C

Concept: According to Molecular Orbital (MO) theory, the stability of a molecule or ion is directly proportional to its bond order.
• Bond Order (B.O.): Calculated as $\frac{1}{2}(N_b - N_a)$, where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number in antibonding orbitals.
• Stability Rule: Stability increases as the bond order increases. If two species have the same bond order, the one with fewer electrons in antibonding orbitals is more stable.

Step 1: Calculate the Bond Order for each species.
• He\textsubscript{2\textsuperscript{+}}: Total electrons = 3. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^1$. B.O. = $(2-1)/2 = 0.5$.
• Li\textsubscript{2: Total electrons = 6. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2$. B.O. = $(4-2)/2 = 1.0$.
• C\textsubscript{2: Total electrons = 12. Configuration: $[KK] (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$. B.O. = $(8-4)/2 = 2.0$.
• O\textsubscript{2\textsuperscript{+}}: Total electrons = 15. Configuration: $[KK] (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. B.O. = $(10-5)/2 = 2.5$.

Step 2: Arrange species by Bond Order. The calculated bond orders are: $0.5 < 1.0 < 2.0 < 2.5$. Stability order: $He_2^+ < Li_2 < C_2 < O_2^+$. This corresponds to Option (C).