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Molecular orbital theory predicts the $F_2^{2+}$ ion has a bond order of

$\begin{array}{1 1}(a)\;0.5&(b)\;1\\(c)\;1.5&(d)\;2\end{array}$

1 Answer

In molecular orbital theory, we calculate bond orders by assuming that two electrons in a bonding molecular orbital contribute one net bond and that two electrons in an antibonding molecular orbital cancel the effect of one bond.
Bond Order = $\large\frac{1}{2}[Σ (bonding\; e^⁻) - Σ (antibonding\; e^⁻)]$
$F2^2+ is \therefore 12e^⁻ = \sigma s(2e^⁻) \sigma s*(2e^⁻) \sigma p(2e^⁻) \pi1(4e^⁻) \pi 2*(2e^⁻) \sigma p*(0)$
Bond Order = $\large\frac{1}{2}[\sigma s(2e^⁻) \sigma p(2e^⁻) \pi1(4e^⁻)- \sigma p*(2e^⁻) \pi2*(2e^⁻)] = \large\frac{1}{2}[8-4] = 2.0$
The bond order of $F_2^{2+}$ is two.
Hence (d) is the correct answer.
answered Dec 6, 2013 by sreemathi.v
edited Feb 27 by sharmaaparna1
 

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