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Question: In which of the following pairs of molecules/ions both the species are not likely to exist? a.) \[...

In which of the following pairs of molecules/ions both the species are not likely to exist?
a.) $$$$ H2−,He22−{{H}_{2}}^{-},H{{e}_{2}}^{2-}
b.) H22+,He2{{H}_{2}}^{2+},H{{e}_{2}}^{{}}
c.) H2−,He22+{{H}_{2}}^{-},H{{e}_{2}}^{2+}
d.) H2+,He22+{{H}_{2}}^{+},H{{e}_{2}}^{2+}

Explanation

Solution

When two atomic orbitals combine they form two new molecular orbitals, one by the additive effect of the atomic orbitals which is called the bonding molecular orbital and the other is formed by the subtractive effect of the atomic orbital which is known as the antibonding molecular orbitals.

Complete step by step answer:
To solve this question we have to find the electronic configuration of given elements:
Electronic configuration of hydrogen atom H2−{{H}_{2}}^{-} (Z of hydrogen= 1) total electrons 2+1 = 3
σ1s2σ∗1s1\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{1}}
Bond order is defined as the difference between the number of bonding and antibonding molecular orbital.
Bond order= 2−12=\dfrac{2-1}{2}=0.5

Electronic configuration of hydrogen atom He2−2H{{e}_{2}}^{-2} (Z of helium= 2) total electrons 4+2 = 6
σ1s2σ∗1s2σ2s2\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}
Bond order= 4−22=\dfrac{4-2}{2}=1

Electronic configuration of hydrogen atom H22+{{H}_{2}}^{2+} (Z of hydrogen= 1) total electrons 2-2 = 0
σ1s0\sigma 1{{s}^{0}}
Bond order= 0−02=\dfrac{0-0}{2}=0

Electronic configuration of hydrogen atom He2H{{e}_{2}} (Z of helium= 2) total electrons 4+0 = 4
σ1s2σ∗1s2\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}
Bond order= 2−22=\dfrac{2-2}{2}=0

Electronic configuration of hydrogen atom He2+2H{{e}_{2}}^{+2} (Z of helium= 2) total electrons 4-2 = 2
σ1s2\sigma 1{{s}^{2}}
Bond order= 2−02=\dfrac{2-0}{2}=1

Electronic configuration of hydrogen atom H2+{{H}_{2}}^{+} (Z of hydrogen= 1) total electrons 2-1 = 1
σ1s1\sigma 1{{s}^{1}}
Bond order= 1−02=\dfrac{1-0}{2}=0.5
The bond order of H22+,He2{{H}_{2}}^{2+},H{{e}_{2}}^{{}} is 0. Hence their existence is not possible.
So, the correct answer is “Option B”.

Note: The inner shell does not take part in the bonding hence these orbitals are called nonbonding orbitals and the electrons present in them are known as nonbonding electrons. So if we draw the energy level diagram this part is usually omitted. If unpaired electrons are present it is known as a paramagnetic compound and if paired electrons are present it is known as a diamagnetic compound.