Question

Calculate bond order of $H{e^{ + 2}}$

Answer 1

Hint: Bond order is a standard parameter to calculate the total number of chemical bonds present between two same or different atoms in a same molecule. Helium is a noble gas and highly inert in nature and does not participate in chemical reactions under normal circumstances.

Complete answer:
Bond order basically calculates the total number of bonds present within the molecule. Bonds between the atoms may be single, double and triple bonds. Helium is the first member of group $\left( {VIIIA} \right)$. Atomic number of helium is $2$and has only one $s - $orbital around the nucleus. Electronic configuration of helium is $1{s^2}$.
When two atoms of helium combine with each other the total number of electrons present in the molecule will become $4$. Now when one helium atom loses one electron from its outer shell it gets converted into $H{e^{ + 2}}$ with $1{s^0}$ configuration.
So the total number of electrons left in the molecule is $3$.
Therefore, out of four electrons of a helium molecule, $2$ electrons are present in the bonding orbital of atoms which take part in bond formation while $1$ electrons are present in antibonding electrons.
Formula to calculate bond order is: -
$BO = \dfrac{1}{2}\left[ {Nb - Na} \right]$
Where, $BO$ bond order is expressed as
$Nb$,is showing the total bonding electrons
$Na$,is showing the total antibonding electrons
Now put all the values in above mentioned formula to calculate bond order of $H{e^{ + 2}}$-
$BO = \dfrac{1}{2}\left[ {2 - 1} \right]$
After solving the above equation, we get
$BO = 0.5$
$ \Rightarrow $ Therefore, the bond order of $H{e^{ + 2}}$is $0.5$.

Note:
Bond order is equivalent in the case of isoelectronic species which have the same number of electrons around the atoms. As the value of bond order increases, bond enthalpy as well as bond length of molecule decreases. Higher bond order shows shorter bond length therefore more energy is required to break the molecule.