Question

Using molecular orbital theory, compare the bond energy and magnetic character of $O_{2}^{+}$ and $O_{2}^{+}$ species.

Answer 1

The given ions are as follows.

${ O }_{ 2 }^{ + }$ and ${ O }_{ 2 }^{ - }$

Let's write electronic configuration of each ion by using molecular orbital diagram.

$\quad { O }_{ 2 }^{ + }:\quad (\sigma 1{ s) }^{ 2 }{ (\sigma  }^{ * }2{ s) }^{ 2 }(\sigma 2{ s) }^{ 2 }{ (\sigma  }^{ * }2{ s) }^{ 2 }(\sigma { 2p }_{ z }{ ) }^{ 2 }(\pi 2p_{ x }^{ 2 }\quad ,\pi 2p_{ y }^{ 2 })\quad ({ \pi  }^{ * }2p_{ x }^{ 1 })$  

$\quad { O }_{ 2 }^{ - }:\quad (\sigma 1{ s) }^{ 2 }{ (\sigma  }^{ * }2{ s) }^{ 2 }(\sigma 2{ s) }^{ 2 }{ (\sigma  }^{ * }2{ s) }^{ 2 }(\sigma { 2p }_{ z }{ ) }^{ 2 }(\pi 2p_{ x }^{ 2 }\quad ,\pi 2p_{ y }^{ 2 })\quad ({ \pi  }^{ * }2p_{ x }^{ 2 },{ \pi  }^{ * }2p_{ y }^{ 1 })$

Bond order of ${ O }_{ 2 }^{ + }\quad \quad =\quad \frac { 10\quad -\quad 5 }{ 2 } \quad =\quad 2.5$

Bond order of ${ O }_{ 2 }^{ - }\quad \quad =\quad \frac { 10\quad -\quad 7 }{ 2 } \quad =\quad 1.5$

Higher bond order of ${ O }_{ 2 }^{ + }$ shows that more stable than ${ O }_{ 2 }^{ - }$ and both ions have unpaired electrons.

Therefore, it have paramagnetic nature.

Answer 2

hey mate.....
here is your answer......

Bond order of O2+ = 2.5
Bond order of O2- = 1.5



BOND ORDER OF O2+>O2-


I HOPE ITS HELPFUL......☺✌☺