Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them:

O_{2}^{-} < N O < C_{2}^{2 -} < H e_{2}^{+}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{2}^{2 -} < H e_{2}^{+} < O_{2}^{-} < N O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H e_{2}^{+} < O_{2}^{-} < N O < C_{2}^{2 -}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

N O < O_{2}^{-} < C_{2}^{2 -} < H e_{2}^{+}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $H e_{2}^{+} < O_{2}^{-} < N O < C_{2}^{2 -}$
Four diatomic species are listed below. Identify the correct order in which the bond order is increasing them,
First we will calculate the bond order for all elements.
Bond order $= \frac{N_{b} - N_{a}}{2}$
$N_{b} =$ Number of electrons in bonding molecular orbital
$N_{a} =$ Number of electrons in anti0bonding molecular orbital
$N O = 15$ electrons
$σ_{1 s^{2}}, σ_{1 s^{2}}^{}, σ_{2 s^{2}}, σ_{2 s^{2}}^{}, σ_{2 p_{z}^{2}}, π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}}, π_{2 p_{x}^{1}}^{}$
B.O $= \frac{8 - 3}{2} = 2.5$
$O_{2}^{-} = 17$ electrons
$σ_{1 s^{2}}, σ_{1 s^{2}}^{}, σ_{2 s^{2}}, σ_{2 s^{2}}^{}, σ_{2 p_{z}^{2}}, π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}},, π_{2 p_{x}^{2}}^{} = π_{2 p_{y}^{1}}^{}$
B.O $= \frac{8 - 5}{2} = 1.5$
$C_{2}^{2 -} = 14$ electrons
$σ_{1 s^{2}}, σ_{1 s^{2}}^{}, σ_{2 s^{2}}, σ_{2 s^{2}}^{}, σ_{2 p_{z}^{2}}, π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}}$
B.O $= \frac{8 - 2}{2} = 3$
${H e}_{2}^{2 +} = 3$ electrons
$σ_{1 s^{2}}, σ_{1 s^{1}}^{}$
B.O $= \frac{2 - 1}{2} = 0.5$
Therefore order is ${H e}_{2}^{+} < O_{2}^{-} < N O < C_{2}^{2 -}$

Suggest Corrections

Similar questions

(i) O_{2}^{+} (i i) C O (i i i) B_{2} (i v) O_{2}^{-}

(v) N O^{+} (v i) H e_{2}^{+} (v i i) C_{2}^{2 +} (v i i i) C N^{-}

(i x) N_{2}^{-}

C_{2}, L i_{2}, O_{2}^{+}, H e_{2}^{+}

H e_{2}

H e_{2}

H e_{2}

N O, N O^{+}, O_{2}, N O^{-}, O_{2}^{+}

Join BYJU'S Learning Program