In which of the following pairs of molecules/ | vedclass

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(C) According to Molecular Orbital Theory,species with a bond order of $0$ or less do not exist.For $H_{2}^{2+}$: Total electrons $= 1 + 1 - 2 = 0$. E

Vedclass · Accepted Answer

$H_{2}^{2+}, He_{2}$

Vedclass · Answer

(C) According to Molecular Orbital Theory,species with a bond order of $0$ or less do not exist.For $H_{2}^{2+}$: Total electrons $= 1 + 1 - 2 = 0$. Electronic configuration $= \sigma 1s^{0}$. Bond order $= 0$.For $He_{2}$: Total electrons $= 2 + 2 = 4$. Electronic configuration $= \sigma 1s^{2}, \sigma^* 1s^{2}$. Bond order $= \frac{2 - 2}{2} = 0$.Since both $H_{2}^{2+}$ and $He_{2}$ have a bond order of $0$,they are not likely to exist.

Pair of species	Identical property
$A$. $B_2 \& O_2$	$P$. Bond order $= 2.5$
$B$. $Be_2 \& H_2^{2-}$	$Q$. Paramagnetic nature
$C$. $N_2^{+} \& N_2^{-}$	$R$. Diamagnetic nature
$D$. $O_2^{+} \& O_2^{-}$	$S$. Doesn't exist

In which of the following pairs of molecules/ions,both the species are not likely to exist?

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Match the following:

Pair of species Identical property

$A$. $B_2 \& O_2$ $P$. Bond order $= 2.5$

$B$. $Be_2 \& H_2^{2-}$ $Q$. Paramagnetic nature

$C$. $N_2^{+} \& N_2^{-}$ $R$. Diamagnetic nature

$D$. $O_2^{+} \& O_2^{-}$ $S$. Doesn't exist

$He_{2}$ molecule is not possible. Explain.

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