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Answer : H_2^+ ion will be more stable as compared to H_2^- ion because of absence of anti-bonding electrons.Explacountry : According to the molecular orbital concept, the basic molecular orbital configuration approximately nitrogen will certainly be, (\sigma_1s),(\sigma_1s^*),(\sigma_2s),(\sigma_2s^*),(\sigma_2p_z),<(\pi_2p_x)=(\pi_2p_y)>,<(\pi_2p_x^*)=(\pi_2p_y^*)>,(\sigma_2p_z^*) As tbelow are 1 electron current in hydrogen.(a) The variety of electrons present in H_2^+ molecule = 2(1) - 1 = 1 The molecular orbital configuration of H_2^+ molecule will certainly be, (\sigma_1s)^1,(\sigma_1s^*)^0,(\sigma_2s)^0,(\sigma_2s^*)^0,(\sigma_2p_z)^0,<(\pi_2p_x)^0=(\pi_2p_y)^0>,<(\pi_2p_x^*)^0=(\pi_2p_y^*)^0>,(\sigma_2p_z^*)^0 The formula of bonding order = \frac12\times (\textNumber of bonding electrons-\textNumber of anti-bonding electrons) The bonding order of H_2^+ = \frac12\times (1-0)=0.5 (b) The number of electrons present in H_2^- molecule = 2(1) + 1 = 3The molecular orbital configuration of H_2^- molecule will be, (\sigma_1s)^2,(\sigma_1s^*)^1,(\sigma_2s)^0,(\sigma_2s^*)^0,(\sigma_2p_z)^0,<(\pi_2p_x)^0=(\pi_2p_y)^0>,<(\pi_2p_x^*)^0=(\pi_2p_y^*)^0>,(\sigma_2p_z^*)^0 The bonding order of H_2^- = \frac12\times (2-1)=0.5 From this we conclude that the both the molecule have exact same bond order. But as we know that the even more the number of electrons are in anti-bonding, the molecule will be less stable. That implies anti-bonding decreases the stcapacity of the molecule.In H_2^-, the even more number of electrons current in anti-bonding orbital but in H_2^+, there is no electrons existing in anti-bonding orbital. So, H_2^+ ion will be even more steady as compared to H_2^- ion.">" data-test="answer-box-list">

*

Since H2- consists of one electron in the antibonding orbital which results in repulsion and also decrease the stcapacity . on the other hand also H2+ does not contain any type of electron in the antibonding molecular orbital.

You are watching: Will h2+ be more or less stable than h2 and why?


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BarrettArcherBarrettArcher
Answer :
*
ion will be even more secure as compared to
*
ion due to absence of anti-bonding electrons.

Explanation :

According to the molecular orbital theory, the basic molecular orbital configuration approximately nitrogen will be,

*
,<(pi_2p_x^*)=(pi_2p_y^*)>,(sigma_2p_z^*)" alt="(sigma_1s),(sigma_1s^*),(sigma_2s),(sigma_2s^*),(sigma_2p_z),<(pi_2p_x)=(pi_2p_y)>,<(pi_2p_x^*)=(pi_2p_y^*)>,(sigma_2p_z^*)" align="absmiddle" class="latex-formula">

As tbelow are 1 electron present in hydrogen.

(a) The number of electrons current in

*
molecule = 2(1) - 1 = 1

The molecular orbital configuration of

*
molecule will be,

*
,<(pi_2p_x^*)^0=(pi_2p_y^*)^0>,(sigma_2p_z^*)^0" alt="(sigma_1s)^1,(sigma_1s^*)^0,(sigma_2s)^0,(sigma_2s^*)^0,(sigma_2p_z)^0,<(pi_2p_x)^0=(pi_2p_y)^0>,<(pi_2p_x^*)^0=(pi_2p_y^*)^0>,(sigma_2p_z^*)^0" align="absmiddle" class="latex-formula">

The formula of bonding order =

*

The bonding order of

*
=
*

(b) The number of electrons existing in

*
molecule = 2(1) + 1 = 3

The molecular orbital configuration of

*
molecule will be,

*
,<(pi_2p_x^*)^0=(pi_2p_y^*)^0>,(sigma_2p_z^*)^0" alt="(sigma_1s)^2,(sigma_1s^*)^1,(sigma_2s)^0,(sigma_2s^*)^0,(sigma_2p_z)^0,<(pi_2p_x)^0=(pi_2p_y)^0>,<(pi_2p_x^*)^0=(pi_2p_y^*)^0>,(sigma_2p_z^*)^0" align="absmiddle" class="latex-formula">

The bonding order of

*
=
*

From this we conclude that the both the molecule have exact same bond order. But as we understand that the even more the variety of electrons are in anti-bonding, the molecule will certainly be less secure. That suggests anti-bonding decreases the stability of the molecule.

See more: Which Element Can Form More Than One Kind Of Monatomic Ion? ?

In

*
, the even more number of electrons existing in anti-bonding orbital however in
*
, tbelow is no electrons current in anti-bonding orbital. So,
*
ion will be more stable as compared to
*
ion.