H2,Li2,B2 each has bond order equal to 1, the order of their stability is

Question

H2,Li2,B2 each has bond order equal to 1, the order of their stability is (A) H2=Li2=B2 (B) H2>Li2>B2 (C) H2>B2>Li2 (D) B2>Li2>H2

Tardigrade Admin · Accepted Answer

H 2, Li 2, B 2 are not equally stable. Li atom is much larger in size than H-atom. The bond order in Li 2 is much larger than bond length in H 2. Moreover, Li 2 molecule has two electrons in the antibonding molecular orbital while H 2 has no electrons in the antiobinding molecular orbital. Thus, L i2 is less sable than H 2(. Bond energy of Li 2=110 kJ mol-2+ Bond energy of H 2=438 k. . mol -1) Boron atom is smaller than lithium atom but larger than hydrogen atom. The B2 is more stable than L i2 but less stable than H 2(. Bond energy of . B 2=290 kJ mol -1)

Q. $H_{2}, L i_{2}, B_{2}$ each has bond order equal to $1$ , the order of their stability is

$H_{2} = L i_{2} = B_{2}$

$H_{2} > L i_{2} > B_{2}$

$H_{2} > B_{2} > L i_{2}$

$B_{2} > L i_{2} > H_{2}$

Q. H2​,Li2​,B2​ each has bond order equal to 1, the order of their stability is

H2​=Li2​=B2​

H2​>Li2​>B2​

H2​>B2​>Li2​

B2​>Li2​>H2​

Q. $H_{2}, L i_{2}, B_{2}$ each has bond order equal to $1$ , the order of their stability is

$H_{2} = L i_{2} = B_{2}$

$H_{2} > L i_{2} > B_{2}$

$H_{2} > B_{2} > L i_{2}$

$B_{2} > L i_{2} > H_{2}$