A) \[{{H}_{2}}\]
B) \[H_{2}^{+}\]
C) \[H{{e}_{2}}\]
D) \[H{{e}_{2}}^{+}\]
Correct Answer: A
Solution :
Key Idea: Write the configuration of each molecule according to molecular orbital theory and then find the bond order. Compare the bond orders to find correct answer. [a] \[{{H}_{2}}=(1+1=2)\,\,\,\sigma 1{{s}^{2}}\] no. of electrons - No. of electrons \[B.O.=\frac{in\text{ }bonding\text{ }M.O.\text{ }in\text{ }and\text{ }bonding\text{ }M.\text{ }0.}{2}\] \[=\frac{2-0}{2}\] = 1 [b] \[H_{2}^{+}(1)=\sigma \,1{{s}^{1}}\] Bond order \[=\frac{1-0}{2}=\frac{1}{2}\] [c] \[H{{e}_{2}}=(2+2)=\sigma 1{{s}^{2}},\,\,\overset{*}{\mathop{\sigma }}\,1\,{{s}^{2}}\] Bond order \[=\frac{2-2}{2}=0\] [d] \[{{H}_{2}}^{+}(4-1=3)=\sigma 1{{s}^{2}},\,\overset{*}{\mathop{\sigma }}\,1{{s}^{2}}\] Bond order \[=\frac{2-1}{2}=\frac{1}{2}\] \[\therefore \] \[{{H}_{2}}\] has highest bond order (1).
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