A) \[{{O}_{2}}>{{O}_{2}}^{-}>{{O}_{2}}\]
B) \[{{O}_{2}}^{+}>{{O}_{2}}>{{O}_{2}}^{-}\]
C) \[{{O}_{2}}>{{O}_{2}}^{-}>{{O}_{2}}^{+}\]
D) \[{{O}_{2}}^{-}>{{O}_{2}}^{+}>{{O}_{2}}\]
Correct Answer: B
Solution :
Key Idea: Write the electronic configuration of given specieses then find the bond order and then arrange them in correct sequence. \[{{O}_{2}}=16=KK\,\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\], \[\pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1},\overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] Bond order \[=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] where, \[{{N}_{b}}=\] no. of electrons in bonding molecular orbitals \[{{N}_{a}}=\] no. of electrons in antibonding molecular orbitals \[=\frac{10-6}{2}\] = 4/2 = 2.0 \[O_{2}^{+}=8+8-1=15\] \[=KK\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{2},\,\pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\] Bonding order \[=\frac{10-5}{2}\] = 5/2 = 2.5 \[O_{2}^{-}=8+8+1=17\] \[=KK\,\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{2},\pi 2p_{y}^{2}\], \[\overset{*}{\mathop{\pi }}\,2p_{x}^{2},\overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] Bonding order \[=\frac{10-7}{2}\] = 3/2 = 1.5 \[\therefore \] The correct order \[O_{2}^{+}>{{O}_{2}}>O_{2}^{-}\]
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