InterviewSolution
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InterviewSolution
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Fill in the blanks with appropriate choice. Bond ordr of `N_(2)^(+) is ul(" "P)`while that of `N_(2) is ul(" "Q)`. Bond order of `O_(2)^(+) is ul(" "R)` while that of `O_(2) is ul(" "S)`. N - N bond distance `ul(" "T)` when `N_(2)` changes to `N_(2)^(+) ` and when `O_(2)` changes to `O_(2)^(+)` , the O - O bond distance `ul(" "U)`.A. `{:(P,Q,R,S," "T," "U),(2,2.5,2.5,1,"increases","decreases"):}`B. `{:(P,Q,R,S," "T," "U),(2.5,3,2,1.5,"decreases","increases"):}`C. `{:(P,Q,R,S," "T," "U),(3,2,1.5,1,"increases","decreases"):}`D. `{:(P,Q,R,S," "T," "U),(2.5,3,2.5,2,"increases","decreases"):}` |
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Answer» Correct Answer - D `N_(2)^(+) : (sigma1s^(2))(sigma^(**)1s^(2))(sigma2s^(2))(sigma^(**)2s^(2))(pi 2p_(x)^(2) = pi 2p_(y)^(2))(sigma2p_(z)^(1))` B.O. ` = 1/2 xx (9-4) = 2.5` `N_(2) : B.O. = 1/2 xx (10 - 4) = 3` `O_(2)^(+) : (sigma1s^(2))(sigma^(**)1s^(2))(sigma2s^(2)) (sigma^(**)2s^(2))(sigma2p_(z)^(2))(pi 2p_(x)^(2) pi 2p_(y)^(2))(pi^(**)2p_(x)^(1))` B.O. ` = 1/2 xx (10 - 5) = 2.5` ` O_(2) : B.O. = 1/2 xx (10 - 6) = 2` Since `N_(2)^(+)` has lower bond order than `N_(2)`, bond length of N - N in `N_(2)^(+)` increases. In `O_(2)^(+)`, bond order increases from 2 to `2.5` hence, bond length decreases. |
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