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When N(2) goes to N(2)^(+), the N-N bond...

When `N_(2)` goes to `N_(2)^(+)`, the `N-N` bond distance …………, and when `O_(2)` goes to `O_(2)^(+)` the `O-O` bond distance…….

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The correct Answer is:
Increases, decrease

Inceases decreases
In `N_(2)` to `N_(2)^(o+)` bond order decreases fro m3 to 2.5 thus bond length increases) In case of `O_(2)` to `O_(2)^(o+)` bond order increases from thus bond length decreases ) .
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Write the molecular orbital electron distribution of oxygen (O_(2)) Specify its bond order and magnetic property Fill in the blanks When N_(2) goes to N_(2)^(o+) , the N-N bond distance___ and when O_(2) goes to O_(2)^(o+) the O-O bond distance ____ .

When N_(2) goes N_(2)^(o+) the N-N bond distance________and when O_(2) goes to O_(2)^(o+) the O-O bond distance _______.

In N_2O , the N - N distance pertains to

0.Changing N_(2) to N_(2)^(+) ,the dissociation energy of N-N bond.....and on changing O_(2) to O_(2)^(+) the dissociation energy of O-O bond..... (A) increases,decreases (B) decreases,increases (C) decrease in both cases (D) increases in both cases

Do N_(2)^(+) and O_(2)^(+) have same bond order?

How does molecular orbital theory account for the following? (a) Bond order of N_(2) is greater than N_(2)^(+) but the bond order of O_(2) is less than that of O_(2)^(+) . (b ) Be_(2) does not exist.

CENGAGE CHEMISTRY-CHEMICAL BONDING AND MOLECULAR STRUCTURE-Exercises Fill In The Blanks
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