Fill in the blanks with appropriate choice .
Bond order of `N_(2)^(+) " is" underline (P)` while that of ` N_(2) " is " underline (Q)` .
Bond order of ` O_(2)^(+) " is " underline(R)` while that of ` O_(2) " is " underline (S)`.
N-N bond distance ` underline(T)` when ` N_(2) " changes to " N_(2)^(+) `
and when ` O_(2)` changes to ` O_(2)^(+)` , the O-O bond distance ` underline(U)`.

Options

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 , "decrases" , "increases"):}`
C. `{:(P,Q,R,S,T,U),(3,2,1.5,1 , "incrases" , "decreases"):}`
D. `{:(P,Q,R,S,T,U),(2.5,3,2.5,2 , "incrases" , "decreases"):}`

Answer

d

Explanation

`N_(2)^(+) : (sigma 1s^(2)) (sigma^(**) 1s^(2)) (sigma 2s^(2)) (sigma^(**) 2s^(2)) (pi2p_(x)^(2) = pi 2p_(y)^(2)) (sigma 2p_(z)^(1))`
B.O. `= 1/2 xx(9-4 ) = 2.5`
` N_(2) : B.O. = (1)/(2) xx(10 - 4) = 3`
` O_(2)^(+) : (sigma 1s^(2)) (sigma^(**) 2s^(2)) (sigma 2s^(2)) (sigma 2p_(z)^(2)) (pi2py_(x)^(2) = pi2p_(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 form 2 to 2.5 hence , bond length decreases.