Amongst the following, the correct statement(s) is /are

1.	Amongst the following, the correct statement(s) is /are
Answer» NO has one unpaired electron in the antiboding molecular orbital `NO^(+)` is more stable than `O_(2)^(+)` `OF^(+)` is more paramagnetic than `Ne_(2)^(+)` In a `pi` bond, the electron density is conncentrated along the bond axis. Solution :`NO(15):KK(SIGMA2S)^(2)(sigma*2s)^(2)(sigma2p_(Z))^(2)(pi2p_(x))^(2)` `=(pi2p_(y))^(2)(pi2p_(x))^(1)` `NO^(+)(14):KK(sigma2s)^(2)(sigma2s)^(2)(sigma2p_(z))^(2)(pi2p_(x))^(2)=(pi2p_(y))^(2)` B.O. =3.0 `O_(2)^(+)(15):KK(sigma2s)^(2)(sigma2s)^(2)(sigma2p_(z))^(2)(pi2p_(x))^(2)=(pi2p_(y))^(2)(pi2p_(x))^(1)` B.O. =2.5 `OF^(+)(16):KK(sigma2s)^(2)(sigma2s)^2)(sigma2p_(z))^(2)(pi2p_(x))^(2)=(pi2p_(y))^(2)(pi2p_(x))^(1)=(pi2p_(y))^(1)` `Ne_(2)^(+)(19):KK(sigma2s)^(2)(sigma2s)^(2)(sigma2p_(z))^(2)(pi2p_(x))^(2)=(pi2p_(y))^(2)` `=(pi2p_(x))^(2)=(pi2p_(y))^(2)(sigma*2p_(z))^(1)` In a `pi`- bond, the electron density is concentrated in the region PERPENDICULAR tothe bond axis.

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