1. Tardigrade
  2. Question
  3. Chemistry
  4. Fill in the blanks with appropriate choice. Bond order of N2+ is P while that of N2 is Q . Bond order of O2+ is R while that of O2 is S . N— N bond distance T, when N2 changes to N2+ and when O2 changes to O2+, the O — O bond distance U . P Q R S T U (a) 2 2.5 2.5 1 increases decreases (b) 2.5 3 2 1.5 decreases increases (c) 3 2 1.5 1 increases decreases (d) 2.5 3 2.5 2 increases decreases

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Q. Fill in the blanks with appropriate choice.
Bond order of $N_2^+$ is P while that of $N_2$ is Q .
Bond order of $O_{2}^+$ is R while that of $O_2$ is S .
$ N— N$ bond distance T, when $N_2$ changes to $N_{2}^{+}$ and when $O_2$ changes to $O_2^+$, the $O — O$ bond distance U .
$\quad$ P$\quad$ Q$\quad$ R$\quad$ S$\quad$ T$\quad$ U$\quad$
(a)$\quad$ 2$\quad$ 2.5$\quad$ 2.5$\quad$ 1$\quad$ increases$\quad$ decreases$\quad$
(b) 2.5 3 2 1.5 decreases increases
(c) 3 2 1.5 1 increases decreases
(d) 2.5 3 2.5 2 increases decreases

Chemical Bonding and Molecular Structure

Solution:

$N_{2}^{+} :\left(\sigma 1s^{2}\right)\left(\sigma^* 1s^{2}\right)\left(\sigma 2s^{2}\right)\left(\sigma^* 2s^{2}\right)\left(\pi 2p_{x}^{2} = \pi2p_{y}^{2}\right)\left(\sigma 2p_{z}^{1}\right)$
$B .O. = \frac{1}{2} \times (9 - 4) = 2.5$
$N_{2} : B. O. = \frac {1}{2} \times (10-4) = 3$
$O_{2}^{+} : \left(\sigma 1s^{2}\right)\left(\sigma^* 1s^{2}\right)\left(\sigma 2s^{2}\right)\left(\sigma^* 2s^{2}\right)\left(\sigma 2 p_{z}^{2}\right) \left(\pi 2p_{x}^{2} = \pi2p_{y}^{2}\right) $
$\left(\pi^ * 2p_{x}^{1}\right)$
$B .O. = \frac{1}{2}\times (10-5 )= 2.5$
$ O_{2} : B .O. = \frac{1}{2}\times (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.