Question

When $N_2$ changes to $N_2^{+}$, the N - N bond distance increases and bond order decreases.
if true enter 1, else enter 0.

• A. 1

Answer 1

The correct option is A 1
$N_2$ has 14 electrons
$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi {)}^4,(2p_z{)}^2$
So Bond Order$=\frac{1}{2}(a-b)$
a is the number of electrons in bonding molecular orbitals and
b is the number of electrons in antibondng molecular orbitals.
Bond Order$=\frac{1}{2}(10-4)$
$=3$
now
$N_2^{+}$ has 13 electrons
$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi {)}^4,(2p_z{)}^1$
So Bond Order$=\frac{1}{2}(a-b)$
Bond Order$=\frac{1}{2}(9-4)$
$=2.5$
hence bond order decreases and bond distance increases as bond distance is inversely proportional bond order