Question

Compare and arrange the following in the increasing order of stability: N2,N2+,N2-,N22-​

Answer 1

Answer:

Explanation:

According to it bond order of N2 is 3, N2(+) & N2(-) is 2.5 and N2(2+) & N2(2-) is 2.

Stability is directly proportional to bond order.

Moreover positive charge molecule is more stable than negative charge molecule.

.

Therefore, the order of stability is : N2 > N2(+) > N2(-) > N2(2+) > N2(2-)

Answer 2

Arrange the increasing order in the equation or $N2^{-}$<$N2^{+}$<$N2^{-}$<$N2^{2-}$<$N2^{2+}$ using Bond order method.

Explanation:

Bond order:

Bond order is the number of bonding pairs of electrons between two atoms. In a covalent bond between two atoms.

    Types of Bond order:

• A single bond has a bond order of one.
• A double bond has a bond order of two.
• A triple bond has a bond order of three.

                         Bond order  =  $\frac{Nb-Na}{2}$

                                                 = $\frac{10-6}{2}$

                                                 =2

• In N2(14) Bond order=3.

• In N2+(15) Bond order=2.5.

• In N2+(15) Bond order=2.5.

Bond order ∞ stability.  

Order by the inceasing order :

$N2^{-}$<$N2^{+}$<$N2^{-}$<$N2^{2-}$<$N2^{2+}$