Given below are two statements : **Statement (I) :** The correct sequence of bond lengths in the following species is : $$O_2^+ < O_2 < O_2^- < O_2^{2-}$$ **Statement (II) :** The correct sequence of number of unpaired electrons in the following species is : $$O_2 > O_2^+ > O_2^- > O_2^{2-}$$ In the light of the above statements, choose the correct answer from the options given below :

The four species $$O_2^{2-},\;O_2^{-},\;O_2,\;O_2^{+}$$ all have the same sequence of molecular orbitals because they are homonuclear di-atomic molecules of the second period. For $$2p$$ orbitals, the energy order is

$$\sigma(2p_z)\lt \pi(2p_x)=\pi(2p_y)\lt \pi^{*}(2p_x)=\pi^{*}(2p_y)\lt \sigma^{*}(2p_z)$$

The inner $$\sigma(2s)$$ and $$\sigma^{*}(2s)$$ levels are completely filled in every case, so only the $$2p$$ block decides bond order and magnetism.

Let us fill the $$2p$$ molecular orbitals for each species and compute the bond order $$\bigl(\text{B.O.}= \tfrac{1}{2}(n_b-n_a)\bigr)$$.

Case 1: $$O_2^{2-}\;(18\;e^-)$$
$$\sigma(2p_z)^2\;\pi(2p_x)^2\;\pi(2p_y)^2\;\pi^{*}(2p_x)^2\;\pi^{*}(2p_y)^2$$
$$n_b=8,\;n_a=4\;\Longrightarrow\;\text{B.O.}= \tfrac{1}{2}(8-4)=1$$
All electrons are paired ⇒ 0 unpaired electrons. Case 2: $$O_2^{-}\;(17\;e^-)$$
$$\sigma(2p_z)^2\;\pi(2p_x)^2\;\pi(2p_y)^2\;\pi^{*}(2p_x)^2\;\pi^{*}(2p_y)^1$$
$$n_b=8,\;n_a=3\;\Longrightarrow\;\text{B.O.}= \tfrac{1}{2}(8-3)=1.5$$
One electron remains unpaired in the $$\pi^{*}$$ level ⇒ 1 unpaired electron. Case 3: $$O_2\;(16\;e^-)$$
$$\sigma(2p_z)^2\;\pi(2p_x)^2\;\pi(2p_y)^2\;\pi^{*}(2p_x)^1\;\pi^{*}(2p_y)^1$$
$$n_b=8,\;n_a=2\;\Longrightarrow\;\text{B.O.}= \tfrac{1}{2}(8-2)=2$$
Two electrons singly occupy degenerate $$\pi^{*}$$ orbitals ⇒ 2 unpaired electrons. Case 4: $$O_2^{+}\;(15\;e^-)$$
$$\sigma(2p_z)^2\;\pi(2p_x)^2\;\pi(2p_y)^2\;\pi^{*}(2p_x)^1\;\pi^{*}(2p_y)^0$$
$$n_b=8,\;n_a=1\;\Longrightarrow\;\text{B.O.}= \tfrac{1}{2}(8-1)=2.5$$
Only one electron is present in the antibonding $$\pi^{*}$$ set ⇒ 1 unpaired electron.

Collecting the results:

Bond orders: $$O_2^{+}(2.5)\; \gt \; O_2(2)\; \gt \; O_2^{-}(1.5)\; \gt \; O_2^{2-}(1)$$
Since bond length is inversely proportional to bond order, the bond-length sequence is $$O_2^{+} \lt O_2 \lt O_2^{-} \lt O_2^{2-}$$ This is exactly Statement (I) ⇒ Statement (I) is true.

Number of unpaired electrons: $$O_2\;(2) \gt O_2^{+}\;(1) = O_2^{-}\;(1) \gt O_2^{2-}\;(0)$$ Statement (II) claims the strict order $$O_2 \gt O_2^{+} \gt O_2^{-} \gt O_2^{2-}$$, implying $$O_2^{+}$$ has more unpaired electrons than $$O_2^{-}$$, which is incorrect because both have one. Hence Statement (II) is false.

Therefore, the correct choice is:
Option C which is: Statement I is true but Statement II is false