Which of the following options represents the correct bond order?

Solution :
$${\text{Bond order of}}\,\,O_2^ - $$
$$O_2^ - = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},$$      $$\mathop \sigma \limits^* 2{s^2}\,\sigma 2p_z^2\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$     $$\left( {{\pi ^*}2p_x^1 = {\pi ^*}2p_y^1} \right)$$
$$\eqalign{ & {\text{Bond}}\,\,{\text{order}} \cr & = \frac{{{\text{number}}\,\,{\text{of}}\,\,{\text{electron}}\,\,{\text{in}}\,\,BMO - {\text{number}}\,\,{\text{of}}\,\,{\text{electrons}}\,\,ABMO}}{2} \cr & = \frac{{10 - 7}}{2} \cr & = \frac{3}{2} \cr & = 1.5 \cr} $$
$$O_2^ + = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},$$      $$\mathop \sigma \limits^* 2{s^2}\,\sigma 2p_z^2\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$     $$\left( {{\pi ^*}2p_x^1 = {\pi ^*}2p_y^0} \right)$$
$$\eqalign{ & BO = \frac{{10 - 5}}{2} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{5}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 2.5 \cr} $$
$${O_2} = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},$$      $$\mathop \sigma \limits^* 2{s^2}\,\sigma 2p_z^2\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$     $$\left( {{\pi ^*}2p_x^1 = {\pi ^*}2p_y^1} \right)$$
$$\eqalign{ & BO = \frac{{10 - 6}}{2} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{4}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 2 \cr & {\text{So, the correct sequence is}} \cr & O_2^ - < {O_2} < O_2^ + \cr} $$