IIT JEE Advanced 2009 Paper 1

(A) $\mathrm{B}_2$ → $\sigma1s^2\sigma^*1s^2\sigma2s^2\sigma^*2s^2 2p_x^1$ = $\pi2p_y^1$ It is paramagnetic due to two unpaired electrons. The bond order is $\frac{N_b-N_a}{2}$ = $\frac{6-4}{2}$ = 1 The gain of electron increases bond order, so reduction is possible. (B) $\mathrm{N}_2$ → $\sigma1s^2\sigma^*1s^2\sigma2s^2\sigma^*2s^2\pi2p_x^2$ = $\pi2p_y^2\sigma2p_z^2$ There are no unpaired electrons. Bond order is $\frac{_{b}-N_a}{2}$ = $\frac{10-4}{2}$ = $\frac{6}{2}$ = 3 Mixing of 2s and 2 p orbitals is possible because of similar energies. (C) $\mathrm{O}_2^{-}$ → $\sigma1s^2\sigma1s^2\sigma2s^2\sigma^*2s^2\sigma2p_z^2\pi2p_x^2$ = $\pi2p_y^2\pi^*2p_x^2\pi^*2p_y^1$ The molecule is paramagnetic due to presence of unpaired electron. Bond order is less than 2. $\frac{N_b-N_a}{2}$ = $\frac{10-7}{2}$ = $\frac{3}{2}$ Loss of electron increases the bond order so oxidation is possible. (D) $\mathrm{O}_2$ → $\sigma1s^2\sigma^*1s^2\sigma2s^2\sigma^*2s^2\sigma2p_z^2\pi2p_x^2$ The molecule is paramagnetic due to presence of unpaired electrons. Bond order is $\frac{N_b-N_a}{2}$ = $\frac{10-6}{2}$ = $\frac{4}{2}$ = 2 Loss of electron causes increase in bond order, so it undergoes oxidation.