Question

The bond order of dioxygen is \(m\). The bond order values of \(N_2^{+}\) and \(C_2^{2-}\) are respectively

• A. \(\dfrac{5m}{4},\ \dfrac{3m}{2}\)
• B. \(\dfrac{3m}{2},\ \dfrac{5m}{4}\)
• C. \(\dfrac{m}{2},\ \dfrac{m}{3}\)
• D. \(\dfrac{2m}{3},\ \dfrac{m}{2}\)

Hint:

Bond order is calculated using \[ \text{B.O.}=\frac{N_b-N_a}{2}. \] Removing an electron from a bonding orbital decreases bond order by \(0.5\), while adding an electron to a bonding orbital increases bond order by \(0.5\).

Answer 1

The Correct Option is A

Step 1: Find the bond order of dioxygen.
For molecular oxygen, \[ O_2 \] The molecular orbital configuration gives bond order \[ \text{B.O.}=\frac{N_b-N_a}{2} \] For \(O_2\), \[ \text{B.O.}=2 \] Given that bond order of dioxygen is \(m\), \[ m=2 \]

Step 2: Calculate bond order of \(N_2^{+}\).
For \(N_2\), \[ \text{B.O.}=3 \] Removal of one electron from \(N_2\) forms \(N_2^{+}\). The electron is removed from a bonding molecular orbital.
Hence, \[ \text{B.O.}(N_2^{+}) =3-\frac{1}{2} =\frac{5}{2} \] Expressing in terms of \(m\), \[ \frac{5}{2} = \frac{5}{4}(2) = \frac{5m}{4} \]

Step 3: Calculate bond order of \(C_2^{2-}\).
For \(C_2\), \[ \text{B.O.}=2 \] Addition of two electrons produces \[ C_2^{2-} \] The added electrons enter bonding \(\pi\)-orbitals.
Therefore, \[ \text{B.O.}(C_2^{2-}) =2+1 =3 \] Expressing in terms of \(m\), \[ 3 = \frac{3}{2}(2) = \frac{3m}{2} \]

Step 4: Final conclusion.
Hence, \[ \boxed{\text{B.O.}(N_2^{+})=\frac{5m}{4}, \qquad \text{B.O.}(C_2^{2-})=\frac{3m}{2}} \] Therefore, the correct option is (1).