Q5 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Chemistry
Observe the following molecules: \(\mathrm{C}_2, \mathrm{~N}_2, \mathrm{O}_2, \mathrm{~F}_2\) Which one of the following statements is correct for the above molecules?
- AThey exhibit same magnetic property
- Correct answerBThe have same number of bonding molecular orbtials and same number of antibonding molecular orbitals
- CThe sequence of molecular orbitals is as follows \(\sigma 2 p_z < \left(\pi 2 p_x=\pi 2 p_y\right) < \left(\pi 2 p_x^*=\pi 2 p_y^*\right) < \sigma 2 p_z^*\)
- DThey have same bond order
Solution
The electronic configuration, magnetic property and bond order of given molecules are as follows:
(i)
\(\begin{aligned}
& \mathrm{C}_2=12 \\
& =\sigma s^2 < \sigma^* 1 s^2 < \sigma 2 s^2 < \sigma * 2 s^2 < \pi 2 p_x{ }^2=\pi 2 p_y{ }^2
\end{aligned}\)
It is diamagnetic in nature.
\(\begin{aligned}
\mathrm{BO} & =\frac{1}{2}\left[N_b-N_a\right] \\
& =\frac{1}{2}[8-4]=2
\end{aligned}\)
(ii)
\(\begin{aligned}
& \mathrm{N}_2=14 \\
& =\sigma s^2 < \sigma^* 1 s^2 < \sigma 2 s^2 < \sigma^* 2 s^2 < \pi 2 p_x{ }^2 \\
& =\pi 2 p_y{ }^2 < \sigma 2 p_z{ }^2
\end{aligned}\)
It is diamagnetic in nature.
\(\text{BO }=\frac{1}{2}[10-4]=3\)
(iii)
\(\begin{aligned}
\mathrm{O}_2= & 16=\sigma s^2 < \sigma^* 1 s^2 < \sigma 2 s^2 < \sigma^* 2 s^2 < \\
& \sigma 2 p_z{ }^2 < \sigma 2 p_x{ }^2 < \pi 2 p_y{ }^2 < \pi^* 2 p_x{ }^1=\pi^* 2 p_y{ }^1
\end{aligned}\)
It is paramagnetic in nature.
\(\mathrm{BO}=\frac{1}{2}[10-6]=2\)
(iv)
\(\begin{array}{r}
\mathrm{F}_2=18=\sigma s^2 < \sigma^{\star} 1 s^2 < \sigma 2 s^2 < \sigma^{\star} 2 s^2 < \sigma 2 p z^2 \\
\pi 2 p_x{ }^2=\pi 2 p_y{ }^2 < \pi^{\star} 2 p_x{ }^2=\pi^{\star} 2 p_y{ }^2
\end{array}\)
It is diamagnetic in nature.
\(\mathrm{BO}=\frac{1}{2}[10-8]=1\)
Here, the filling of electrons in bonding and antibonding orbital is different but they all have same number of molecular and antimolecular orbital (even are empty).
Thus, statement (b) is correct.
Hence, the correct answer is option: 'B'.