Solution

1. $$Be_2$$:
valence electrons $$=4$$
$${ \left( \sigma 2s \right) }^{ 2 }{ \left( \sigma ^*2s \right) }^{ 2 }$$

2. $$B_2$$:
valence electrons $$=6$$
$${ \left( \sigma 2s \right) }^{ 2 }{ \left( \sigma *2s \right) }^{ 2 }{ \left( \sigma 2p_z \right) }^{ 2 }$$

3.$$C_2$$:
valence electrons $$=8$$
$${ \left( \sigma 2s \right) }^{ 2 }{ \left( \sigma ^*2s \right) }^{ 2 }{ \left( \sigma 2p _z\right) }^{ 2 }{ \left( \pi 2p_x \right) }^{ 1 }{ \left( \pi 2p_y \right) }^{ 1 }$$

4. $$N_2$$:
valence electrons=$$10$$
$${ \left( \sigma 2s \right) }^{ 2 }{ \left( \sigma *2s \right) }^{ 2 }{ \left( \sigma 2p_z \right) }^{ 2 }{ \left( \pi 2p_x \right) }^{ 2 }{ \left( \pi 2p_y \right) }^{ 2 }$$

$$C_2$$ has two unpaired electrons, thus its paramagnetic

Therefore, the option is C.