The existence of a diatomic molecule can be predicted using its bond order, which is calculated from its molecular orbital electronic configuration. A molecule is stable and can exist if its bond order is greater than zero.

Step 1: Determine the electronic configuration and calculate bond order for each option.

• Bond Order (B.O.) $=\frac{1}{2}(N_b-N_a)$, where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.

Step 2: Analyze Option A) ${\mathrm{He}}_2$

• Total electrons = 4.
• Configuration: $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2$
• Bond Order $=\frac{1}{2}(2-2)=0$.
• Since the bond order is 0, ${\mathrm{He}}_2$ does not exist.

Step 3: Analyze Option B) ${\mathrm{Be}}_2$

• Total electrons = 8.
• Configuration: $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2$
• Bond Order $=\frac{1}{2}(4-4)=0$.
• Since the bond order is 0, ${\mathrm{Be}}_2$ does not exist.

Step 4: Analyze Option C) ${\mathrm{Ne}}_2$

• Total electrons = 20.
• Configuration: $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\sigma 2p_z{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2({\pi }^{\ast }2p_x{)}^2({\pi }^{\ast }2p_y{)}^2({\sigma }^{\ast }2p_z{)}^2$
• Bond Order $=\frac{1}{2}(10-10)=0$.
• Since the bond order is 0, ${\mathrm{Ne}}_2$ does not exist.

Step 5: Analyze Option D) ${\mathrm{C}}_2$

• Total electrons = 12.
• Configuration: $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2$
• Bond Order $=\frac{1}{2}(8-4)=2$.
• Since the bond order is 2 (greater than 0), ${\mathrm{C}}_2$ can exist.

Correct Answer: (D)