Chemistry MCQs for NEET — Practice Questions with Answers

The boiling points of compounds in groups 14, 15, 16, and 17 generally increase with increasing molecular weight and size. However, due to hydrogen bonding, there are exceptions. For example, $H_2O$ has a higher boiling point than $H_2S$, $HF$ has a higher boiling point than $HCl$, and $NH_3$ has a higher boiling point than $PH_3$. But $CH_4$ has a lower boiling point than $SiH_4$. So, the correct pair that is not in order is $CH_4 extgreater SiH_4$.

Ionic bonds are formed between metals and non-metals due to the transfer of electrons. Among the given options, $MgCl_2$ (magnesium chloride) is an ionic compound because it is formed from magnesium (a metal) and chlorine (a non-metal). The other options are covalent compounds.

The bond energy of a hydrogen bond is typically around 40 kJ/mol. This value indicates the amount of energy required to break the hydrogen bond and is significantly stronger than van der Waals forces but weaker than covalent bonds. Hence, option o2 is correct.

Intermolecular hydrogen bonding is possible in molecules where hydrogen is directly bonded to highly electronegative atoms such as nitrogen, oxygen, or fluorine. In the given options, $NH_3$ (ammonia) has hydrogen atoms bonded to nitrogen, making intermolecular hydrogen bonding possible. Therefore, option o4 is correct.

Theorectical dipole momentum $ H = q \times d [ 1 D = 1 \times 10 ^ {-18} esa cm ] $ $ = 4.8 \times 10 ^ {-10} \times 1.275 \times 10 ^ {-8} $ $ = 6.12 \times 10 ^ {-10} e.s.u cm $ $ = 6.12 D $ $ % of ionic character = { Practical ( \mu ) \over Theoretical (\mu) } $ $ = { 1.03 \over 6.12} \times 100 $ $ = 16.83 $ $ \approx 17 \% $

Bond enthalpy is affected by the number of nonbonding electron pairs on bonded atoms. Nonbonding electron pairs can cause repulsion, which weakens the bond and thus lowers the bond enthalpy.

KO₂ (Potassium superoxide) is paramagnetic due to the presence of unpaired electrons in its molecular orbital configuration. The presence of a single unpaired electron in the antibonding Ï€* orbital makes KO₂ paramagnetic.