NEET Practice Questions (MCQs) with Answers & Solutions

In molecular orbital theory, if the number of electrons in bonding molecular orbitals (BMO) is greater than the number of electrons in anti-bonding molecular orbitals (ABMO), a bond will be formed. This is because the net stabilization energy is greater than the destabilization energy, leading to a stable bond.

The ammonium ion (NH₄⁺) has a tetrahedral geometry due to sp³ hybridization, which results in bond angles of approximately 109.28°.

The bond length of a molecule is inversely proportional to the bond order. The bond orders for $O_2$, $H_2O_2$, and $O_3$ are 2, 1, and 1.5 respectively. Therefore, the bond length order is: $H_2O_2 extgreater O_3 extgreater O_2$.

A sigma (σ) bond is generally stronger than a pi (π) bond. The sigma bond is formed by the end-to-end overlap of atomic orbitals, which allows for a stronger overlap compared to the side-to-side overlap of the pi bond. Therefore, the statement 'σ bond is weaker than π bond' is incorrect.

$S^{2-}$ (sulfide ion) is not paramagnetic because it has no unpaired electrons. Paramagnetism arises due to the presence of unpaired electrons, which are absent in $S^{2-}$. Hence, $S^{2-}$ is diamagnetic.

Definition of the rate of reaction

the activation energy of the reaction is large

$ 0.2 \times 10 ^ {-4} $ $ { d [H_2 ] \over dt } = - 0.3 \times 10 ^ {-4} Ms^{-1} $ $ But Rate = - { 1 \over 3 } { d [ H_2 ] \over dt } = + { 1 \over 2 } { d [ NH_3 ] \over dt } $ $ Hence { d [NH_3 ] \over dt } = - { 2 \over 3} { d [H_2 ] \over dt }= - { 2 \over 3 } ( -0.3 \times 10 ^ {-4} ) = 0.2 \times 10 ^ {-4 } $

$ \triangle $ H of reaction