Grade 11 : Chemistry - Chemical Bonding (Set 1)

Explanation:An ionic bond is formed by the complete transfer of valence electrons from a electropositive atom (metal) to an electronegative atom (non-metal), creating oppositely charged ions held together by electrostatic forces.

Explanation:In $CH_4$, the central carbon atom has 4 bonding electron pairs and 0 lone pairs ($AX_4$ system). According to VSEPR theory, four bonding pairs orient themselves to minimize repulsion, resulting in a tetrahedral geometry with bond angles of $109.5^\circ$.

Explanation:Oxygen has 6 valence electrons. It forms 2 single covalent bonds with two hydrogen atoms using 2 electrons. The remaining 4 electrons form 2 non-bonding lone pairs on the central oxygen atom.

Explanation:In $O=C=O$, the central carbon atom forms two double bonds (two $\sigma$ bonds and two $\pi$ bonds) and has no lone pairs. Possessing 2 steric regions, carbon undergoes sp hybridization leading to a linear molecule.

Explanation:$BF_3$ has a symmetrical trigonal planar geometry ($120^\circ$ bond angles). The individual polar B-F bond dipoles cancel each other out completely vectorially, resulting in a net dipole moment of zero.

Explanation:Ethyne structure is $H-C\equiv C-H$. There are 2 C-H single bonds ($2\sigma$) and 1 C$$\equiv$$C triple bond ($1\sigma + 2\pi$). Total count = 3 $\sigma$ bonds and 2 $\pi$ bonds.

Explanation:Formal charge = $V - N - \frac{B}{2}$. For central oxygen: Valence $V=6$, Non-bonding $N=2$ (1 lone pair), Bonding $B=6$ (one double bond + one single bond = 3 shared pairs). Formal charge = $6 - 2 - \frac{6}{2} = +1$.

Explanation:In the ammonium ion ($NH_4^+$), ammonia ($NH_3$) donates its lone pair of electrons on nitrogen to an empty 1s orbital of $H^+$, forming a coordinate covalent bond.

Explanation:For $N_2$ (14 electrons): bonding electrons $N_b = 10$, antibonding electrons $N_a = 4$. Bond Order = $\frac{N_b - N_a}{2} = \frac{10 - 4}{2} = 3$.

Explanation:A linear arrangement has two electron domains arranged at maximum distance apart on opposite sides of the central atom, giving a bond angle of $180^\circ$.

Explanation:Fluorine is the most electronegative element on the periodic table with a value of 4.0 on the Pauling scale due to its small atomic radius and high effective nuclear charge.

Explanation:$SF_6$ has 6 bonding pairs and 0 lone pairs around the central sulfur atom ($AX_6$ system). This leads to an octahedral arrangement with bond angles of $90^\circ$.

Explanation:Hydrogen bonding occurs when H is bonded to highly electronegative elements (F, O, N). Among the choices, hydrogen fluoride (HF) forms strong hydrogen bonds.

Explanation:In $PCl_5$, the central phosphorus atom shares 10 electrons (5 bonding pairs) using available d-orbitals, thus expanding its valence octet beyond 8 electrons.

Explanation:The central boron atom forms 3 single $\sigma$ bonds with fluorine atoms and has 0 lone pairs. Three electron domains require $sp^2$ hybridization.

Explanation:According to MOT, the last 2 electrons in $O_2$ occupy the degenerate $\pi^* 2p_x$ and $\pi^* 2p_y$ antibonding orbitals singly with parallel spins (Hund's rule), making the molecule paramagnetic.

Explanation:$NH_3$ has 3 bonding pairs and 1 lone pair ($AX_3E$ system). The electron geometry is tetrahedral, but the molecular geometry (shape of atoms) is trigonal pyramidal.

Explanation:Higher bond order corresponds to stronger electrostatic attraction between nuclei and bonding electrons, resulting in a shorter bond length. Triple bonds ($C\equiv C$) are shorter than double and single bonds.

Explanation:In $CO_3^{2-}$, there are 4 total bonds (1 double, 2 single) shared equally across 3 resonance positions: $\text{Bond Order} = \frac{\text{Total bonding pairs}}{\text{Resonance positions}} = \frac{4}{3} \approx 1.33$.

Explanation:Lattice enthalpy is proportional to $\frac{|q_1 q_2|}{r_+ + r_-}$. MgO contains divalent ions ($Mg^{2+}$ and $O^{2-}$), giving a charge product of 4, whereas monovalent alkali halides have a charge product of 1.