Chemical bonding & molecular structure practice · Mixed difficulty

Option A (Released and ionic bond is formed) is correct because the reaction between sodium and chlorine is highly exothermic, releasing energy, and forms an ionic bond due to electron transfer.

Option B ($s{p^2}$-hybridization) is correct because equilateral triangles have bond angles of 120 degrees, which corresponds to $s{p^2}$ hybridization.

Option B ($\mathrm{CsF}$) is correct because it is an ionic compound with strong electrostatic forces, requiring more energy to break compared to the other options which are held together by weaker intermolecular forces.

Option A (4) is correct because each water molecule in ice forms hydrogen bonds with four other water molecules in a tetrahedral arrangement.

Option B ($1 \times {10^{ - 18}}\,esu.\,cm$) is correct because 1 debye is defined as $1 \times {10^{ - 18}}$ electrostatic units times centimeter.

Option B ($Mg\mathrm{ }\, and \mathrm{ }\,F$) is correct because magnesium (Mg) readily loses two electrons to form Mg²⁺, and fluorine (F) readily gains one electron to form F⁻. The higher charges lead to a stronger electrostatic attraction and a more stable ionic bond compared to the other options.

Option B ($ICl_2^ - ,\,BeC{l_2}$) is correct because $ICl_2^-$ has $sp^3d$ hybridization and linear geometry, while $BeCl_2$ has $sp$ hybridization and linear geometry.

Option D (${\mathrm{H}_2}\mathrm{O}$) is correct because water has a bent shape due to the presence of two lone pairs on the oxygen atom, making it non-linear.

Option A ($\mathrm{\theta } = 90^\circ$) is correct because the resultant dipole moment is maximum when the angle between the two bond dipoles is the smallest, which is $90^\circ$ in this case.

Option C (5) is correct because the formula of the compound is A_(2)B_(3), so x+y = 2+3 = 5.

Option C ($N{O_2}^ +$) is correct because it has a linear structure with a bond angle of 180 degrees, which is the largest among the given options.

Option A (Na) is correct because sodium (Na) only forms one stable oxidation state (+1) and thus only forms one chloride (NaCl), while the other elements can have multiple oxidation states.

Option A (Tetrahedral) is correct because the silicate anion ($\mathrm{SiO}_4^{4 - }$) has four oxygen atoms surrounding the central silicon atom, forming a tetrahedral structure.

Option C (Lone pair-lone pair repulsion and lone pair-bond pair repulsion) is correct because lone pairs occupy equatorial positions to minimize both types of repulsions.

Option B (5) is correct because after the excitation of two electrons, chlorine can form five covalent bonds by using one s, three p, and one d orbitals.

Option C ($109^\circ$) is correct because methane ($CH_4$) has four bonding pairs and no lone pairs around the central carbon atom, resulting in a tetrahedral geometry with bond angles of approximately 109 degrees.

Option C (${P_2}L,RL,PQ$ and $R{Q_2}$) is correct because it correctly predicts the ionic compounds formed based on the electronic configurations and valencies of the elements.

Option C (136 pm) is correct because the bond length of H-Cl is approximately the sum of the covalent radii of H and Cl, which can be estimated from the given internuclear distances.

Option B (-1,+1,0) is correct because it represents the formal charges calculated based on the number of valence electrons, lone pairs, and bonds for each atom.

Option C ($M$) is correct because element M has the electronic configuration $1{s^2},2{s^2}2{p^4}$, meaning it needs two more electrons to complete its octet, thus forming a double bond.

Option A (16 shared and 8 unshared valence electrons) is correct because acetic acid has 8 bonding pairs (16 shared electrons) and 4 lone pairs (8 unshared electrons).

Option B is correct. Langmuir introduced the term \"covalent bond.\"

Option B (45%) is correct because, based on Fajan's rules and electronegativity differences, CsCl exhibits approximately 45% covalent character.

Option A (Bond order three and isoelectronic) is correct because all three species have 14 electrons and a bond order of 3.

Option B (Hydrogen bonding) is correct because methanol (CH3OH) has a strong hydrogen bonding network that must be overcome to convert it to a gas.

Option C (-0.75) is correct because the formal charge on each oxygen atom in the phosphate ion is -0.75.

Option C ($HF_2^ -$) is correct because in $HF_2^-$, the bonding is primarily ionic and covalent, without a coordinate bond.

Option B ($BeS{O_4}$) is correct because beryllium compounds tend to have more covalent character due to the small size and high polarizing power of $Be^{2+}$, leading to lower melting points compared to other alkaline earth metal sulfates.

Option D (Greatest in acetylene) is correct because the percentage of s-character is highest in acetylene (50%), leading to a stronger and shorter C-H bond, and thus higher bond energy.

Option D ($\mathrm{Cl}{\mathrm{F}_3}$) is correct because Cl has 7 valence electrons, and in ClF3, it forms 3 bonds and has 2 lone pairs, resulting in 10 electrons around Cl.