WBJEE Inorganic Chemistry Syllabus

Why does the $\ce{He2}$ molecule not exist according to MO theory?

$\ce{He2}$ has 4 electrons: configuration $\sigma_{1s}^{2}\,\sigma^{*}_{1s}^{2}$. Bond order $=\frac{1}{2}(2-2)=0$, so no net bond forms and the molecule is unstable.

Use MO theory to explain why $\ce{O2}$ is paramagnetic.

$\ce{O2}$ (16 electrons) has the configuration with two unpaired electrons in the degenerate $\pi^{*}_{2p}$ antibonding orbitals: $\pi^{*}_{2p_x}^{1}\pi^{*}_{2p_y}^{1}$. Unpaired electrons make it paramagnetic; bond order $=2$.

For the second-period diatomics up to $\ce{N2}$, what is the MO energy ordering, and how does it change at $\ce{O2}$?

Up to $\ce{N2}$: $\sigma_{2s} < \sigma^{*}_{2s} < \pi_{2p_x}=\pi_{2p_y} < \sigma_{2p_z} < \pi^{*} < \sigma^{*}_{2p}$ ($\pi$ below $\sigma_{2p}$ due to s-p mixing). For $\ce{O2}$ and $\ce{F2}$, $\sigma_{2p_z}$ drops below the $\pi_{2p}$ orbitals.

Define hydrogen bonding and distinguish intermolecular from intramolecular hydrogen bonding.

A hydrogen bond is the attraction between an H atom bonded to a highly electronegative atom ($\ce{F}$, $\ce{O}$, $\ce{N}$) and a lone pair on another electronegative atom. Intermolecular H-bonds form between molecules (e.g. water); intramolecular H-bonds form within a single molecule (e.g. o-nitrophenol).

Why does o-nitrophenol have a lower boiling point than p-nitrophenol?

o-Nitrophenol forms intramolecular hydrogen bonds (chelation), so molecules do not associate. p-Nitrophenol forms intermolecular hydrogen bonds, increasing molecular association and raising its boiling point.

Define dipole moment, give its formula and SI/conventional unit.

Dipole moment is the product of the magnitude of charge and the distance between the charges: $$\mu = q \times d$$ It is a vector (points from positive to negative). Commonly expressed in debye (D); $1\,\text{D} = 3.336 \times 10^{-30}\ \text{C m}$.

Why is the dipole moment of $\ce{CO2}$ zero but that of $\ce{H2O}$ nonzero?

$\ce{CO2}$ is linear and symmetric, so the two bond dipoles cancel ($\mu = 0$). $\ce{H2O}$ is bent, so the two O-H bond dipoles do not cancel and add to a net dipole ($\mu \approx 1.85\,\text{D}$).

Why is the dipole moment of $\ce{NH3}$ greater than that of $\ce{NF3}$?

In $\ce{NH3}$ the bond dipoles and the lone-pair dipole point in the same direction (add up). In $\ce{NF3}$ the N-F bond dipoles point opposite to the lone-pair dipole and partly cancel, giving a smaller net moment ($\ce{NH3}=1.46\,\text{D}$, $\ce{NF3}=0.24\,\text{D}$).

Justify placing hydrogen with both group 1 and group 17.

Like alkali metals (group 1), $\ce{H}$ has one valence electron ($1s^{1}$), forms $\ce{H+}$, and shows $+1$ oxidation state. Like halogens (group 17), it needs one electron to complete its shell, forms $\ce{H-}$ (hydride), and exists as a diatomic gas $\ce{H2}$.

Name the three isotopes of hydrogen with their mass numbers and composition.

Protium $\ce{^1_1H}$ (1p, 0n), Deuterium $\ce{^2_1H}$ or D (1p, 1n), and Tritium $\ce{^3_1H}$ or T (1p, 2n). Tritium is radioactive ($\beta$-emitter).

What is water gas (synthesis gas) and how is it produced industrially?

Water gas is a mixture of $\ce{CO}$ and $\ce{H2}$. It is produced by passing steam over red-hot coke: $$\ce{C(s) + H2O(g) ->[1270\,K] CO(g) + H2(g)}$$

Distinguish temporary and permanent hardness of water, including the responsible salts.

Temporary hardness is due to dissolved bicarbonates $\ce{Ca(HCO3)2}$ and $\ce{Mg(HCO3)2}$ and is removed by boiling or Clark's method (lime). Permanent hardness is due to chlorides and sulphates ($\ce{CaCl2}$, $\ce{MgSO4}$) and is removed by washing soda or ion-exchange.

What is the structure and key property of hydrogen peroxide ($\ce{H2O2}$) regarding redox behavior?

$\ce{H2O2}$ has a non-planar open-book structure. It acts as both an oxidizing and a reducing agent in acidic and basic media. Example oxidation: $\ce{2I- + H2O2 + 2H+ -> I2 + 2H2O}$; reduction (toward stronger oxidant): $\ce{2MnO4- + 5H2O2 + 6H+ -> 2Mn^2+ + 5O2 + 8H2O}$.

Classify hydrides into three types with an example of each.

Ionic/saline hydrides (s-block metals, e.g. $\ce{NaH}$, $\ce{CaH2}$); covalent/molecular hydrides (p-block, e.g. $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$); metallic/interstitial hydrides (many d- and f-block metals, often non-stoichiometric, e.g. $\ce{LaH_{2.87}}$).

List the general physical and chemical characteristics of alkali metals (group 1).

They are soft, low-density, low-melting silvery metals with one valence electron ($ns^{1}$). They are highly electropositive, the most reactive metals, strong reducing agents, form $+1$ ions, and impart characteristic flame colors. Reactivity increases down the group.

Why do alkali metals impart characteristic colors to a flame, and give the colors of $\ce{Li}$, $\ce{Na}$, and $\ce{K}$.

Flame heat excites valence electrons to higher levels; on returning they emit visible-light photons. $\ce{Li}$ = crimson red, $\ce{Na}$ = golden yellow, $\ce{K}$ = lilac/violet.

What is washing soda, give its formula, and the industrial process for its manufacture.

Washing soda is sodium carbonate decahydrate, $\ce{Na2CO3.10H2O}$. It is made by the Solvay (ammonia-soda) process, the key step being $\ce{NaCl + NH3 + CO2 + H2O -> NaHCO3 + NH4Cl}$, followed by heating: $\ce{2NaHCO3 ->[\Delta] Na2CO3 + H2O + CO2}$.

Why is $\ce{LiF}$ much less soluble in water than other alkali metal halides like $\ce{LiCl}$?

$\ce{LiF}$ has a very high lattice enthalpy (both ions are very small), which is not compensated by the hydration enthalpy, so it is sparingly soluble. With larger anions (e.g. $\ce{Cl-}$) lattice enthalpy falls and solubility rises.

State the general properties of alkaline earth metals (group 2) and how they compare to group 1.

They have $ns^{2}$ configuration, form $+2$ ions, are harder, denser, higher-melting, and less reactive than the corresponding alkali metals. They are good reducing agents but weaker than group 1; reactivity increases down the group.

Compare the solubility trends of group 2 hydroxides and sulphates down the group.

Solubility of hydroxides increases down the group ($\ce{Mg(OH)2}$ sparingly soluble, $\ce{Ba(OH)2}$ fairly soluble) because hydration enthalpy dominates. Solubility of sulphates decreases down the group ($\ce{BaSO4}$ insoluble) because lattice enthalpy decreases slowly while hydration enthalpy falls sharply for the large cations.

What is dead burnt plaster and Plaster of Paris? Give the formula and setting reaction of Plaster of Paris.

Plaster of Paris is calcium sulphate hemihydrate, $\ce{CaSO4.\tfrac{1}{2}H2O}$, made by heating gypsum to $\sim 393\,\text{K}$. On adding water it sets to gypsum: $\ce{CaSO4.\tfrac{1}{2}H2O + \tfrac{3}{2}H2O -> CaSO4.2H2O}$. Heating gypsum above $473\,\text{K}$ gives anhydrous 'dead burnt' plaster.

Why is $\ce{BeCl2}$ covalent and a Lewis acid, unlike other group 2 chlorides?

$\ce{Be^2+}$ is very small with high charge density (high polarizing power, Fajans' rules), so $\ce{BeCl2}$ is essentially covalent. Be has an incomplete octet (electron-deficient), making it a Lewis acid; in the solid/vapour it polymerizes via chloride bridges.

Why does boron show an anomalously high first ionization enthalpy trend break and form only covalent compounds like $\ce{BF3}$ with an incomplete octet?

Boron is very small with high ionization enthalpy, so it cannot easily lose its 3 valence electrons to form $\ce{B^3+}$; it forms covalent bonds instead. In $\ce{BF3}$ boron has only 6 valence electrons (incomplete octet), making it a strong Lewis acid that accepts a lone pair to complete its octet.

Explain the inert pair effect and its consequence in group 14 (e.g. $\ce{Pb}$).

The inert pair effect is the reluctance of the $ns^{2}$ electrons to participate in bonding for heavier p-block elements (poor shielding by d/f electrons). In group 14, the $+2$ state becomes more stable than $+4$ down the group, so $\ce{Pb^2+}$ ($\ce{PbCl2}$) is more stable than $\ce{Pb^4+}$.

Why is $\ce{CO2}$ a gas while $\ce{SiO2}$ is a high-melting solid?

$\ce{CO2}$ is a discrete linear molecule ($\ce{O=C=O}$) with weak intermolecular forces, so it is a gas. $\ce{SiO2}$ forms a giant 3D covalent network of $\ce{Si-O}$ single bonds (Si does not readily form $p\pi$-$p\pi$ bonds), giving a hard, high-melting solid.

Why does carbon form strong multiple bonds and long chains (catenation) far more than silicon?

Carbon is small and forms strong $\ce{C-C}$ ($\sigma$) and effective $p\pi$-$p\pi$ ($\pi$) bonds. Silicon is larger; its $\ce{Si-Si}$ bonds are weaker and it forms poor $p\pi$-$p\pi$ bonds, so its catenation and multiple-bond tendency are much lower.

Why is the catenation order $\ce{N} \ll \ce{P}$ and why does only nitrogen form $\ce{N2}$ with a triple bond?

Nitrogen, being small, forms strong $p\pi$-$p\pi$ triple bonds, giving the very stable $\ce{N#N}$ molecule, but its $\ce{N-N}$ single bond is weak (lone-pair repulsion), so it catenates poorly. Phosphorus forms stronger $\ce{P-P}$ single bonds and exists as $\ce{P4}$, with greater catenation than nitrogen.

Describe the industrial manufacture of ammonia (Haber process) with optimum conditions.

$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \quad \Delta H = -92\ \text{kJ mol}^{-1}$$ Optimum conditions: pressure $\sim 200\,\text{atm}$, temperature $\sim 700\,\text{K}$, with finely divided iron catalyst promoted by $\ce{K2O}$ and $\ce{Al2O3}$ (Mo as promoter).

State the trend in thermal stability and reducing character of group 15 hydrides ($\ce{NH3}$ to $\ce{BiH3}$).

Down the group the $\ce{M-H}$ bond weakens, so thermal stability decreases ($\ce{NH3} > \ce{PH3} > \ce{AsH3} > \ce{SbH3} > \ce{BiH3}$) and reducing character increases ($\ce{BiH3}$ is the strongest reducing agent). Basicity also decreases: $\ce{NH3} > \ce{PH3} > \ce{AsH3} > \ce{SbH3}$.

Describe the manufacture of sulphuric acid by the Contact process, including the key catalytic step.

Key steps: burn sulphur/pyrites to $\ce{SO2}$; catalytically oxidize $$\ce{2SO2 + O2 <=>[V2O5] 2SO3}$$ (vanadium pentoxide catalyst, $\sim 720\,\text{K}$, $2\,\text{atm}$); absorb $\ce{SO3}$ in conc. $\ce{H2SO4}$ to form oleum $\ce{H2S2O7}$, then dilute: $\ce{H2S2O7 + H2O -> 2H2SO4}$.

Why does fluorine show anomalous behavior such as a low bond dissociation enthalpy of $\ce{F2}$ and only the $-1$ oxidation state?

$\ce{F2}$ has a low bond enthalpy due to strong repulsion between non-bonding lone pairs on the two small F atoms. Fluorine is the most electronegative element and has no available d-orbitals, so it shows only the $-1$ oxidation state (never positive).

Explain why noble gases are largely inert and which one forms compounds; give an example.

Noble gases (group 18) have completely filled valence shells ($ns^{2}np^{6}$, He is $1s^{2}$), giving very high ionization enthalpies and near-zero electron gain enthalpy, hence chemical inertness. Xenon, with the lowest ionization enthalpy among the reactive ones, forms fluorides, e.g. $\ce{Xe + F2 ->[673\,K,\,1\,atm] XeF2}$ ($\ce{XeF4}$, $\ce{XeF6}$ under other conditions).

Give the shapes of $\ce{XeF2}$, $\ce{XeF4}$, and $\ce{XeF6}$ from VSEPR.

$\ce{XeF2}$: linear ($sp^{3}d$, 3 lone pairs). $\ce{XeF4}$: square planar ($sp^{3}d^{2}$, 2 lone pairs). $\ce{XeF6}$: distorted octahedral ($sp^{3}d^{3}$, 1 lone pair).