JEE Chemistry Inorganic Chemistry Mixed Topics MCQs Set 01

Answer: (d) \(\ce{K3[Fe(CN)6]}\)

Explanation:
1. Ferric alum and ferric sulphate are ionic salts that release free \(\ce{Fe^{3+}}\) in water.
2. In \(\ce{K3[Fe(CN)6]}\) the iron is bound inside the stable complex ion \(\ce{[Fe(CN)6]^{3-}}\), so almost no free \(\ce{Fe^{3+}}\) is available.
3. \(\ce{Fe4[Fe(CN)6]3}\) (Prussian blue) is insoluble and does not form a true solution at all, so the soluble complex salt is the intended answer.

Teacher's Note:
1. Complex ions hold the metal strongly, so ordinary ion tests fail.
2. The ferricyanide ion gives a blue precipitate with ferrous ions, not the red colour with thiocyanate.

Answer: (c) -3, +5

Explanation:
1. In \(\ce{NH4+}\): \(x + 4(+1) = +1\), so \(x = -3\).
2. In \(\ce{NO3-}\): \(x + 3(-2) = -1\), so \(x = +5\).
3. The two nitrogen atoms are at \(-3\) and \(+5\).

Teacher's Note:
1. Work out the oxidation number separately for the cation and the anion.
2. The same element can show two different oxidation numbers in one compound.

Answer: (b) \(\ce{Pb3O4}\)

Explanation:
1. Heating PbO in air or oxygen at about 400 to 500 degrees C gives red lead: \(\ce{6PbO + O2 -> 2Pb3O4}\).
2. \(\ce{PbO2}\) is not formed by simple heating, it is made by oxidising with strong oxidising agents.

Teacher's Note:
1. Red lead, \(\ce{Pb3O4}\), is a mixed oxide, \(\ce{2PbO.PbO2}\).
2. It on stronger heating loses oxygen and goes back to PbO.

Answer: (d) \(\ce{Ag2O + H2O2 -> 2Ag + H2O + O2}\)

Explanation:
1. As a reducing agent, \(\ce{H2O2}\) loses electrons and its oxygen goes from \(-1\) to 0 in \(\ce{O2}\).
2. In (a), (b) and (c) the oxygen of \(\ce{H2O2}\) goes from \(-1\) to \(-2\), so it is the oxidising agent.
3. In (d) \(\ce{O2}\) is liberated from \(\ce{H2O2}\) while \(\ce{Ag2O}\) is reduced to Ag, so \(\ce{H2O2}\) reduces it.

Teacher's Note:
1. Oxygen is released from \(\ce{H2O2}\) only when it acts as a reductant.
2. Look for \(\ce{O2}\) among the products.

Answer: (d) \(\ce{[SiCl6]^{2-}}\)

Explanation:
1. Silicon can expand its octet using d orbitals and accept six fluoride ions, giving the stable \(\ce{[SiF6]^{2-}}\).
2. Six large chloride ions cannot be accommodated around the small silicon atom because of steric repulsion.
3. Hence \(\ce{[SiCl6]^{2-}}\) is not known, while the Ge and Sn complexes are stable.

Teacher's Note:
1. Fluoride is small and fits six around the heavier group 14 atoms.
2. Chloride is larger, so the hexachloro species of Si does not form.

Answer: (c) B

Explanation:
1. The \(\ce{MF6^{3-}}\) ion needs six fluoride ions around the metal, which means the metal must have vacant d orbitals to expand its coordination number beyond 4.
2. Boron is in the second period and has no d orbitals, so its maximum covalency is 4 (as in \(\ce{BF4-}\)).
3. Al, Ga and In can form \(\ce{AlF6^{3-}}\), \(\ce{GaF6^{3-}}\) and \(\ce{InF6^{3-}}\).

Teacher's Note:
1. Elements of the second period cannot exceed a covalency of four.
2. This explains why \(\ce{BF3}\) forms \(\ce{BF4-}\) but not \(\ce{BF6^{3-}}\).

Answer: (b) \(\ce{M2O}\)

Explanation:
1. The configuration has 11 electrons, so M is sodium with one valence electron.
2. It forms \(\ce{M+}\) and combines with \(\ce{O^{2-}}\) in the ratio 2 : 1.
3. The oxide is \(\ce{M2O}\).

Teacher's Note:
1. Valency equals the number of electrons in the outermost shell for s-block metals.
2. Criss-cross the valencies to get the formula.

Answer: (c) LiOH

Explanation:
1. Alkali metal hydroxides are thermally stable, but LiOH is less so because the small \(\ce{Li+}\) ion polarises the hydroxide and the lattice is less stable.
2. On heating, LiOH decomposes: \(\ce{2LiOH -> Li2O + H2O}\).
3. NaOH, KOH and RbOH are stable.

Teacher's Note:
1. Lithium behaves differently from the other alkali metals.
2. Its compounds resemble those of magnesium.

Answer: (b) Square planar, \(sp^3d^2\)

Explanation:
1. Xenon has 8 valence electrons; four are used in bonds with fluorine and the other four form two lone pairs.
2. There are 6 electron pairs, so the hybridization is \(sp^3d^2\) and the electron geometry is octahedral.
3. The two lone pairs occupy opposite (axial) positions, so the molecule is square planar.

Teacher's Note:
1. Count bond pairs and lone pairs to get the hybridization first, then place lone pairs as far apart as possible.
2. \(\ce{XeF4}\) is non-polar because the dipoles cancel.

Answer: (c) \(\ce{BCl3}\)

Explanation:
1. \(\ce{BCl3}\) has three bond pairs and no lone pair on boron, so it is trigonal planar with a bond angle of \(120^\circ\).
2. \(\ce{NCl3}\) and \(\ce{PH3}\) are pyramidal (lone pair on the central atom), and \(\ce{ClF3}\) is T-shaped.

Teacher's Note:
1. Trigonal planar means \(sp^2\) with no lone pair.
2. Lone pairs squeeze bond angles below the ideal values.

Answer: (d) Different, with 1, 0 and 2 lone pairs respectively

Explanation:
1. \(\ce{SF4}\) has 4 bond pairs and 1 lone pair: see-saw shape.
2. \(\ce{CF4}\) has 4 bond pairs and 0 lone pairs: tetrahedral.
3. \(\ce{XeF4}\) has 4 bond pairs and 2 lone pairs: square planar.
4. The shapes are different, with 1, 0 and 2 lone pairs respectively.

Teacher's Note:
1. Same number of atoms attached does not mean same shape.
2. Lone pairs change the geometry.

Answer: (c) \(\ce{XeF4}\)

Explanation:
1. \(\ce{NH4+}\), \(\ce{BF4-}\) and \(\ce{CCl4}\) all have four bond pairs and no lone pairs, so they are tetrahedral.
2. \(\ce{XeF4}\) has two lone pairs on the central atom in an octahedral arrangement, which makes it square planar.

Teacher's Note:
1. Four bond pairs with no lone pair means tetrahedral.
2. Square planar needs two lone pairs on opposite sides.

Answer: (d) \(\ce{CH3OCH3}\)

Explanation:
1. Hydrogen bonding needs a hydrogen atom attached to N, O or F.
2. Ethanol, acetic acid and ethylamine have O-H or N-H bonds.
3. Dimethyl ether has no hydrogen attached to oxygen, so it cannot form hydrogen bonds with itself.

Teacher's Note:
1. The molecule must contain an H on N, O or F to be a hydrogen-bond donor.
2. That is why ethanol boils much higher than its isomer dimethyl ether.

Answer: (b) \(\ce{N2}\)

Explanation:
1. Bond order \(= \tfrac{1}{2}(N_b - N_a)\).
2. \(\ce{N2}\) has 10 bonding and 4 antibonding electrons, so the bond order is \(\tfrac{1}{2}(10 - 4) = 3\).
3. \(\ce{H2}\) is 1, \(\ce{O2}\) is 2 and \(\ce{He2+}\) is 0.5.

Teacher's Note:
1. \(\ce{N2}\) has a triple bond and is very inert.
2. \(\ce{CO}\) and \(\ce{CN-}\) are isoelectronic with it.

Answer: (b) Covalent and coordinate

Explanation:
1. \(\ce{N2O5}\) is formed between two non-metals, so it has no ionic bond.
2. The structure \(\ce{O2N-O-NO2}\) has N=O double bonds, N-O single bonds and coordinate (dative) bonds from nitrogen to oxygen.
3. So both covalent and coordinate bonds are present.

Teacher's Note:
1. Oxoacids and their anhydrides of nitrogen contain coordinate bonds.
2. The answer cannot be "only covalent" because of the dative bonds.

Answer: (c) \(\ce{B2}\), \(\ce{C2}\), \(\ce{N2}\)

Explanation:
1. For molecules up to \(\ce{N2}\) (including \(\ce{Li2}\), \(\ce{Be2}\), \(\ce{B2}\), \(\ce{C2}\)) the \(\pi 2p\) orbitals lie below \(\sigma 2p_z\) because of s-p mixing.
2. For \(\ce{O2}\) and \(\ce{F2}\) the order is reversed: \(\sigma 2p_z\) lies below \(\pi 2p\).
3. The only option with all three from the first group is \(\ce{B2}\), \(\ce{C2}\), \(\ce{N2}\).

Teacher's Note:
1. The cross-over of \(\sigma 2p\) and \(\pi 2p\) happens between \(\ce{N2}\) and \(\ce{O2}\).
2. Remember this to write the electronic configuration correctly.

Answer: (d) hydrogen bonding

Explanation:
1. Nitrogen is highly electronegative and small, so the N-H bonds are polar.
2. Ammonia molecules form intermolecular hydrogen bonds, and extra energy is needed to break them.
3. This is why its boiling point is higher than that of \(\ce{PH3}\).

Teacher's Note:
1. Hydrogen bonding is found in the hydrides of N, O and F.
2. The heavier hydrides of the group do not show it.

Answer: (a) Lattice energy

Explanation:
1. The Born-Haber cycle applies Hess's law to the formation of an ionic solid from its elements through atomisation, ionisation, electron gain and lattice formation.
2. All the other quantities can be measured, so the cycle is used to find the lattice energy, which cannot be measured directly.

Teacher's Note:
1. Lattice energy is the energy released when gaseous ions form one mole of the solid.
2. The cycle is a closed energy balance.

Answer: (b) \(\ce{He2+} < \ce{O2-} < \ce{NO} < \ce{C2^{2-}}\)

Explanation:
1. \(\ce{He2+}\) has 3 electrons: 2 bonding and 1 antibonding, so the bond order is 0.5.
2. \(\ce{O2-}\) (17 electrons) has bond order 1.5.
3. \(\ce{NO}\) (15 electrons) has bond order 2.5.
4. \(\ce{C2^{2-}}\) (14 electrons, isoelectronic with \(\ce{N2}\)) has bond order 3.
5. Increasing order: \(\ce{He2+} < \ce{O2-} < \ce{NO} < \ce{C2^{2-}}\).

Teacher's Note:
1. Count electrons, fill the MO diagram, then use \(\tfrac{1}{2}(N_b - N_a)\).
2. Isoelectronic species have the same bond order.