JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 18

Answer: (a) \( \ce{N2} > \ce{O2} > \ce{O2-} \)

Explanation:
1. Bond orders are \( \ce{N2} \) 3, \( \ce{O2} \) 2 and \( \ce{O2-} \) 1.5.
2. Bond dissociation energy follows bond order.

Teacher's Note:
1. Higher bond order means a stronger bond.
2. Adding an electron to an antibonding orbital weakens the bond.

Answer: (a) HCN

Explanation:
1. HCN is H-C#N, with one C-H sigma bond and a triple bond (one sigma and two pi).
2. That gives two sigma and two pi bonds in total.
3. The other molecules have more sigma bonds.

Teacher's Note:
1. Triple bond is one sigma and two pi.
2. Count all bonds, including those to hydrogen.

Answer: (a) NO

Explanation:
1. NO has an odd number of electrons (15), so it has one unpaired electron.
2. \( \ce{O3} \), \( \ce{N2} \) and CO have all electrons paired.

Teacher's Note:
1. Odd electron molecules are paramagnetic.
2. NO is a common example.

Answer: (b) ICl

Explanation:
1. The I-Cl bond is polar and weaker than the Cl-Cl and Br-Br bonds.
2. Interhalogen compounds are generally more reactive than the halogens, except \( \ce{F2} \).

Teacher's Note:
1. Interhalogens are more reactive than the corresponding halogens.
2. Polar, weaker bonds react more easily.

Answer: (b) Xe

Explanation:
1. Noble gases are held only by London dispersion forces.
2. Xe has the most electrons and the largest polarisability, so its dispersion forces are the strongest.
3. So Xe has the highest boiling point.

Teacher's Note:
1. Boiling points rise down the group.
2. Dispersion forces increase with atomic size.

Answer: (b) square pyramidal

Explanation:
1. Xe has five bond pairs (four F and one O) and one lone pair.
2. Six electron pairs give an octahedral arrangement, and the shape is a square pyramid.

Teacher's Note:
1. AX5E is square pyramidal.
2. Same as \( \ce{BrF5} \) and \( \ce{IF5} \).

Answer: (a) \( \ce{X2Cl3} \)

Explanation:
1. \( \ce{X2O3} \), \( \ce{X2(SO4)3} \) and \( \ce{XPO4} \) all correspond to X in the +3 state.
2. With chlorine the formula for X(III) should be \( \ce{XCl3} \), so \( \ce{X2Cl3} \) is incorrect.

Teacher's Note:
1. Balance charges to test a formula.
2. Use the common oxidation state from the other compounds.

Answer: (b) 0.05

Explanation:
1. Charge q = mu / d = 0.38 x 10^-18 esu cm / 1.617 x 10^-8 cm = 2.35 x 10^-11 esu.
2. Fraction = 2.35 x 10^-11 / 4.802 x 10^-10 = 0.049.
3. So the fractional charge is about 0.05.

Teacher's Note:
1. Dipole moment = charge x distance.
2. Compare the charge with the electronic charge.

Answer: (d) \( \ce{NO2+} \)

Explanation:
1. \( \ce{NO2+} \) has two bond pairs and no lone pair on nitrogen, so it is linear and sp.
2. \( \ce{NO2-} \) and \( \ce{NO3-} \) are sp2, and \( \ce{NO2} \) is also sp2 with an odd electron.

Teacher's Note:
1. Linear AX2 means sp.
2. Steric number 2 corresponds to sp.

Answer: (b) \( \ce{ClF3} \), \( \ce{XeOF2} \), \( \ce{XeF3+} \)

Explanation:
1. \( \ce{ClF3} \), \( \ce{XeOF2} \) and \( \ce{XeF3+} \) each have three bond pairs and two lone pairs, so all are T-shaped.
2. The other groups mix different shapes.

Teacher's Note:
1. Same steric number and lone pairs give the same shape.
2. Count carefully for ions.

Answer: (a) \( \ce{CH4} \)

Explanation:
1. \( \ce{CH4} \) has no lone pair and a bond angle of 109.5 degrees.
2. The lone pairs in \( \ce{NH3} \) (107) and \( \ce{H2O} \) (104.5) reduce the angle, and \( \ce{PH3} \) is only about 93.

Teacher's Note:
1. Lone pairs squeeze bond angles.
2. Third period hydrides have angles near 90 degrees.

Answer: (b) CO

Explanation:
1. CO has 14 electrons, all paired in molecular orbitals, so it is diamagnetic.
2. NO has an unpaired electron, \( \ce{O2} \) has two and \( \ce{B2} \) has two.

Teacher's Note:
1. CO is isoelectronic with \( \ce{N2} \).
2. Diamagnetic means no unpaired electrons.

Answer: (b) \( \ce{CO3^2-} \), \( \ce{NO3-} \), \( \ce{SO3} \)

Explanation:
1. \( \ce{CO3^2-} \), \( \ce{NO3-} \) and \( \ce{SO3} \) each have three bond pairs and no lone pair on the central atom, so all are trigonal planar.
2. \( \ce{NF3} \), \( \ce{NH3} \) and \( \ce{NCl3} \) are pyramidal.

Teacher's Note:
1. AX3 with no lone pair is trigonal planar.
2. Nitrogen trihalides are pyramidal.

Answer: (b) \( \ce{H2S2O7} > \ce{H2SO4} > \ce{H2SO3} \)

Explanation:
1. The number of S=O bonds gives the number of pi bonds.
2. \( \ce{H2SO3} \) has 1, \( \ce{H2SO4} \) has 2 and \( \ce{H2S2O7} \) has 4.
3. So the order is \( \ce{H2S2O7} > \ce{H2SO4} > \ce{H2SO3} \).

Teacher's Note:
1. Draw the Lewis structure with S=O bonds.
2. Each S=O has one pi bond.

Answer: (c) \( \ce{PF5} \)

Explanation:
1. \( \ce{SF6} \) and \( \ce{BrF5} \) have six electron pairs, so they use \( sp^3d^2 \).
2. \( \ce{[CrF6]^3-} \) is an octahedral complex using \( d^2sp^3 \) (an inner orbital form, equivalent octahedral hybridisation).
3. \( \ce{PF5} \) has five electron pairs and is \( sp^3d \).

Teacher's Note:
1. Five pairs mean \( sp^3d \) and six pairs mean \( sp^3d^2 \).
2. Count pairs on the central atom.

Answer: (a) It decreases the dipole moment, because the resonance shifts charge opposite to the inductive pull of Cl

Explanation:
1. Chlorine withdraws electrons through the sigma bond (inductive effect) but donates a lone pair into the double bond (resonance).
2. The two effects act in opposite directions, so the net dipole moment is lowered.
3. The measured value for vinyl chloride (about 1.4 D) is less than that of ethyl chloride (about 2.0 D).

Teacher's Note:
1. Compare vinyl chloride with ethyl chloride.
2. The lone pair donation partly cancels the C-Cl bond dipole.

Answer: (a) S < Cl < N < O < F

Explanation:
1. Hydrogen bond strength depends on the electronegativity and small size of X.
2. F is the most electronegative and smallest, followed by O and N.
3. Cl is large for its electronegativity, and S is much weaker, so the order is S < Cl < N < O < F.

Teacher's Note:
1. Strength follows electronegativity and small size.
2. Cl is more electronegative than N but forms much weaker H-bonds.

Answer: (a) oxygen is in the -1 state, which can go down to -2 or up to 0

Explanation:
1. In \( \ce{H2O2} \) oxygen has oxidation number -1, intermediate between -2 (in water) and 0 (in \( \ce{O2} \)).
2. It can be reduced to -2 (oxidising agent) or oxidised to 0 (reducing agent).

Teacher's Note:
1. An element in an intermediate oxidation state can both gain and lose electrons.
2. The same is true of sulphur in \( \ce{SO2} \).

Answer: (a) 80

Explanation:
1. Dipole moment for a pure ionic bond = e x d = 1.602 x 10^-19 C x 2.6 x 10^-10 m = 4.17 x 10^-29 C m.
2. Percentage ionic character = 3.336 / 4.17 x 100.
3. This is about 80 percent.

Teacher's Note:
1. Compare the observed moment with the full-charge value.
2. KCl is highly ionic.

Answer: (a) Fajans' rules: the small F- is hardly polarised so LiF is ionic, while the large I- is easily polarised so LiI has covalent character

Explanation:
1. \( \ce{Li+} \) is small with high polarising power.
2. The small fluoride ion resists polarisation, so LiF is largely ionic.
3. The large iodide ion is easily distorted, so LiI has considerable covalent character.

Teacher's Note:
1. Same cation with different anions: larger anion means more covalent.
2. This is a standard Fajans' rule example.

Answer: (b) \( \ce{I2} \)

Explanation:
1. \( \ce{I-} \) donates a lone pair to the iodine molecule, so it is the Lewis base.
2. \( \ce{I2} \) accepts the electron pair into its antibonding orbital, so it is the Lewis acid.

Teacher's Note:
1. Lewis acid = electron pair acceptor.
2. Polyhalide ions form by this kind of donation.

Answer: (a) \( \ce{Ag+} \), because of its higher effective nuclear charge and pseudo-noble gas core

Explanation:
1. \( \ce{Ag+} \) has 18 electrons in its outer core, which shield the nucleus poorly.
2. So its effective nuclear charge is greater and it attracts electron pairs more strongly.
3. \( \ce{Na+} \) has a noble gas core and is a much weaker acceptor.

Teacher's Note:
1. Poor shielding by d electrons raises the Lewis acidity.
2. Compare Cu+ and K+ in the same way.

Answer: (a) \( \ce{Al^3+} < \ce{Mg^2+} < \ce{Li+} < \ce{K+} \)

Explanation:
1. Radii in pm: \( \ce{Al^3+} \) 54, \( \ce{Mg^2+} \) 72, \( \ce{Li+} \) 76, \( \ce{K+} \) 138.
2. The sequence is therefore \( \ce{Al^3+} < \ce{Mg^2+} < \ce{Li+} < \ce{K+} \).

Teacher's Note:
1. Diagonal relationship: Li+ and Mg2+ are similar in size.
2. Higher charge on the same electron count means a smaller ion.

Answer: (a) \( sp^3 \), +2 and -1

Explanation:
1. Oxygen has two bond pairs and two lone pairs, so it is \( sp^3 \) and V-shaped.
2. Fluorine is more electronegative, so it takes -1 each and oxygen is +2.

Teacher's Note:
1. F is always -1 in compounds.
2. This is one of the few compounds in which oxygen is positive.