JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 32

Answer: (d) 1, 3, 4, 5 and 6

Explanation:
1. In \( \ce{O2F2} \) oxygen is +1 because F is more electronegative, so (2) is out
2. The others contain a peroxide group O-O: \( \ce{H2O2} \), \( \ce{H2SO5} \), \( \ce{CrO5} \), \( \ce{H2S2O8} \) and the perborate \( \ce{Na2[B2(O2)2(OH)4]} \).

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) In \( \ce{F2} \), the energy of \( \sigma 2p_z \) is higher than that of \( \pi 2p_x \) and \( \pi 2p_y \)

Explanation:
1. For \( \ce{F2} \) (no s-p mixing) the sigma 2pz orbital lies below the pi 2px and pi 2py orbitals, so the first statement is wrong
2. Bond orders 2.5, 2 and 1.5 support the stability order
3. \( \ce{He2} \) has bond order zero
4. \( \ce{C2} \), \( \ce{O2^2-} \) and \( \ce{Li2} \) have all their electrons paired.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) During \( \ce{O2^-} \) formation, one electron is added to a bonding molecular orbital of \( \ce{O2} \)

Explanation:
1. The extra electron in \( \ce{O2^-} \) enters the pi* antibonding orbital, so the statement that it goes to a bonding orbital is wrong
2. In \( \ce{N2} \) the sigma 2p bonding electron is lost, in \( \ce{O2} \) a pi* electron is lost, and the odd electron of CN is in a bonding sigma 2p orbital so the added one pairs there.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (d) I, II and III

Explanation:
1. Each has 2.5 as bond order: \( \ce{O2^+} \) (bonding 10, antibonding 5), \( \ce{NO} \) (10, 5) and \( \ce{N2^+} \) (9, 4)
2. Each has one unpaired electron, so all are paramagnetic.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (a) increases, decreases

Explanation:
1. \( \ce{N2^+} \) has lost a bonding electron, so the bond order falls from 3 to 2.5 and the bond lengthens
2. \( \ce{O2^+} \) has lost an antibonding electron, so the bond order rises from 2 to 2.5 and the bond shortens.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (c) \( \ce{ClO2} \)

Explanation:
1. \( \ce{ClO2} \) has 7 + 12 = 19 valence electrons, an odd number, so it has an unpaired electron
2. The others have an even number of electrons and all are paired.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (a) I, III and IV only

Explanation:
1. (I) is true as \( \ce{PCl4^+} \) is sp³. (II) is false: in \( \ce{H2O} \), an AB2L2 species, the angle is 104.5 degrees, less than 109.5. (III) each is sp³ with one lone pair. (IV) the tetrahedron has six edges and each P has one lone pair.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (d) a, c and d are true, b is false

Explanation:
1. \( \ce{B2H6} \) has 12 valence electrons, all used in bonds, so (a) is true
2. Borazine \( \ce{B3N3H6} \) is a stable electron-precise ring, so (b) is false. (c) is true and (d) is true as in pyrosilicate.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) +4

Explanation:
1. In \( \ce{H4P2O6} \) the two P atoms are equivalent and joined by a P-P bond
2. The sum of oxidation states is 2x + 4 - 12 = 0, so x = +4 for each.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) 31.25%

Explanation:
1. For 100% ionic character, \( \mu = e \times d = 4.8 \times 10^\( \ce{-10} \)\ \text\( \ce{esu} \) \times 2.67 \times 10^\( \ce{-8} \)\ \text\( \ce{cm} \) = 12.8 \times 10^\( \ce{-18} \)\) esu cm = 12.8 D. The ionic character is 4.0/12.8 = 0.3125, that is 31.25%.
2. Only option (b) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (a) all of (A), (B) and (C) are correct

Explanation:
1. \( \ce{D2O} \) boils at 101.4 degrees C, higher than \( \ce{H2O} \), because the O-D...O hydrogen bonds are slightly stronger
2. Solid \( \ce{D2O} \) is denser than liquid \( \ce{H2O} \), so it sinks
3. All three statements are correct.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (c) All the species have a dipole moment

Explanation:
1. Each has 10 valence electrons (14 electrons in all), so they are isoelectronic and diamagnetic, and being diatomic all are linear
2. \( \ce{N2} \) is homonuclear and has no dipole moment, so the statement that all have a dipole moment is incorrect.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (c) Increasing strength of hydrogen bonding: Cl-H...Cl < N-H...N < O-H...O < F-H...F

Explanation:
1. Hydrogen bond strength rises with the electronegativity and smallness of the donor atom, so it follows Cl < N < O < F. The Xe-F bonds get shorter from \( \ce{XeF2} \) to \( \ce{XeF6} \), \( \ce{PH5} \) does not exist and sp³d is trigonal bipyramidal, and \( \ce{CH3Cl} \) has the larger dipole moment (1.87 D against 1.85 D).
2. Only option (c) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (a) hydrogen bonds

Explanation:
1. The two strands are linked through hydrogen bonds between complementary bases: A-T (two) and G-C (three)
2. The sugar-phosphate backbone is covalent and runs along each strand.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (b) \( \ce{PbCl2} \)

Explanation:
1. Ionic character rises with a low oxidation state and a large cation
2. \( \ce{Pb^2+} \) in \( \ce{PbCl2} \) is large and less polarising than \( \ce{Pb^4+} \)
3. So \( \ce{PbCl2} \) is the most ionic of these.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (b) \( \ce{BaCO3} \) is much less soluble than \( \ce{MgCO3} \)

Explanation:
1. Fajans' rules explain the greater covalent character and lower solubility of \( \ce{Ag2S} \), \( \ce{Fe(OH)3} \) and the low melting \( \ce{AlCl3} \)
2. The lower solubility of \( \ce{BaCO3} \) comes from the lattice and hydration energy trend of the large \( \ce{Ba^2+} \), not from polarisation.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (d) S1, S2 and S3

Explanation:
1. AgI is covalent and insoluble, but AgF is ionic and soluble, so S1 is true
2. \( \ce{BeCl2} \) is covalent and \( \ce{BaCl2} \) ionic, so S2 is true
3. A smaller, more highly charged ion attracts more water, so the hydrated radius is largest for \( \ce{Al^3+} \), so S3 is true.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (a) W > Fe > Li > K

Explanation:
1. Metallic bonding strength rises with the number of electrons contributed (d electrons included) and with smaller atomic size
2. Tungsten uses the most d electrons, then iron, then lithium (smaller) and finally potassium.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (c) Q, S, T, R

Explanation:
1. The elements are P = Si, Q = K, R = Al, S = Cr and T = Kr
2. \( \ce{K^+} \) is isoelectronic with \( \ce{P^3-} \) (i)
3. Cr shows +6 and its compounds are coloured (ii)
4. Kr, a noble gas, has the largest radius and highest ionisation energy in its period (iii)
5. Al has an amphoteric oxide (iv).

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (a) 16

Explanation:
1. The bonds are three Si-H (3), one Si-N (1), the N=C double bond (2) and the C=O double bond (2)
2. That is 8 bonds, so 8 x 2 = 16 bonding electrons.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (d) B and D only

Explanation:
1. (A) and (C) are true
2. Lone pairs and multiple bonds need more space, so they go to equatorial positions, making (B) false
3. Bigger atoms also go to the roomier equatorial positions, so (D) is false.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (b) A and B only

Explanation:
1. The angles are 109.5, 107 and about 104 degrees, so (A) is correct
2. Boron in \( \ce{(CH3)3B} \) has 3 bond pairs, so (B) is correct
3. N in \( \ce{NH4^+} \) is sp³, so (C) is false
4. Each S in \( \ce{S8} \) has two lone pairs (4 electrons), which is 32 electrons in all, so (D) is false.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) A and D only

Explanation:
1. In \( \ce{BF3} \) the filled p orbital of F back-donates to the empty p orbital of B. In \( \ce{SiH3NCO} \) the lone pair of N goes to the empty d orbitals of Si
2. Neither \( \ce{(CH3)3N} \) nor \( \ce{CH3NCO} \) has a suitable empty orbital on a neighbouring atom.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (d) B and C only

Explanation:
1. \( \ce{(CH3)2O} \) is a good Lewis base and reacts with \( \ce{BF3} \)
2. In \( \ce{(SiH3)2O} \) and \( \ce{(SiH3)3N} \) the lone pairs are delocalised into the empty d orbitals of silicon, so these compounds are very weak bases and do not form adducts with \( \ce{BF3} \).

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.