Question 1

Bromine trifluoride autoionizes to form BrF _2^ and BrF _4^ . The shapes of the cation and anion are respectively ________, and ________.

Accepted Answer

The autoionization of bromine trifluoride is given by the reaction: 2 BrF _3 BrF _2^ + BrF _4^ For the cation BrF _2^ : The central atom Br has 7 valence electrons. Due to the positive charge, it has 7 - 1 = 6 electrons available. It forms 2 single bonds with F atoms, leaving 4 electrons, which constitute 2 lone pairs. Steric number = 2 (bond pairs) + 2 (lone pairs) = 4. The hybridization is sp^3. With 2 bond pairs and 2 lone pairs, the shape is bent (or V-shaped). For the anion BrF _4^ : The ce

Question 2

Given below are two statements : Statement I : The number of compounds among SO_2, SO_3, SF_4, SF_6 and H_2S in which sulphur does not obey the Octet rule is 3. Statement II : Among [H_2O, ClF_3, SF_4

Accepted Answer

Statement I: Number of compounds in which sulphur does not follow the octet rule (from SO_2, SO_3, SF_4, SF_6 and H_2S). SO_2: The formal-charge-minimised structure O=S=O places 10 electrons around sulphur (expanded octet). SO_3: With three S=O double bonds, sulphur has 12 electrons around it (expanded octet). SF_4: 4 bond pairs + 1 lone pair = 10 electrons around S (expanded octet). SF_6: 6 bond pairs = 12 electrons around S (expanded octet). H_2S: 2 bond pairs + 2 lone pairs = 8 electrons arou

Question 3

Given below are two statements : Statement I : F_2O Statement II : SiF_4, SnF_4 and PbF_4 are ionic in nature. In the light of the above statements, choose the correct answer from the options given be

Accepted Answer

In Statement I, the bond angle of F_2O is 103^ , H_2O is 104.5^ , and Cl_2O is 110.9^ . The increasing order of bond angle is F_2O In Statement II, SiF_4 is a covalent compound (it is a gas at room temperature), whereas SnF_4 and PbF_4 are ionic solids. Thus, Statement II is false. Answer: Statement I is true but Statement II is false

Question 4

The pairs among A = [ SO _3^ 2- , CO _3^ 2- ], B = [ O _2^ 2- , F _2], C = [ CN ^-, CO ], D = [ NH _3, H _3 O ^+] and E = [ MnO _4^ 2- , CrO _4^ 2- ] that do not have similar Lewis dot structure are

Accepted Answer

Species with the same number of valence electrons and the same skeletal structure generally have similar Lewis dot structures (isoelectronic species). For pair A: SO _3^ 2- has 6 + 3(6) + 2 = 26 valence electrons. CO _3^ 2- has 4 + 3(6) + 2 = 24 valence electrons. Since they have different numbers of valence electrons, they do not have similar Lewis dot structures ( SO _3^ 2- has a lone pair on the central atom and is pyramidal, while CO _3^ 2- has no lone pair on the central atom and is trigona

Question 5

The covalent radii of atoms A and B are r_A and r_B, respectively. The covalent bond length and total length of AB molecule are respectively

Accepted Answer

The covalent bond length of the AB molecule is the internuclear distance between atom A and atom B. Bond length = r_A + r_B The total length of the AB molecule is the distance between the outermost boundaries of the two atoms along the internuclear axis. Total length = Radius of A + Bond length + Radius of B Total length = r_A + (r_A + r_B) + r_B = 2(r_A + r_B) Answer: (r_A + r_B), 2(r_A + r_B)

Question 6

Consider the following species: BrF_5, XeF_5^-, BF_4^-, ICl_4^-, XeF_4, SF_4, NH_4^+, ClF_3, XeF_2, ICl_2^- Number of species having sp^3d hybridized central atom is __________.

Accepted Answer

The steric number H is calculated as H = 1 2 (V + M - C + A), where V is the number of valence electrons on the central atom, M is the number of monovalent atoms, C is the cationic charge, and A is the anionic charge. For sp^3d hybridization, H = 5. BrF_5: H = 1 2 (7 + 5) = 6 (sp^3d^2) XeF_5^-: H = 1 2 (8 + 5 + 1) = 7 (sp^3d^3) BF_4^-: H = 1 2 (3 + 4 + 1) = 4 (sp^3) ICl_4^-: H = 1 2 (7 + 4 + 1) = 6 (sp^3d^2) XeF_4: H = 1 2 (8 + 4) = 6 (sp^3d^2) SF_4: H = 1 2 (6 + 4) = 5 (sp^3d) NH_4^+: H = 1 2 (

Question 7

Which of the following pictorial diagram most correctly represents the ^* ( -antibonding) molecular orbital between two atoms if the internuclear axis is taken to be in the z-direction ( z-axis )?

Accepted Answer

To determine the correct pictorial representation of a ^ * ( -antibonding) molecular orbital, we need to consider how it is formed. 1. A molecular orbital is formed by the sideways (lateral) overlap of atomic orbitals (like p-orbitals) whose axes are perpendicular to the internuclear axis (z-axis). 2. An antibonding molecular orbital ( ^ * ) is formed by the out-of-phase (subtractive) overlap of these atomic orbitals. This means the positive lobe of one orbital overlaps with the negative lobe of

Question 8

According to Lewis theory, the total number of bond-pairs and lone pair of electrons around the central atom of XeO_6^ 4- ion is _____.

Accepted Answer

The central atom in XeO_6^ 4- is Xenon (Xe). The number of valence electrons in Xe is 8. Let the oxidation state of Xe in XeO_6^ 4- be x. x + 6(-2) = -4 x = +8 Since the oxidation state of Xe is +8, all 8 of its valence electrons are involved in bonding with the 6 oxygen atoms. There are no non-bonding electrons left on the central atom. Number of lone pairs on Xe = 0 Xe forms 6 bonds with the 6 oxygen atoms (and 2 bonds to satisfy its +8 oxidation state and minimize formal charge). Number of bo

Question 9

SF_4 is isostructural with: A. BrF_4^ B. CH_4 C. IF_4^ D. XeF_4 E. XeO_2 F_2 Choose the correct answer from the options given below:

Accepted Answer

In SF_4, the central sulphur atom has 6 valence electrons. It forms 4 sigma bonds with fluorine atoms and has 1 lone pair. The steric number is 5, corresponding to sp^3d hybridization, and the shape is see-saw. In BrF_4^ , the central bromine atom has 8 valence electrons (7 + 1). It forms 4 sigma bonds and has 2 lone pairs. The steric number is 6, corresponding to sp^3d^2 hybridization, and the shape is square planar. In CH_4, the central carbon atom has 4 valence electrons. It forms 4 sigma bon

Question 10

Given below are two statements : Statement (I) : The correct sequence of bond lengths in the following species is : O_2^+ Statement (II) : The correct sequence of number of unpaired electrons in the f

Accepted Answer

The bond order of the given oxygen species can be calculated using molecular orbital theory: For O_2 (16 electrons), the configuration is _ 1s ^2 ^ * _ 1s ^2 _ 2s ^2 ^ * _ 2s ^2 _ 2p_z ^2 _ 2p_x ^2 = _ 2p_y ^2 ^ * _ 2p_x ^1 = ^ * _ 2p_y ^1. Bond order = 10 - 6 2 = 2. Number of unpaired electrons = 2. For O_2^+ (15 electrons), one electron is removed from a ^ * orbital. Bond order = 10 - 5 2 = 2.5. Number of unpaired electrons = 1. For O_2^- (17 electrons), one electron is added to a ^ * orbital.

Question 11

Match List - I with List - II according to shape. \( array llll & List-I & & List-II \ A. & XeO_3 & I. & BrF_5 \ B. & XeF_2 & II. & NH_3 \ C. & XeO_2F_2 & III. & [ I_3 ]^- \ D. & XeOF_4 & IV. & SF

Accepted Answer

Determine molecular shapes using VSEPR theory: A. XeO₃ (3 bonds, 2 lone pairs on Xe): pyramidal B. XeF₂ (2 bonds, 3 lone pairs on Xe): linear C. XeO₂F₂ (4 bonds, 2 lone pairs on Xe): see-saw D. XeOF₄ (5 bonds, 1 lone pair on Xe): square pyramidal Matching with shapes: I. BrF₅: square pyramidal → D-I II. NH₃: pyramidal → A-II III. [I₃]⁻: linear → B-III IV. SF₄: see-saw → C-IV

Question 12

Given below are two statements: Statement I : The number of species among BF _ 4 ^ - , SiF _ 4 , XeF _ 4 and SF _ 4 , that have unequal E - F bond lengths is two. Here, E is the central atom. Statemen

Accepted Answer

Statement I: Let's analyze the geometry of the given species. BF_4^-: sp^3 hybridization, tetrahedral geometry. All B-F bond lengths are equal. SiF_4: sp^3 hybridization, tetrahedral geometry. All Si-F bond lengths are equal. XeF_4: sp^3d^2 hybridization, square planar geometry. All Xe-F bond lengths are equal. SF_4: sp^3d hybridization, see-saw geometry. It has axial and equatorial bonds. Axial bonds are longer than equatorial bonds due to greater repulsion. Thus, only SF_4 has unequal bond len

Question 13

Pair of species among the following having same bond order as well as paramagnetic character will be-

Accepted Answer

Bond orders and magnetic character: O₂⁻ (BO=1.5, paramagnetic), O₂⁺ (BO=2.5, paramagnetic), O₂²⁻ (BO=1, diamagnetic); N₂⁺ (BO=2.5, paramagnetic), N₂⁻ (BO=2.5, paramagnetic), N₂²⁻ (BO=2, diamagnetic) 1 : BO 1.5 vs 2.5 (different); 2 : BO 2.5 vs 2 (different); 3 : BO 1.5 vs 2.5 (different); 4 : O₂⁺ (BO=2.5, paramagnetic) and N₂⁻ (BO=2.5, paramagnetic) - both same bond order and paramagnetic character

Question 14

Among the following, the CORRECT combinations are A. IF _ 3 ~T -shaped (sp . ^ 3 ~d ) B. IF _ 5 Square pyramidal ( sp ^ 3 ~d ^ 2 ) C. IF _ 7 Pentagonal bipyramidal ( sp ^ 3 ~d ^ 3 ) D. ClO _ 4 ^ - Squ

Accepted Answer

Analyzing geometry and hybridization: A. IF₃: Iodine has 7 valence electrons, forms 3 bonds with F, leaving 2 lone pairs. Total electron pairs = 5 (trigonal bipyramidal geometry). Molecular shape is T-shaped. Hybridization = sp³d. CORRECT. B. IF₅: Iodine forms 5 bonds with F and has 1 lone pair. Total electron pairs = 6 (octahedral geometry). Molecular shape is square pyramidal with 1 lone pair in axial position. Hybridization = sp³d². CORRECT. C. IF₇: Iodine forms 7 bonds with F and has 0 lone

Question 15

Given below are statements about some molecules/ions. Identify the CORRECT statements. A. The dipole moment value of NF _ 3 is higher than that of NH _ 3 . B. The dipole moment value of BeH _ 2 is zer

Accepted Answer

Statement A: In NH_3, the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of the N-H bonds. In NF_3, the orbital dipole is in the opposite direction to the resultant dipole moment of the N-F bonds. Thus, dipole moment of NH_3 (1.47\ D) is higher than NF_3 (0.23\ D). Statement A is incorrect. Statement B: BeH_2 is a linear molecule (sp hybridization). The two Be-H bond dipoles are equal and opposite, so they cancel each other out, making the net dipole

Question 16

Which statements are NOT TRUE about XeO _ 2 ~F _ 2 ? A. It has a see-saw shape. B. Xe has 5 electron pairs in its valence shell in XeO _ 2 ~F _ 2 . C. The O - Xe - O bond angle is close to 180^ . D. T

Accepted Answer

In XeO_2F_2: Xe has 8 valence electrons forming 4 bonding pairs (2 with O, 2 with F) and 2 lone pairs, giving a trigonal bipyramidal electron geometry with a see-saw molecular shape, making statement A TRUE. Statement B is NOT TRUE because the total electron pairs around Xe is 6 (4 bonding + 2 lone), not 5. For statement C, the O-Xe-O angle in a see-saw geometry is approximately 120°, not close to 180°, making it NOT TRUE. Statement D is TRUE as the F-Xe-F angle is approximately 180° when F atom

Question 17

Identify the molecule (X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among HNO _ 3 , H _ 2 SO _ 4 , NF _ 3 and O _ 3 . Choose the correct bond angle made by th

Accepted Answer

Analyzing lone pairs on central atoms: HNO₃: N is central with 3 bonds and 0 lone pairs H₂SO₄: S is central with 4 bonds and 0 lone pairs NF₃: N is central with 3 bonds and 1 lone pair O₃: Central O with 2 bonding regions and 1 lone pair NF₃ has maximum lone pairs (1) with the clearest structure. N undergoes sp³ hybridization with 1 lone pair and 3 bonding pairs giving trigonal pyramidal geometry. The F-N-F bond angle is reduced from ideal tetrahedral 109.5° due to lone pair repulsion, resulting

Question 18

Among H _ 2 ~S , H _ 2 O , NF _ 3 , NH _ 3 and CHCl _ 3 , identify the molecule ( X ) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule

Accepted Answer

Comparing dipole moments: H_2S ≈ 0.97 D (bent, 2 lone pairs on S), H_2O ≈ 1.85 D (bent, 2 lone pairs on O), NF_3 ≈ 0.24 D (pyramidal, 1 lone pair on N), NH_3 ≈ 1.42 D (pyramidal, 1 lone pair on N), CHCl_3 ≈ 1.04 D (tetrahedral, 0 lone pairs on C). NF_3 has the lowest dipole moment because the highly polar N-F bonds are oriented in such a way that their contributions nearly cancel with the effect of the lone pair. The central atom in NF_3 is N, which has 1 lone pair of electrons.

Question 19

Two p-block elements X and Y form fluorides of the type EF _ 3 . The fluoride compound XF _ 3 is a Lewis acid and YF _ 3 is a Lewis base. The hybridizations of the central atoms of XF _ 3 and YF _ 3 r

Accepted Answer

XF₃ is a Lewis acid, indicating the central atom X lacks a complete octet and can accept electron pairs. This is characteristic of Boron (B). Boron in BF₃: Valence electrons: 3 Steric number: 3 (three B-F bonds, no lone pairs) Geometry: Trigonal planar Hybridization: sp² YF₃ is a Lewis base, indicating the central atom Y has lone pair(s) available to donate. This is characteristic of Nitrogen (N). Nitrogen in NF₃: Valence electrons: 5 Steric number: 4 (three N-F bonds + one lone pair) Geometry:

Question 20

The formal charges on the atoms marked as (1) to (4) in the Lewis representation of HNO _ 3 molecule respectively are

Accepted Answer

The formal charge (F.C.) of an atom in a Lewis structure is calculated using the formula: F.C. = [Total valence electrons] - [Total non-bonding (lone pair) electrons] - 1 2 [Total bonding (shared) electrons] For atom (1) Oxygen: Valence e^- = 6, Lone pair e^- = 4, Bonding e^- = 4. F.C. = 6 - 4 - 1 2 (4) = 0 For atom (2) Nitrogen: Valence e^- = 5, Lone pair e^- = 0, Bonding e^- = 8. F.C. = 5 - 0 - 1 2 (8) = +1 For atom (3) Oxygen: Valence e^- = 6, Lone pair e^- = 4, Bonding e^- = 4. F.C. = 6 - 4

Chemical Bonding and Molecular Structure — Previous Year Questions & Solutions