Q: Given below are two statements : Statement I : A hypothetical diatomic molecule with bond order zero is quite stable. Statement II : As bond order increases, the bond length increase. In the light of

. array l He _2 \\ Be _2 \\ Ne _2 array \ aligned & Bond order =0 (does not exist) \\ & Statement I is false aligned Bond order 1 Bond length Statement II is false

Question 1

The correct formal charges on oxygen atoms numbered 2, 1 and 3 respectively are :

Accepted Answer

The formal charge (FC) on an atom in a Lewis structure is given by the formula: FC = V - L - S 2 where V is the number of valence electrons, L is the number of non-bonding electrons, and S is the number of shared (bonding) electrons. For oxygen, the number of valence electrons is V = 6. For oxygen atom 1 (central atom): It has 1 lone pair (2 non-bonding electrons) and 3 bonds (6 shared electrons). FC_ 1 = 6 - 2 - 6 2 = +1 For oxygen atom 2 (left atom): It has 2 lone pairs (4 non-bonding electron

Question 2

Match List I with List II : List I List II A. C _2 H _4 I. 3\ bonds, 2\ bonds B. C _2 H _2 II. 3\ bonds, one lone pair C. CH _4 III. 4\ bonds D. NH _3 IV. 5\ bonds, 1\ bond Choose the correct answer f

Accepted Answer

For C _2 H _4 (Ethene), the structure is H _2 C = CH _2. It contains 4 C - H single bonds and 1 C = C double bond. A single bond is a bond, and a double bond consists of 1 and 1 bond. Total bonds: 5\ bonds and 1\ bond. Thus, A matches IV. For C _2 H _2 (Ethyne), the structure is HC CH . It contains 2 C - H single bonds and 1 C C triple bond. A triple bond consists of 1 and 2 bonds. Total bonds: 3\ bonds and 2\ bonds. Thus, B matches I. For CH _4 (Methane), the central carbon atom is bonded to 4

Question 3

Match the species in List I with their geometry in List II List I List II A. PCl_5 I. Tetrahedral B. BrF_5 II. Square Planar C. BF_4^- III. Trigonal bipyramidal D. [Ni(CN)_4]^ 2- IV. Square pyramidal

Accepted Answer

For PCl _5, the central atom P has 5 valence electrons. It forms 5 bond pairs with Cl atoms and has 0 lone pairs. The hybridization is sp^3d, and the geometry is trigonal bipyramidal. (A III) For BrF _5, the central atom Br has 7 valence electrons. It forms 5 bond pairs with F atoms and has 1 lone pair. The hybridization is sp^3d^2, and the geometry is square pyramidal. (B IV) For BF _4^-, the central atom B has 3 valence electrons plus 1 from the negative charge, giving 4 electrons. It forms 4

Question 4

The highest occupied molecular orbital for Ne_2 is

Accepted Answer

The atomic number of Ne is 10. The total number of electrons in Ne_2 is 20. The molecular orbital electronic configuration of Ne_2 is: _ 1s ^ 2 \ ^ * _ 1s ^ 2 \ _ 2s ^ 2 \ ^ * _ 2s ^ 2 \ _ 2p_z ^ 2 \ _ 2p_x ^ 2 \ _ 2p_y ^ 2 \ ^ * _ 2p_x ^ 2 \ ^ * _ 2p_y ^ 2 \ ^ * _ 2p_z ^ 2 The highest occupied molecular orbital is ^ * _ 2p . Answer: ^*_ 2p

Question 5

Identify the correct statement about ClF _3 from the following options :

Accepted Answer

The central atom in ClF _3 is Chlorine (Cl). Number of valence electrons of Cl = 7. Number of monovalent atoms (F) attached = 3. Steric number = 1 2 (7 + 3) = 5. The hybridization of the central atom is sp^3d. Number of bond pairs = 3. Number of lone pairs = 5 - 3 = 2. The electron geometry is trigonal bipyramidal. According to VSEPR theory, to minimize repulsion, the two lone pairs occupy the equatorial positions. This results in a T-shaped molecular geometry. Therefore, ClF _3 has a T-shaped g

Question 6

Identify the correct orders against the property mentioned (A) H _2 O NH _3 CHCl _3 - dipole moment (B) XeF _4 XeO _3 XeF _2- number of lone pairs on central atom (C) O - H C - H N - O - bond length (

Accepted Answer

(A) H _2 O NH _3 CHCl _3 - correct (C) O - H C - H N - O - Incorrect According to size of element. (D) N _2 O _2 H _2 - correct N N O = O H - H

Question 7

Given below are two statements : Statement I : A hypothetical diatomic molecule with bond order zero is quite stable. Statement II : As bond order increases, the bond length increase. In the light of

Accepted Answer

. array l He _2 \ Be _2 \ Ne _2 array \ aligned & Bond order =0 (does not exist) \ & Statement I is false aligned Bond order 1 Bond length Statement II is false

Question 8

Arrange the following compounds in increasing order of their solubilities in chloroform: NaCl , CH _3 OH , cyclohexane, CH _3 CN

Accepted Answer

Since CHCl _3 is an organic solvent so, covalent (non-polar) compounds will be more soluble in it. As the dipole moment of solute increases, solubility in chloroform decreases. Hence increasing order of solubility. NaCl CH _3 CN CH _3 OH Cyclohexane

Question 9

Identify the incorrect statement about PCl _5.

Accepted Answer

It is s p^3 d hybridised with axial to equatorial angle of 90^ and equatorial bond angles of 120^ . It has five P - Cl sigma bonds. Axial bonds are longer than equatorial bonds.

Question 10

Match List I with List II. Choose the correct answer from the options given below:

Accepted Answer

Answer option index: 1

Question 11

Match List I with List II. Choose the correct answer from the options given below:

Accepted Answer

NH _3 s p^3 hybridised with 1 lone pair. Structure will be Trigonal Pyramidal. BrF _5 s p^3 d^2 hybridised with 1 lone pair. Structure will be Square Pyramidal. XeF _4 s p^3 d^2 with two lone pairs. Structure will be Square Planar. SF _6 s p^3 d^2 with no lone pair. Structure will be Octahedral. A-I, B-IV, C-II, D-III

Question 12

Identify the correct answer.

Accepted Answer

(1) (2) (3) (4) In ozone; there are two resonating structures.

Question 13

Which of the following molecules has "NON ZERO" dipole moment value?

Accepted Answer

Dipole moment of a molecule depends both on shape and bond dipole.

Question 14

Intramolecular hydrogen bonding is present in

Accepted Answer

In o-nitrophenol intramolecular H -bonding is present.

Chemical Bonding and Molecular Structure — Previous Year Questions & Solutions